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Unit-III, Effects of Current, , 5, , Electricity, , Topics Covered, 5.1, 5.2, 5.3, , Circuit Diagram, Ohm’s Law, Factors on which Resistance Depends, Resistance of a System of Resistors, Heating Effect of Current, , C hapter map, Electric Current, (I) and its unit, , Ohm’s Law, , Resistance, Factors, affecting, resistance, , Resistivity, Factors affecting, resistivity, , Advantage, Disadvantage, , Electric, Power, Safety Devices, , Combination of Resistances, In series, RS = R1 + R2 + R3, , Heating, Effect, , In parallel, , Electric, Fuse, , Earth, Wire, , 1= 1, 1, 1, + +, RP R1 R2 R3, Advantage, Disadvantage, , Topic 1. Circuit Diagram, Ohm’s Law, Factors on which Resistance Depends, , Circuit diagram: The schematic diagram in which different components of the circuit are represented by the, symbols conveniently used., Sl. No., Components, 1., Electric cell, 2., , A battery, , 3., , Plug key (switch open), , 4., , Plug key (switch closed), , 5., , A wire joint, , Symbols

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6., , Wire crossing without joining, , 7., , Electric bulb, , 8., , A resistor of resistance R, , 9., , Variable resistor or Rheostat, , 10., , Ammeter, , 11., , Voltmeter, , R, , Ohm’s law: It states ‘the potential difference across the given metallic wire in an electric circuit is directly, proportional to the electric current flowing through it, if the temperature remains constant’., V ∝I, V, V, = R ⇒ V = IR, = constant ⇒, I, I,                , ‘R’ is a constant for a given metallic wire at a given temperature and is called resistance., Resistance: It is the property of a conductor to resist the flow of charges through it. Its S.I. unit is Ω (ohm)., Ohm: The resistance of a conductor is 1 ohm if 1 ampere current is passed across the two ends of conductor, having a potential difference of 1 V., 1 volt, V, 1 ohm =, R=, 1 ampere   , I, •, , I=, , V, , The current through a resistor is inversely proportional to its resistance., R, , • If resistance is doubled, current gets halved if ‘V’ remains the same., Variable resistance (Rheostat): It is a device which is used to change the resistance in the circuit due to, which current flowing through the circuit will change., • Different components of an electric circuit, ammeter, voltmeter, torch bulb, Nichrome wire have different, resistance, therefore flow of current will be different in different components., • Motion of electrons in an electric circuit constitutes an electric current. The electrons are not completely, free to move within a conductor. They are held by attractive forces of the atoms in which they move., • Motion of electrons is retarted by resistance., • A good conductor offers low resistance., • A component of electricity that offers high resistance is a poor conductor., • An insulator offers maximum resistance and does not allow the current to flow., Factors on which resistance of a conductor depend:, (i) Length: Resistance of a wire is directly proportional to its length. If length of the wire is doubled,, resistance will also become double., (ii) Area of cross-section: Resistance is inversely proportional to the area of cross section of a conductor., A thin wire will have more resistance than a thick wire., (iii) Temperature: When we increase the temperature, resistance of metallic conductor will increase., (iv) Nature of material: A material may be a conductor with low resistance, semiconductor with medium, resistance, or an insulator having high resistance., R ∝l R∝, , l, l, 1, ⇒ ⇒ R = ρ., A, A, A, , , r(rho) is a constant of proportionality called resistivity of the material of conductor., Resistivity (r): Resistivity is defined as the resistance of a metallic wire whose area of cross section is 1 m2, and of length 1 m

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If A = 1 m2,, , , l = 1 m,, , then r = R, , ρ=R×, , A, l, , 2,    ρ = ohm × m = ohm m, m, Its S.I. unit is Ω m (Ohm –metre)., Resistivity is a characteristic property of a material., Metals have low resistivity (10–8 to 10–6 Ω m), therefore they are good conductors., Insulators like rubber and glass have high resistivity, 1012 to 1017 Ω m., Resistance and resistivity of a material vary with temperature., Resistivity of an alloy is generally higher than the constituent metals. Alloys do not oxidise (burn) readily at, high temperature, therefore these are used in making heating devices like electric iron, toaster, immersion, rod, heaters, etc., Tungsten is used for filaments of electrical bulbs due to their high resistivity and high melting points., Copper and aluminium are used for electrical transmission lines due to their low resistance and high, conductance., , , •, •, •, •, •, •, •, •, , I. Multiple Choice Questions, , Exercise 5.1, , (1 Mark), , Choose the correct answer from the given options., 1. Electrical resistivity of a given metallic wire depends upon, (a) its length , (b) its thickness, (c) its shape , (d) nature of the material, 2. Identify the circuit (Fig) in which the electrical components have been properly connected., +, , –, , +, , –, , –, , V, , R, , A, , +, –, , +, , 4., 5., , 6., , +, –, , V, , (iii), , 3., , A, , +, , R, , –, , (iv), , (a) (i), (b) (ii), (c) (iii), (d) (iv), Electric current flows from ‘A’ to ‘B’ in metallic conductor. The point is at higher potential is, (a) A , (b) B, (c) Both have equal potential, (d) Both have lower potential, If electrons flow from A to B, current will flow from, (a) A to B, (b) B to A, (c) It will not flow (d) None of these, When a 4V battery is connected across an unknown resistor there is a current of 100 mA in the circuit., The value of the resistance of the resister is:, (a) 4 Ω, (b) 40 Ω, (c) 400 Ω, (d) 0.4 Ω, [CBSE Sample Paper 2019-2020], A cell, a resistor, a key and ammeter are arranged as shown in the circuit given below. The current, recorded in the ammeter will be, K – +, R, R, –, + –, + –, A+, K, –, K, R, A, –, +, A+, (i), (ii), (iii), (a) maximum in (i) , (c) maximum in (iii) , , (b) maximum in (ii), (d) the same in all the cases, , II. Assertion-Reason Type Questions, (1 Mark), , For question numbers 1 and 2 two statements are given-one labeled as Assertion (a) and the other labeled, Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:, (a) Both ‘A’ and ‘R’ are true and ‘R’ is correct explanation of the Assertion., (b) Both ‘A’ and ‘R’ are true but ‘R’ is not correct explanation of the Assertion.

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(c) ‘A’ is true but ‘R’ is false., (d) ‘A’ is false but ‘R’ is true., 1. Assertion: Resistance depends on length and area of cross-section of material and temperature., , Reason: Resistivity is a characteristic property of material., 2. Assertion: Fuse wire is thin and has high resistance., , Reason: Alloys do not oxidise readily at high temperature., , III. Very Short Answer Type Questions, , (1 Mark), , 1. Let the resistance of an electrical component remain constant, while the potential difference across the, two ends of the component decreases to half of its former value. What change will occur in the current, through it?, [NCERT] [HOTS], 2. Which among iron and mercury is a better conductor?, , (Given: rFe = 10.0 × 10–8 W – m, rHg = 95 × 10–8 W – m, 3. Write the S.I. unit of resistivity., [Delhi 2015, 16], 4. Mention two reasons why tungsten is used for making filaments of electric lamps., [Delhi 2016], 5. How does resistivity of alloys compared with those of pure metals from which they have been, formed? , [CBSE 2014], 6. State the difference between a wire used in the element of electric heater and in fuse wire., [Delhi 2013] [CBSE 2014], 7. Name a device which helps to maintain the potential difference across a conductor., [CBSE 2014], 8. Draw a circuit diagram having the following components:, (a) Bulb (b) A two cell battery (c) Ammeter (d) A closed key, 9. Name the factors on which the resistance of a conductor depend?, [NCERT], OR, List the factors on which the resistance of a conductor in terms of shape of a wire depends., 10. What happens to the resistance of a conductor when its temperature is increased?, [CBSE 2010], 11. Nichrome is used to make the element of an electric heater, why?, [CBSE 2010], 12. Mention two special features of the material to be used as an element of an electric iron.[CBSE 2014], 13. What happens to resistance of a conductor when its area of cross section is increased?, [Delhi 2011], 14. What does electric circuit mean?, [NCERT], 15. List two factors on which resistance depends., [CBSE 2015], 16. In an electric circuit, state the relationship between the direction of conventional current and the, direction of flow of electrons., [CBSE 2014], 17. A given length of wire is doubled on itself and the process is repeated once again. By what factor does, the resistance of wire will change? , [CBSE 2011], , IV. Short Answer Type Question-I, , (2 Marks), , 1. Out of the two wires ‘X’ and ‘Y’ of the same material as shown below, which one has greater resistance?, Justify your answer., [Delhi 2012], , V. Short Answer Type Questions-II, , (3 Marks), , 1. Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will, be the resistivity of the metal at that temperature?, [NCERT] [HOTS], 2. The figure below shows three cylindrical copper conductors along their face areas and lengths. Discuss, in which geometrical shape the resistance will be the highest., [Delhi 2013]

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3. Calculate the resistance of 1 km long wire of copper of radius 1 mm. (Resistivity of copper is, 1.72 × 10–8 Ω m.)[Delhi 2013], 4. The resistance of a wire of 0.01 cm radius is 10 Ω. If resistivity of the material of wire is 50 × 10–8 ohm, metre, find the length of the wire. , [Delhi 2016], 5. The resistance of a wire of length 250 m is 1 ohm. If the resistivity of the material of wire is, 1.6 × 10–8 ohm metre, find the area of cross-section of the wire. How much does the resistance change, if the diameter of the wire is doubled?, [Delhi 2016], 6. The electrical resistivity of three materials A, B, C are given below:, ‘A’ = 2.3 × 103 ohm metre   ‘B’ = 2.63 × 10–8 ohm metre   ‘C’ = 1.0 × 1015 ohm metre, Which material will you use for making (i) electric wires, (ii) handles for soldering iron and (iii) solar, cells? Give reason to support your answer., 7. Suppose the ammeter or voltmeter you are using in Ohm’s law experiment do not have +ve and –ve, terminal markings, how will you use such an ammeter or voltmeter in the circuit?, [Delhi 2015], 8. Study the V–I graph for a resistor as shown in the figure and prepare, a table showing the values of I (in amperes) corresponding to four, different values of V (in volts). Find the value of current for V = 10, volts. How can we determine the resistance of the resistor from this, graph?, 9. Name and define SI unit of resistance. Calculate the resistance of a, resistor if current flowing through it is 200 mA, when the applied, [CBSE 2014], potential difference is 0.8 V., 10. V–I graph for two wires A and B are shown in the figure. If both the wires are of same, length and same thickness, which of the two is made of a material of high resistivity?, Give justification for your answer., 11. The figure below shows three cylindrical copper conductors along with their face, areas and lengths. Compare the resistance and the resistivity of the three conductors. Justify your, answer., , , 12. The potential difference between the terminals of an electric heater is 60 V when it draws a current of, 4 A from the source. What current will the heater draw if the potential difference is increased upto 120, V?[NCERT], 13. (i) How much current will an electric bulb draw from a 220 V source, if resistance of the filament of a, bulb is 1200 Ω?, , (ii) How much current will an electric heater coil draw from a 220 V source, if the resistance of the, heater coil is 100 Ω? , [NCERT] [Delhi 2012], 14. A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would, be the resistance of another wire of the same material having length l/2 and area of cross-section 2A?, [NCERT] [HOTS], 15. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this, wire to make its resistance 10 Ω? How much does the resistance change if the diameter of the wire is, doubled?[NCERT]

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16. The values of current I flowing in a given resistor for the corresponding values of potential difference, V across the resistor are given below :, I (amperes), , 0.5, , 1.0, , 2.0, , 3.0, , 4.0, , V (volts), , 1.6, , 3.4, , 6.7, , 10.2, , 13.2, , Plot a graph between V and I and calculate the resistance of that resistor. [NCERT] [HOTS], 17. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the, circuit. Find the value of the resistance of the resistor., [NCERT], 18. A piece of wire of resistance 20 Ω is drawn out of a material so that its length, B, is increased to twice of its original length. Calculate the resistance in the, new situation., [HOTS], A, 19. V-I graph for two wires ‘A’ and ‘B’ are shown in the figure. If both the wires, are of same length and are of same thickness, which of these two is made of, the material of higher resistivity? Give justification for your answer., 20. (i) What are the values of mA and mA in amperes?, , (ii) Draw the symbols of battery and rheostat?, [CBSE 2014], , VI. Long Answer Type Questions, , (5 Marks), , 1. (i) Write an expression for the resistivity of a substance., , (ii) State the SI unit of resistivity., , (iii) Distinguish between resistance and resistivity., , (iv) Name two factors on which the resistivity of a substance depends and two factors on which it does, not depend. , [Delhi 2015], 2. List two distinguishing features between the resistance and resistivity of a conductor. A wire is, stretched so that its length becomes 6/5 times of its original length. If its original resistance is 25 W,, find its new resistance and resistivity. Give justification for your answer in each case. [Delhi 2016], 3. State Ohm’s law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a, graph between the potential difference and current flowing through a metallic conductor, what kind of, curve will you get? Explain how would you use this graph to determine the resistance of the conductor., [Delhi 2016], 4. You have been assigned a duty to create awareness in your school, about saving electricity., , (i) Write any two ways by which you will create awareness, among your schoolmates about saving electricity., , (ii) Explain how saving electricity is important at individual level, as well as at national level.             [Delhi 2015], 5. (i) Draw a closed circuit diagram consisting of a 0.5 m long, nichrome wire XY, and ammeter, a voltmeter, four cells of 1.5, V each and a plug key., , (ii) Following graph was plotted between V and I values:, What would be the values of V/I ratios when the potential, difference is 0.8 V, 1.2 V and 1.6 V respectively? What, conclusion do you draw from these values?, , Answers 5.1, I. 1. (d) It depends only upon nature of the material. Resistance depends on the length of wire and area of, cross section., , 2. (b) It is correct because Ammeter should be connected in series and voltmeter should be connected in, parallel., , 3. (a) ‘A’ because current flow from higher to lower potential., , 4. (b) Current flows in opposite direction to the direction of flow of electrons, B to A., , 5. (b) V = IR, V = 4V, I = 100 mA = 0.1 A; Hence, R = V/I = 4/0.1 = 40., , 6. (d) It will be same in all cases, II. 1. (b) Both ‘A’ and ‘R’ are true but ‘R’ is not correct explanation of the assertion., , 2. (b) Both ‘A’ and ‘R’ are true but ‘R’ is not correct explanation of the assertion.

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The current flowing through the electrical component will get halved., Iron, Ω m (ohm metre), (i) High resistivity, (ii) High melting point, Alloys have higher resistivity as compared to pure metals., The wire used in the elements of an electric heater has high melting point whereas a fuse wire has low, melting point., 7. Cell or battery eliminator., 8., , III. 1., 2., 3., 4., 5., 6., , 9. (i) Nature of material, (ii) Length of conductor [or R µ l], , (iii) Area of cross section of the conductor [or R µ 1/A], (iv) Temperature, 10. Its resistance increases., 11. It is an alloy with high resistivity and high melting point., 12. The material should have (i) high resistivity, (ii) high melting point., l, 13. Resistance decreases, ... R ∝, , Resistance is directly proportional to length but inversely proportional, A, to area of cross-section., 14. It is a continuous and closed path in which current flows with the help of conducting wires. It consists, of a cell, an ammeter, voltmeter, plug key, bulb, etc., 15. (i) Length, (ii) Area of cross section, It is directly proportional to the length and inversely proportional to the area of cross section of a, resistor., 16. The direction of conventional current is opposite to the direction of flow of electrons., 1, 17. Length becomes 4 times whereas area becomes, times of the original value. Since,, 4, L, 4L, 16 L, L, R=ρ =ρ, =ρ, = 16 ρ, where r (rho) is resistivity, L is length and ‘A’ is area of cross section., , A, A, A, A, 4, , \ Resistance will become 16 times of the original value., IV. 1. ‘Y’ has more resistance because resistance is directly proportional to the length of the wire., V. 1. R = 26 Ω, l = 1 m, d = 0.3 mm = 3 × 10–4 m, d 3 × 10 −4, r= =, = 1.5 × 10 −4 m, 2, 2, 22, A = pr2 ⇒ A =, × (1.5 × 10–4)2, 7, R × A 26 × 22 × 2.25 × 10 −8, = 1.84 × 10–6 W m, , ρ=, =, l, 7 × 1m, L, A, 2L, L, R2 = ρ ×, = 4ρ × = 4 R 1, A/2, A, L/2 1, L 1, R3 = ρ ×, = × ρ × = R1, 2A 4, A 4, , 2. R1 = ρ ×, , (i), (ii), (iii), , Thus in figure (b) the resistance is maximum., So, from equations (i), (ii) and (iii), we get R2 > R1 > R3.

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Hence, the cylinder having length (2L) and area of cross-section (A/2) has the highest resistance, followed by the cylinder having length (L) and area of cross section (a), while cylinder with length, (L/2) and area of cross-section (2A) has the least resistance. This is because resistance of a substance, is directly proportional to the length and is inversely proportional to the area of cross-section., 3. R = ρ ×, , l, A, , 22, × 1 × 10–6, , r = 1.72 × 10–8 Ω m; r = 1 mm = 1 × 10–3 m; A = pr2 =, 7, l = 1 km = 1000 m, ⇒ R=, , 4. R = r ×, , 1.72 × 10 −8 × 1000 × 7 12.04, 120.4, =, × 10 −8 + 3 + 6 =, = 5.473 Ω, 22 × 1 × 10 −6, 22, 22, l, A, , R = 10 W,, r = 50 × 10–8 W m, , \ l=, 5. R = r, , r = 0.01 × 10–2 m = 1 × 10–4 m, A = pr2 =, , 22, × 1 × 10–8, 7, , R × A 10 × 22 × 10 −8 220, = 0.628 m = 62.8 cm, =, =, ρ, 50 × 10 −8 × 7, 350, , ρl, l, ⇒ A=, R, A, , 1.6 × 10 −8 Ω m × 250 m, = 4 × 10–6 m2, 1Ω, If diameter is doubled then area of cross-section becomes four times and the resistance will become one, fourth, i.e., 0.25 W., 6. (i) ‘B’ will be used for electric wires because it has lowest resistivity, therefore current can flow easily., , (ii) ‘C’ will be an insulator, so it can be used for handles of soldering iron. Since its resistivity is very, high, therefore current cannot pass through it., , (iii) ‘A’ will be used in solar cells because it is a semiconductor which can convert solar energy into, electric energy., 7. (i) Connect the device in the circuit with battery., , (ii) Close the circuit and notice the deflection of the pointer., , (iii) If the deflection is in the opposite direction i.e. below zero, then interchange the terminals., , A =, , 8., , I (in amp), , 2, , 4, , 6, , 8, , V (in Volt), , 1, , 2, , 3, , 4, , When V = 10 Volts, I = 5 amperes, then, , R=, , V 10 V, =, =2Ω, I, 5A, , 9. (i) The SI unit of resistance is Ohm (W)., , (ii) If the current of 1 ampere flows through a wire on applying a potential difference of one volt across, it, then the resistance of the wire is said to be 1 W., , (iii) I = 200 mA = 200 × 10–3 A, V, 0 .8, =, = 4 ohms, 0 .2, I, 10. Greater the slope of V–I graph, greater will be the resistance of the given metallic wire. In the given, graph, wire A has greater slope than B. Hence, wire A has greater resistance., For the wires of same length and same thickness, resistance depends on the nature of the material of, wire, i.e., R1, ρ, l, l, , R1 = ρ1 and R 2 = ρ2, ⇒, = 1 or R ∝ ρ, A, A, R2, ρ2, , R =, , Since, R1 > R2. Hence, wire ‘A’ is made of a material of high resistivity.

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11. For the geometrical shapes shown in figure:, , l, ρL,  L / 3  1 ρL 1,  3L , Ra = ρ A ; Rb = ρ  A / 3  = 9 A = 9R a ; Rc = ρ  3 A  = 9 A = 9 R a, , , Hence Rb > Ra > Rc, ρ = ρb = ρc because all the three conductors are of same material., a, , , 12. V = 60 V, I = 4 A, V 60, =, = 15Ω [ By Ohm’s Law: V = IR], R=, , I, 4, Now, V = 120 V, I = ? R = 15 Ω, V 120, I=, =, =8A, R, 15, , 13. (i) V = 220 V, R = 1200 Ω, V, 220, I=, =, = 0.18 A [ By Ohm’s Law: V = IR], , R 1200, , (ii) Now, V = 220 V, R = 100 Ω, 220, I=, = 2 .2 A,    , , 100, 14. R1 = ρ ×, , l, = 4 Ω, A, , Also, R 2 = ρ ×, , l/2, 1 l, = ρ× ×, 2A, 4 A, , 1, 1, , R 2 = R1 = × 4 = 1 Ω, 4, 4, 15. ρ = 1.6 × 10 −8Ω m ; R = 10 Ω, d = 0.5 mm = 5 × 10–4 m [ 1 mm = 10–3 m], 22, 5, × 10–4 m = 2.5 × 10–4 m; A = pr2 =, × (2.5 × 10–4)2 m2, 7, 2, A, , ρ=R×, l, 22 6.25 × 10 −8, 10 × 22 × 6.25 × 10 −8 1375, ×, 1.6 × 10–8 = 10 ×, ⇒ l=, =, = 122.77 m, 7, l, 7 × 1.6 × 10 −8, 11.2, r =, , 16. The slope of the graph is equal to the resistance., Change in V ∆V 13.2 − 1.6 ,, , =, =, R=, Change in I, ∆I, 4 .0 − 0 .5, 11.6, , R=, = 3.314 Ω, 3 .5, 17. V = 12 V, I = 2.5 mA = 2.5 × 10–3 A, , (1 mA = 10–3 A), 12, 12000, V, R=, =, =, = 4800 Ω = 4.8 k Ω, , I 2.5 × 10 −3, 2 .5, , ( By Ohm’s Law: V = IR), l, 2l, 18. R1 = r ×, ; R2 = r ×, A, A, 2l, ρ×, R2, A, , \, =, l, R1, ρ×, A, , ⇒ R2 = 2R1 = 2 × 20 = 40 W, , A, , ∵ R = ρ × l , , 

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19. ‘A’ has greater resistivity than ‘B’. It depends upon the nature of material. High resistivity means, lower value of electric current flowing through the wire., 20. (i) 1mA = 10–3A, 1mA = 10–6A., , (ii), , is the symbol of battery, and, , or, , is the symbol of rheostat., , RA, , (ii) Its SI unit is W m., l, (iii) Resistivity is a characteristic property of a material that does not depend upon the dimensions, of the material whereas resistance depends upon the dimensions of the material., VI. 1. (i) Resistivity (r) =, , , (iv) Resistivity does not depend on the:, , (a) length of conductor,, Resistivity depends on the:, , (a) nature of material of conductor, 2., , (b) area of cross section of conductor, (b) temperature of conductor, , Resistance, , Resistivity, , (i) It is the ratio of voltage over current at a, particular temperature., , (i) It is the resistance of a wire of length 1 m and, area of cross section 1 m2., , (ii) Its S.I. unit is ohm., , (ii) Its S.I. unit is ohm metre., , (iii) It is the property of a conductor., , (iii) It is the property of the material of the conductor., , , (Any two), Since resistance is directly proportional to the length and is inversely proportional to the area of cross, section., 25, 150, 216, 6, \ New Resistance: R = ×, × 36 =, = 43.20 ohms, 2 =, 5  5, 5 × 25, 5,  6 , When the wire is stretched, its length becomes 6/5 times of its original length, and area of cross-section, will become 5/6. Resistivity will remain the same because it does not depend upon the length and area, of cross section. It depends on the nature of material of the substance and the temperature., 3. Ohm’s Law: It states that ratio of potential difference and current is constant and is equal to the, resistance of the conductor at a particular temperature. In other words, the current flowing through a, conductor is directly proportional to the potential difference at a constant temperature., The graph between V and I will be a straight line., Slope = tan θ = ‘R’, BC, Slope =, =R, AC, 4. (i) Speaking over it in the morning assembly in school., , • Putting posters over it on the school notice board., , (ii) If we save electricity, it can be used by those villages which do not have electricity., , • It can be used in industries, agriculture and for other useful purposes., , • It improves the national economy because high speed trains, industries, development in villages, depends upon electricity., 5. (i)

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(ii) The V/I ratio for, 0.8, V, V, 1 .2, = 2.67 ; • V = 1.2 V,, is, =, • V = 0.8 V,, = 2.67, 0.3, I, I, 0.45, V 1 .6, • V = 1.6 V,, =, = 2.67, I, 0 .6, It shows that V a I or, , V, = constant., I, , Topic 2. Resistance of a System of Resistors, There are two methods of joining the resistors together., , Resistors in Series, When resistors are joined end to end, it is called that resistors are connected in series., (i) The value of current in the ammeter remains the same, independent of its position in the electric circuit., It means that in a series combination of resistors, the current remains the same in every part of the circuit, or same current flows through every resistor., , (ii) In the above circuit in V is equal to the sum of V1, V2, V3, that is total potential differences (V) across a, combination of resistors in series is equal to sum of potential difference across the individual combinations,, i.e.,                  , V = V 1 + V2 + V 3, (iii) The current through each resistor and all the three resistors is I.,               V = IR, , \    , V1 = IR1, V2 = IR2 and V3 = IR3,         So,, V = V1 + V2 + V3 ⇒ IR = IR1 + IR2 + IR3,        or      R = R1 + R2 + R3, Thus we can conclude that when several resistors are joined in series combination, then equivalent, resistance of the combination is equal to the sum of their individual resistances and is thus greater than, any of the individual resistance.,               RS = R1 + R2 + R3, , Resistors in Parallel, (i) Suppose we connect three resistances R1, R2, R3 in parallel combination as shown in the diagram., (ii) Let us connect it with the battery, a plug key, an ammeter as shown in the diagram., (iii) Also connect voltmeter in parallel combination with the resistors as shown.

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By applying a potential difference V, the current flowing in the, circuit be I. This circuit is distributed in different branches., Suppose I1, I2, I3 be the current in each resistor., Then we observe, I = I1 + I2 + I3, According to Ohm’s Law, ,            , ∵,        , , I=, , V, Rp, , I1 =, , V, ;, R1, , I2 =, , V, ;, R2, , I3 =, , V, R3, , V, V, V, V, =, +, +, R p R1 R 2 R 3, , I = I1 + I2 + I3 ⇒, , 1, 1, 1, 1, =, +, +, R, R, R, R, 1, 2, 3,              p, , Circuit in series vs Circuit in parallel (Disadvantages of series circuit), , (i) We have observed in series circuit, the current is constant throughout the electric circuit. Thus, it is not, practical to connect an electric bulb and electrical heater in series, because they need current of widely, different values to operate properly. Total resistance increases in a series connection., , (ii) Another disadvantage of series circuit is that when one component fails, the entire circuit is broken and, current stops flowing in the circuit. You must have seen in ‘fairy lights’ used in diwali, if one bulb fuses,, the circuit is broken, none of the small bulbs work or light up., , Parallel Circuit Advantages, , (i) It divides the current through the electric gadgets., , (ii) The total resistance in a parallel circuit decreases. It is helpful when gadgets have different resistances, and they require different current to operate., , Exercise 5.2, I. Multiple Choice Questions, , (1 Mark), , Choose the correct answer from the given options., 1. When the current passing through each resistor is same, the circuit is called, (a) series circuit, (b) parallel circuit, (b) Both (a) and (b) (c) None of these, 2. In parallel circuit, which of the following is same across resistors, (a) Current , (b) Potential difference, (c) Resistance , (d) Resistivity, 3. When resistance is doubled at constant voltage, current will become., 1, 1, (b) 2 times, (c), (d) 4 times, (a), 2, 4, 4. Resistance in a circuit if current flowing is doubled will become, (a) double, (b) half, (c) four times, (d) remains same, 5. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel., If the equivalent resistance of this combination is R′, then the ratio R/R′ is, (a) 1/25, (b) 1/5, (c) 5, (d) 25, [CBSE 2020, NCERT], 6. Two conducting wires of the same material and of equal lengths and equal diameter are first connected, in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced, in series and parallel combinations would be, (a) 1:2, (b) 2:1, (c) 1:4, (d) 4:1, [NCERT], , II. Assertion-Reason Type Questions, (1 Mark), For question numbers 1 and 2 two statements are given-one labeled as Assertion (a) and the other labeled, Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given ahead:, (a) Both ‘A’ and ‘R’ are true and ‘R’ is correct explanation of the Assertion., (b) Both ‘A’ and ‘R’ are true but ‘R’ is not correct explanation of the Assertion.

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(c) ‘A’ is true but ‘R’ is false., (d) ‘A’ is false but ‘R’ is true., 1. Assertion: In parallel circuit, the current is divided into electrical gadgets, therefore, used in domestic, circuit., , Reason: The resistance in parallel circuit decreases, therefore, useful., 2. Assertion: Fairy lights circuit is connected in series connection., , Reason: In series combination, if one component fails then complete circuit is broken., , III. Very Short Answer Type Questions, , (1 Mark), , 1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 6 V each, a, 5 W resistor, an 8 W resistor and a 12 W resistor and a plug key, all connected in series., [NCERT], 2. Judge the equivalent resistance when the following are connected in parallel:, (i) 1 W and 106 W, (ii) 1W, 103 W and 106 W[NCERT], 3. What are the advantages of connecting electrical devices in parallel with the battery instead of, connecting them in series?, [NCERT], 4. Two unequal resistances are connected in parallel. If you are not provided with any other parameters, (eg. numerical values of I and R), what can be said about the voltage drop across the two resistors?, [CBSE Sample Paper 2020-21], , IV. Short Answer Type Questions-I, , (2 Marks), , 1. If you connect three resistors having values 2 W, 3 W and 5 W in parallel, calculate the equivalent, resistance., 2. B1, B2, B3 are three identical bulbs connected as shown in the figure., When all the three bulbs glow, a current of 3 A is recorded by the, ammeter A., , (i) What happens to the glow of the other two bulbs when bulb B1 gets, fused?, , (ii) What happens to the reading of A1, A2, A3 and A when bulb B2 gets, fused.                 [CBSE 2010], 3. Redraw the circuit in the above question, putting an ammeter to, measure the current through the resistors and a voltmeter to measure, I, →, the voltage across the 12 W resistor. What would be reading in the, A, ammeter and the voltmeter?, [NCERT] [HOTS], 4. A student has two resistors- 2 W and 3 W. She has to put one of them, R1 = 4W, 12V, R2, in place of R2 as shown in the circuit. The current that she needs in, the entire circuit is exactly 9A. Show by calculation which of the two, B, resistors she should choose., , V. Short Answer Type Questions-II, , (3 Marks), , 1. What is (a) the highest, (b) the lowest total resistance that can be secured by combination of four coils, of resistances 4 W, 8 W, 12 W, 24 W?, [NCERT], 2. Show how you would connect three resistors, each of resistance 6 W, so that the combination has a, resistance of (i) 9 W, (ii) 4 W., [HOTS], 3. (i) What are the disadvantages of resistances connected in a series circuit?, , (ii) Find the resistance between A and B in the following network., [Delhi 2014]

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4. What are the advantages of connecting electrical devices in parallel combination with the battery, rather than in series combination?, 5. Find the current drawn from the battery by the network of four resistors as shown in the figure., , 6. Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the, combination is (i) 13.5 Ω, (ii) 6 Ω?[CBSE 2018], 7. If three resistors of 6 Ω, 9 Ω and 21 Ω are connected in series to a 12 V battery, find:, , (a) The total resistance of the circuit., , (b) The current flowing through the circuit., , (c) The potential difference across the 21 Ω resistor., 8. An electric lamp of 100 W, a toaster of resistance 50 W, and a water filter of resistance 500 W are, connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same, source that takes as much current as all the three appliances, and what is the current flowing through, it?[NCERT], 9. Study the following electric circuit and find the, , (i) current flowing through the circuit and, , (ii) the potential difference across 10 W resistor., , [Delhi 2016], , 10. Figure shows a 2 W resistor and a 6 W resistor connected in series, 2, with a 4 V battery. Calculate:, , (a) the combined resistance of the resistors., , (b) the current in the battery., , (c) the current in 2 W resistor., 11. How can three resistors of resistances 2 W, 3 W, and 6 W be, connected to give a total resistance of (a) 4 W, (b) 1 W?[NCERT], 12. The parallel combination of two 10 W resistors are placed across, 4V, the terminals of a 24 V battery., , (a) What is the effective resistance of the parallel combination of resistances in the circuit?, , (b) What is the current through the entire circuit?, , (c) What is the current through each branch of the circuit?, 13. How many 176 W resistors (in parallel) are required to carry 5 A current on a 220 V line?, 14. A hot plate of an electric oven connected to a 220 V line has two coils A and B, each of 24 W, resistance, which may be used separately, in series, or in parallel. What are the values of current in, the these cases?, [NCERT]

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VI. Long Answer Type Questions, , (5 Marks), , 1. (i) , What is meant by series combination and parallel combination of, resistances?, , (ii) In the circuit diagram given below five resistances of 5W, 20W, 15W, 20W, and 10 W are connected as given in the figure to a 6 V battery., Calculate: (a) Total resistance in the circuit., , (b) Total current flowing in the circuit.     [Delhi 2016], 2. Study the following electric circuit and find (i) the current flowing in the, circuit and (ii) the potential difference across the 10 W resistor., , 20 Ω, 15 Ω, 20 Ω, 5Ω, 6V, , , 3. Find the net current flowing through the following electric circuit:, , , 4. (a) , With the help of suitable circuit diagram prove that the, reciprocal of the equivalent resistance of a group of resistances, joined in parallel is equal to the sum of the reciprocal of the, individual resistances., , (b) In an electric circuit two resistors of 12 W each are joined in, parallel to a 6V battery. Find the current drawn from the, battery., 5. Figure shows a 3 W resistor and a 6 W resistor connected in parallel, across a 1.5 V cell., Calculate the current across: (i) 3 W resistor, (ii) 6 W resistor, (iii), the cell, (iv) calculate the resistance of the parallel combination of, 3 W and 6 W resistors., , Answers 5.2, I. 1. (a) The circuit is called series circuit., , 2. (b) Potential difference is same., V, 1,     I =, R, 2, 4. (d) Resistance does not depend upon current., , 3. (a) It will became, , 5. (d), , , , 1, R, , ′, , =, , R′ =, , 1, 1, 1, 1, 1 25, +, +, +, +, =, R R R R R, R, 5, 5, 5, 5, 5, R, Ω, 25, , ⇒, , R, = 25 . So option (d) is correct., R′, , 1.5 V, , 10 Ω

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6. (c) Resistances of identical conducting wires are equal. Let R be the resistance of each resistor in series, combination.,     \ Rs = R + R = 2R, V2, 2R, 1, 1 1, 2,     When resistances are connected in parallel,, =, +, =, RP R R R, V2, H, =,     P R / 2, ,    Hs = Heat produced in series. Hence Hs =, , RP =, , R, , 2, , HS V 2, 1, R, =, ×, = . So option (c) is correct., HP 2R 2V 2 4, II. 1. (a) Both ‘A’ and ‘R’ are true and ‘R’ is correct explanation of the assertion., ,    , , 2. (b) Both ‘A’ and ‘R’ are true but ‘R’ is not correct explanation of the assertion., III. 1., , 2. Equivalent resistance in parallel combination of resistances is always less than the lowest individual, resistance i.e. less than 1 Ω in both the cases (i) and (ii)., 3. Parallel circuit advantages:, , (a) Voltage is same across all the components when connected in parallel., , (b) Parallel combination of resistances helps us to obtain an overall lower potential than the resistances, used., , (c) In household circuit, parallel connection is useful because each appliance has its own ON/OFF, switch., , (d) A fault in one appliance will not affect the working of other appliances., (Any two), 4. Voltage-drop is same across both resistors., 1, 1, 1, 1, 1 1 1, 15 + 10 + 6 31, +, +, = + + =, =, IV. 1., =, R, R1 R 2 R 3 2 3 5, 30, 30, 30, = 0.967 Ω, 31, 2. (i) The glow of bulbs B2 and B3 will remain same because they are in parallel connection with bulb B1., , (ii) A 1 shows 1 ampere reading, A 2 shows zero and A 3 show 1 ampere reading. ‘A’ will show, 2 A reading., 3. RS = R1 + R2 + R3 = 5 + 8 + 12 = 25 Ω; V = 6 V, , , , R=, , V, 6V, =, = 0.24 A, , I=, R 25 Ω, In series combination current in all resistances is same., Here, I = 0.24 A across 12 Ω resistance, , \ Voltmeter reading, V = IR = 0.24 × 12 = 2.88 V, , \, I = 0.24 A,, V = 2.88 V, 4. The overall current needed = 9A. The voltage is 12V, Hence by Ohm’s law V=IR,, 4, 12, =, Ω. = R, The resistance for the entire circuit =, 3, 9, R1 and R2 are in parallel., (R1R 2 ) = 4R 2 = 4 , Hence, R =, (R1+R 2 ) (4 + R 2 ) 3, R2 = 2Ω

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V. 1. (a) The highest resistance can be obtained by connecting all resistances in series, i.e., Rs = R1 + R2 + R3 + R4 = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω, (b) The lowest resistance can be obtained by parallel combination:, 1, 1, 1, 1, 1, 1, 1, 6 + 3 + 2 + 1 12, 1 1, =, =, +, +, +, = + +, +, =, R p R1 R 2 R 3 R 4 4 8 12 24, 24, 24, , Rp = 2 Ω, (i) When two 6 Ω resistances are connected in parallel and connected to another 6 Ω in series, we get, 9 Ω resistance., 1, 1 1 2 1, = + = = ⇒ R p = 3 Ω ⇒ RNet = R3 + RP = 6 + 3 = 9 Ω, Rp 6 6 6 3, , (ii) When two resistors are connected in series and third one is in parallel, we get 4 Ω resistance., Rs = R1+ R2 = 6 + 6 = 12 Ω, 2., , 1, 1, 1, 1, 1 1+2, 3, , =, +, =, + =, =, R p R s R 3 12 6, 12, 12, , fi Rp = 4 Ω, , 3. (i) Circuit in series vs Circuit in Parallel (Disadvantages of series circuit):, , (a) If one electrical appliance stops working due to some factor, then all the other appliances also stop, working., , (b) All the electrical appliances have only one switch due to which they can’t turned on or off individually., , (c) The appliances do not get the same voltage as of the input power supply., , (d) Overall resistance increases due to which the current from the input power supply becomes low., , (ii) Two resistances of 2 Ω are in series \ Rs = R1 + R2 = 2 + 2 = 4 Ω, , The third resistance of 4 Ω is parallel to Rs \, , 1, 1, 1, 1 1 2, =, +, = + =, Rp Rs R3 4 4 4, , , ⇒ RP = 2 Ω, 4. (1) The current required by each device is different which is possible only in parallel combination., , (2) If one device fails others can still work., , (3) Total resistance in the circuit is decreased., 5. R1, R2, R3 are in series combination and R4 is in parallel combination to these combination of resistors., Equivalent resistance of the given network is given as:, 1, 1, 1, 1, 1, 1, 1, 3+1, 4, +, +, =, +, =, =, =, =, R, R 4 R1 + R 2 + R 3, 10 10 + 10 + 10 10 30, 30, 30, , , , 30, = 7 .5 Ω, 4, 30 2, V, 3, =, =, =, Current drawn from the battery I =, R 7.5 75 5, , \R=, , 6., , (i) , , ⇒ I = 0.4 A, , , , Two 9 ohm resistors in parallel combination are connected to one 9 ohm resistor in series;, 1, 1 1 2, 9,   , = + =    \     R p = Ω, Rp 9 9 9, 2, R = 9 Ω +, , 9, Ω = 13.5 Ω, 2, , , (ii) Two 9 ohm resistors in series connected are connected to one 9 ohm resistor, in parallel;

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Rs = 9 Ω + 9 Ω = 18 Ω,    , , 1, 1 1, 3, =, + =, R 18 9 18, (b) I =, , 7. (a) 36 W, , \    , , R=6Ω, , V 12 V, =, = 0.33 A, R 36 Ω, , (c) V = R × I ⇒ V = 21 × 0.33 = 6.93 V, , 8. R1 = 100 Ω, R2 = 50 Ω, R3 = 500 Ω, V = 220 V, , , 1, 1, 1, 1, 1, 1, 1, 5 + 10 + 1, 16, =, =, +, +, +, +, =, =, R p R1 R 2 R 3, 100 50 500, 500, 500, , RP =, , 500 125, =, Ω = 31.25 Ω, 16, 4, , V 220 × 4 880, =, =, = 7.04 A, R, 125, 125, Current will remain the same if the electric iron is connected with the same source, i.e., , I = 7.04 A, 125, = 31.25 Ω across the electric iron., R =, 4, 9. (i) Rs = R1 + R2 = 10 Ω + 20 Ω = 30 Ω, , Current through all appliances =, , I =, , V, 3V, =, = 0 .1 A, R 30 Ω, , , (ii) V = I × R = 0.1 × 10 = 1 V, 10. (a) Rs = R1 + R2 = 2 W + 6 W = 8 W, V 4, = = 0 .5 A, R 8, (c) Current across the 2 W resistor = 0.5 A, 11. (a) In order to get 4 W resistance, 2 W resistance should be, connected in series with a combination of 3 W and 6 W, resistances connected in parallel with each other., , , (b) I =, , 1, 1, 1, 1 1 3, =, +, = + = ⇒ Rp = 2 Ω, , R p R1 R 2 3 6 6, ,  Rs = R3 + Rp = 2 W + 2 W = 4 W, , (b) In order to get 1 W resistance all the three resistors should be, connected in parallel., 1, 1, 1, 1, =, +, +, R p R1 R 2 R 3, =, , 1 1 1 3+2+1 6, + + =, = =1, 2 3 6, 6, 6, , ,  Rp = 1 W, 12. (a), , 1, 1, 1, 1, 1, 2 1, +, =, +, =, =, ⇒ RP = 5 Ω, =, RP, R1 R 2 10 10 10 5, , , (b) From Ohm’s law, I =, , (c) I1 =, , V 24, =, = 2 .4 A, R 10, , V 24, =, = 4 .8 A, R, 5, V 24, = 2 .4 A, I2 = =, R 10, , 13. When ‘N’ resistors each of resistance ‘R’ are connected in parallel, then R p =, , R, N

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Current drawn from the cell, I =, , , ⇒, , V, V V×N, =, =, R, Rp, R, N, 220 × N, 5A =, 176 Ω, , , 14. I =, , N=, , 5 × 176 880, =, =4, 220, 220, , V 220, =, = 9.17 A, R, 24, , When coils are used in series, Rs = R1 + R2 = 24 + 24 = 48 W ⇒ I =, , V 220, =, = 4.58 W, R, 48, , When coils are used in parallel,, 1, 1, 1, 1, 1, 2, +, =, +, =, =, ⇒ Rp = 12 W, R p R1 R 2 24 24 24, , ⇒, , I=, , V 220, =, = 18.3 A, R, 12, , VI. 1. (i) Resistors in Series: When resistors are joined end to end, it is called, series combination of resistances., In series combination of resistors, the current remains the same through, each resistor. Therefore the value of current in the ammeter remains the, same, independent of its position in the electric circuit., In the above circuit, V is equal to the sum of V1, V2, V3, that is the total, potential difference (V) across a combination of resistors in series is equal, to the sum of potential differences across the individual resistors., V = V1 + V2 + V3, The current through each resistor is I., V = IR, V1 = IR1 V2 = IR2 V3 = IR3, As V = V1 + V2 + V3 ⇒ \ IR = IR1 + IR2 + IR3 ⇒, , RS = R1 + R2 + R3, , \ We can conclude that when several resistors are joined, in series, then equivalent resistance of the combination, is equal to the sum of their individual resistances and, is thus greater than any of the individual resistance., ,  Resistors in Parallel:, When the resistances, says, R1, R2, R3 are connected, in in parallel combination as shown in the diagram,, then the potential difference remains the same, across each resistor in parallel combination of, resistors. In parallel combination the current is, distributed in different resistances in different, branches., As, I = I1 + I2 + I3, V, V, V, V, , I2 =, , I3 =, According to Ohm’s Law, I =, ⇒ I1 =, R1, R2, R3, Rp, , \ I = I1 + I2 + I3, 1, 1, 1, 1, V, V, V V, =, +, +, +, +, Or, ⇒, =, , R, R, R, R, Rp, R1 R 2 R 3, p, 1, 2, 3, (ii) (a) Resistance across parallel combination,, 1, 1, 1, 1, 1 + 1, , +, +, =, +, R3 R4 R5, R p R1 R 2, , Resultant resistance of 5 Ω, 20 Ω, 15 Ω, 10 Ω, and 20 Ω resistances is given as:, , 20 Ω, 15 Ω, 20 Ω, 5Ω, 6V, , 10 Ω

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1, 1 1, 1, 1, 1, = +, +, +, +, R p 5 20 15 10 20, 1, 28, =, R p 60, , ⇒, , Rp =, , , (b) V= 6 V, Resistance =, , ⇒, , 1 12 + 3 + 4 + 6 + 3, =, Rp, 60, , 60, Ω ⇒ R p = 2.14 Ω, 28, 60, Ω, 28, , V 6 × 28, =, ⇒ I = 2.8 A, R, 60, 2. 10 W and 20 W resistances are connected in series, therefore their equivalent resistance is given by: R1, = R1 + R2 = 10 + 20 = 30 W, V, 3, 1, , (i) Current flowing in the circuit, I =, =, =, = 0 .1 A, R S 30 10, , , \I=, , 1, × 10 = 1 volt, 10, 3. Series combination of 1 W and 3 W resistance is in parallel combination with 6 W resistor. Their, equivalent resistance is, , , (iii) Potential difference across the 10 W resistor, V = IR =, , 1, 1, 1, 1 1 2+3, = + =, = +, RP, 6 3+1 6 4, 12, , ⇒ RP =, , 12, = 2 .4 Ω, 5, , Now, 3.6 W, 2.4 W and 3 W resistors are connected in series, therefore their equivalent resistance be, given by: RS =R1 + R2 + R3 = 3.6 + 2.4 + 3 = 9 W, V 4.5 45 1, =, =, = = 0 .5 A, Hence, the current flowing through the circuit is, I =, R, 9, 90 2, 4. (a) Refer to Answer 1(i) — Long Answer Type Questions., 1, 1, 1, 1, 1, 1+1, 2, =, ⇒ R p = 6 Ohms, , (b) R = R – R = 12 + 12 =, 12, 12, p, 1, 2, , , I=, 5., , V, 6V, =, = 1 Ampere, R 6 ohm, , (i) I1 =, , V, 1 .5, ⇒ I1 =, = 0 .5 A ,, R, 3, , (ii) I2 =, , , (iii) ITotal = I1 + I2 = 0.50 + 0.25 = 0.75 A (iv), , Topic 3. Heating Effect of Current, , V, 1 .5, ⇒ I2 =, = 0.25 A, R, 6, , 1, 1, 1, 1 1 3 1, =, +, = + = =, R p R1 R 2 3 6 6 2, , ⇒ R p = 2Ω, , • Battery or cell is a source of electrical energy in any electric circuit., • The chemical energy is converted into electrical energy in a cell or a battery., • Chemical reaction generates the potential difference that sets, the motion of electrons, current flow through the resistors, connected to the battery., • The source keeps on expending its energy in maintaining the, current., • A part of energy is converted into useful work, e.g. rotation of, motor of fan or mixer and grinder or juicer., • When electric current is passed through a conductor, generated, by it heat due to the resistance it offers to the current flow. The, work done in overcoming the resistance is generated as heat. This is called the heating effect of current.

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Heat produced by a heating element (resistor) is directly proportional to the (i) Square of the current (I2), passing through the conductor, (ii) Resistance R of the conductor, (iii) Time it for which the current is, passed through the conductor., • Heating effect of current has its application in appliances like electric iron, microwave, immersion rod,, induction cooker, heater as these are based on heating effect of current., • Let there be a current I flowing through the resistor of resistance (R). Let the potential difference be ‘V’., • Let ‘t’ be the time during which charge ‘Q’ flows. , (W) Work done = Charge (Q) × Potential difference (V), Power is defined as the energy consumed or work done per unit time.,                P =, , QV, t, , ⇒, , P=, , I×t×V, = V ×I, t, , Amount of heat produced in time ‘t’, H = V × I × t, , ⇒, , H = I2 R t , , ( Q = It), ( V = IR), , Joule’s Law of Heating: It states heat produced in a resistor is, , (i) directly proportional to the square of current for a given resistor., (ii) directly proportional to the resistance for a given resistor., , (iii) directly proportional to the time for which the current flows through a resistor., Practical Applications of Heating Effect of Current, , • The unavoidable heating can increase the temperature of electrical appliances and can alter their properties,, or they may be damaged due to excessive heat., • The electric laundry iron, electric toaster, oven, electric kettle, electric heater, all are based on heating, effect of current., • Heating effect of current in filament produces light in a bulb. It does not melt at high temperature due to, its high resistance. Tungsten has melting point 3380°C. So it is used as a filament in an electric bulb., • Electric bulbs are filled with inert gas like Argon to prevent its oxidation or burning of filament., • Fuse in an electric circuit is most important and is based on heating effect of current. It protect electric, circuits from overvoltage and high current., • The fuse is placed in series combination with the electric device. It is a thin wire of an alloy of copper,, aluminium, iron, etc., • If high current flows in a fuse wire, as the temperature of fuse wire increases, it melts and circuit is, broken., • In domestic circuits different type of fuse wires are used, depending upon the load of the circuit., , Electric Power, The rate of consumption of electric energy in a electric circuit is called electric power, e.g. a 100 watt bulb has, power 100 watt or 100 J s–1. The S.I. unit of power is Watt (W)., , P=V×I, , , \     ,     , , fi     , , R=, , V, I, , fi, , I=, , V, R, , V V2, V2, =, ∴ P=, R, R, R, P = V × I ⇒ P = R × I × I = I2 × R, P=V×, , ( V = R × I), , P = I2 R, , Watt: It is the power consumed by a device that carries 1 A of current when operated at a potential difference, of 1 V.,                1 W = 1 volt × 1 ampere = 1 V A, The unit of electric energy is kilowatt hour (kW h). 1 kilowatt (kW) = 1000 watt, 1 kW h = 1000 watt × 3600 seconds = 3.6 × 106 Watt seconds = 3.6 × 106 J, Important Note: Electrons are not consumed in an electric current, only electrical energy is consumed., Direct current (D.C.): The current which flows only in one direction is called direct current (DC)., • It is produced by a cell or battery., • Circuits in which direct current flow are called D.C. circuits. Both AC and DC can give electric shock.

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AC circuit: The current that flows in two directions alternatively is called alternating current., • It is supplied by electricity department., • Voltage is 220 V to 250 Volt for domestic supply, current is 5 A, frequency is 50 cycles per second or 50, Hertz., • Power is transmitted by AC current., • It gives electric shock if live wire is touched., , Earth wire: It is connected to the metal case of an electrical appliance. In case of leakage of electricity, we may, get shock if live wire is connected to the metal case., • The earth wire allows the current to flow through it to the ground. It can protect us from fatal electric, shock., • In a three pin plug, blue is neutral wire, red is live wire. Earth wire is green or yellow. The earth pin is, longer and thicker than live and neutral pins. The sockets are mechanically protected. The earth pin must, enter in the top hole and other two pins in the bottom holes., , Exercise 5.3, I. Multiple Choice Questions, , (1 Mark), , Choose the correct answer from the given options., 1. In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively, are connected in parallel to an electric source. Which of the following is likely to happen regarding, their brightness?, (a) Brightness of all the bulbs will be the same, (b) Brightness of bulb A will be the maximum, (c) Brightness of bulb B will be more than that of A, (d) Brightness of bulb C will be less than that of B, 2. In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery., The heat dissipated by the 4 Ω resistor in 5 s will be, (a) 5 J      (b) 10 J, (c) 20 J       (d) 30 J, 3. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating, must be used for it?, (a) 0.454A    (b) 2 A, (c) 4 A       (d) 5 A, 4. The type of current supplied by cell or battery is, (a) A.C     (b) D.C, (c) Both (a) and (b) (d) None of these, 5. Unit of electric power may also be expressed as:, (a) volt-ampere (b) kilowatt-hour, (c) watt-second   (d) joule-second, [CBSE Sample Paper 2019-2020], II. Assertion-Reason Type Questions, (1 Mark), For question numbers 1 to 3 two statements are given-one labeled as Assertion (a) and the other labeled, Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given ahead:, (a) Both ‘A’ and ‘R’ are true and ‘R’ is correct explanation of the Assertion., (b) Both ‘A’ and ‘R’ are true but ‘R’ is not correct explanation of the Assertion., (c) ‘A’ is true but ‘R’ is false., (d) ‘A’ is false but ‘R’ is true., 1. Assertion: Electric toasters are made up of alloys., , Reason: Alloys have high resistivity and do not get oxidised at high temperature., 2. Assertion: The commercial unit of electrical energy is 1 kWs, , Reason: 1 kWh = 3.6 × 106 J., 3. Assertion: A fuse wire is always connected in parallel with the mainline., , Reason: If a current larger than the specified value flows through the circuit, fuse wire melts., , [CBSE Sample Paper 2019-2020], , III. Very Short Answer Type Questions, 1. What determines the rate at which energy is delivered by a current?, 2. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?, 3. Name the type of current used in household supply., , (1 Mark), [NCERT], [Delhi 2016]

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4. Write the relationship between heat energy produced in a conductor when potential difference V, is applied across its terminals and a current, I flows through it for the time ‘t’. , [Delhi 2016], 5. Why does the cord of an electric heater not glow while the heating element does?, [NCERT], 6. Why heat is produced when current is passed through a conductor?, 7. Explain the role of fuse wire connected in series with any electrical appliance in an electric circuit., Why should a fuse with a defined rating for an electric circuit not be replaced by the one with a larger, rating?, 8. Why are heating elements made of alloys rather than metals?, OR, Why are alloys commonly used in electrical heating devices like toasters and electric iron? Give reason., [CBSE 2019 AI], 9. A bulb gets dimmer when an electric iron or geyser is switched on, why?, , IV. Short Answer Type Questions-I, , (2 Marks), , 1. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a, potential difference of 50 V., [NCERT], 2. An electric iron of resistance 20 W takes a current of 5 A. Calculate the heat developed in 30 s., [NCERT], 3. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy, consumed in 2 hours., [NCERT], 4. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many, lamps can be connected in parallel with each other across the two wires of 220 V line, if the maximum, allowable current is 5 A? , [NCERT], 5. Find the resistance of bulb rated as 100 W at 250 V., [CBSE 2014], 6. Name the commercial unit of energy. Convert it into Joules. What is other name of commercial unit of, energy?, 7. Find the minimum rating of fuse that can be safely used on a line on which two 1.1 kW rating electric, geysers are to be run simultaneously. The supply voltage is 220 V., [Delhi 2016], , V. Short Answer Type Questions-II, , (3 Marks), , 1. An electric heater of resistance 8 W draws 15 A current from the supply mains for 2 hours. Calculate, the rate at which heat is developed in the heater., 2. A potential difference of 4 V derive a current of 3 A through a resistor. How much electrical energy will, be converted into heat during 10 seconds?, 3. An electric fan has a rating of 460 W on the 230 V mains line. What fuse should be fitted in the plug?, 4. What is the cost of running an AC with average power of 1000 W for 8 hours for 30 days. The cost of, electric energy is `4.70 per kW h., 5. Why do we get electric shock in damp conditions?, 6. Compare the power used in the 2 W resistor in each of the following circuits: (i) a 6 V battery in, series with the 1 W and 2 W resistors, and (ii) a 4 V battery in parallel with 12 W and 2 W resistors., [NCERT], 7. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to, electrical mains supply. What current is drawn from the line if the supply voltage is 220 V?, [Delhi 2018] [NCERT], 8. (i) Explain what is the difference between direct current and alternating current. Write one important, advantage of using alternating current., , (ii) An air conditioner of 2 kW is used in an electric circuit having a fuse rating of 10 A. If the potential, difference of the supply is 220 V, will the fuse be able to withstand when the air conditioner is, switched on? Justify your answer., [Delhi 2016], 9. What is meant by ‘electrical resistance’ of a conductor? State how resistance of a conductor is effected, when, , (i) Low current passes through it for a short duration., , (ii) A heavy current passes through it for 30 seconds. , [Delhi 2015], 10. An electric kettle of 2 kW works for 2 hours daily. Calculate the, , (a) Energy consumed in SI unit and commercial unit., , (b) Cost of running it in the month of June at a rate of ` 3.00 per unit., [Delhi 2014], 11. An electric bulb is rated 220 V and 100 W, when it is operated at 110 V. What will be the power, consumed? , [Delhi 2010]

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12. An electric iron has a rating of 750 W, 220 V. Calculate the, , (i) Current flowing through it and (ii) Its resistance when it is in use., 13. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 4 W, in series with a combination of two resistors (8 W each) in parallel and a voltmeter across parallel, combination. Each of them dissipate maximum energy and can withstand a maximum power of 16 W, without melting. Find the maximum current that can flow through the three resistors., , , [CBSE Sample Paper 2017-2018], , 14. Two lamps, one rated 100 W at 220 V and the other 200 W at 220V are connected (i) in series and, (ii) in parallel to electric main supply of 220V. Find the current drawn in each case., 15. How many 40 W; 220 V rating lamps can be safely connected to a 220 V, 5 A line? Justify your answer., [Delhi 2015], , VI. Long Answer Type Questions, , (5 Marks), , 1. An electric circuit two resistors of 20 W and a conductor of resistance, 4W are connected to a 6 V vattery as shown in the figure. Calculate:, , (a) The total resistance of the circuit,, , (b) the current through the circuit,, , (c) the potential difference across the (i) electric lamp and (ii) conductor,, and, , (d) power of the lamp., 2. Explain the following:, , (i) Why is tungsten used almost exclusively for filament of electric lamps?, , (ii) Why are the conductors of electric heating devices, such as bread-toaster and electric iron, made of, an alloy rather than a pure metal?, , (iii) Why is series arrangement not used for domestic circuits?, , (iv) How does the resistance of a wire vary with its area of cross-section?, , (v) Why are copper and aluminium wires usually used for the transmission of electric current?, [NCERT], 3. (a) Define electric power. Express it in terms of potential difference, V and resistance, R., , (b) An electrical fuse is rated at 2 A. What is meant by this statement?, , (c) An electric iron of 1 kW is operated at 220 V. Find which of the following fuse that are respectively, rated at 1 A, 3 A and 5 A can be used for it., 4. (a) Write two points of differences between electric energy and electric power., , (b) Out of 60 W and 40 W lamps, which one has higher electrical resistance when in use., , (c) What is the commercial unit of electric energy? Convert it into the units of Joule., 5. Two identical wires, one of nichrome and other of copper are connected in series and a current (i) is, passed through them. State the change observed in the temperature of the two wires. Justify your, answer. State the law which explains the above observations., 6. (a) State Ohm’s law. Derive the relation and give graphical representation for it., , (b) An electric oven rated at 500 W is connected to a 220 V line and used for 2 hours daily. Calculate, the cost of electric energy per month at the rate of ` 5 per kW h., 7. In the given circuit, A, B, C and D are four lamps connected with a battery of 60V., , 60 V, , 3A, , A, , Analyse the circuit to answer the following questions., , 4A, , B, , 5A, , C, , 3A, , D

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(i) What kind of combination are the lamps arranged in (series or parallel)?, , (ii) Explain with reference to your above answer, what are the advantages (any two) of this combination, of lamps?, , (iii) Explain with proper calculations which lamp glows the brightest?, , (iv) Find out the total resistance of the circuit., [CBSE Sample Paper 2020-21], 8. (a) What is the function of fuse wire in an electric circuit?, , (b) What would be the rating of the fuse for an electric kettle which is operated at 220 V and consumes, 500 W power?, , (c) How is the SI unit of electric energy related to its commercial unit?, 9. (i) Consider a conductor of resistance ‘R’, length ‘L’, thickness ‘d’ and resistivity ‘ρ’., Now this conductor is cut into four equal parts. What will be the new resistivity of each of these, parts? Why?, , (ii) Find the resistance if all of these parts are connected in: (a) Parallel (b) Series, , (iii) Out of the combinations of resistors mentioned above in the previous part, for a given voltage which, combination will consume more power and why? [CBSE Sample Paper 2019-2020], , Answers 5.3, I. 1. (c) Higher the wattage (power), more will be brightness, 2. (c) H = I2× R × t = 1 × 4 × 5 = 20 J. I = V = 6 V = 1A, R 6Ω, P 1000 W, =, 3. (a) I =, = 0.454 A, V, 220 V, 4. (b) D.C (Direct current) which flows only in one direction., , 5. (a) volt-ampere,  Power = Voltage × Current., II. 1. (a) Both ‘A’ and ‘R’ are true and ‘R’ is correct explanation of the assertion., , 2. (d) ‘A’ is false and ‘R’ is true because commercial unit is kW h and not kW s., , 3. (d) ‘A’ is false and ‘R’ is true., III. 1. Electric power is the rate at which energy is delivered by a current., 2. Energy consumed by 250 W TV in 1 h = 250 × 1 = 250 W h, Energy consumed by 1200 W toaster in 10 minutes = 1200 × 10 = 200 W h, 60, Here total energy consumed by the TV is more than toaster., 3. Alternate current, 4. H = VIt, 5. The cord of the electric heater is made up of copper or aluminium which has lower resistance and, higher conductance, therefore very less heat energy is produced., Heating element is made up of alloys which have high resistance and lot of heat is produced, and only, a little amount of electric energy makes it to glow., 6. The electrons collide with each other while moving and loses some kinetic energy which is converted, into heat energy., 7. Fuse wire is a safety device connected in series with the live wire of the circuit, since it has high, resistivity and low melting point. It melts when a sudden urge of large current passes through it and, disconnects the entire circuit from the electrical supply. But in case if we use a larger rating fuse wire, instead of a defined rating fuse wire then it will not protect the circuit as high current will easily pass, through it and it will not melt., 8. Alloys have high resistivity/high melting point/alloys do not oxidise (or burn readily at high, temperatures)., 9. It is because electric iron or geyser draws heavy current from the circuit, takes heavy power and, therefore bulb becomes dimmer due to decrease in voltage., IV. 1. Q = 96000 C, t = 1 hr = 3600 s, V = 50 volt, H = V × I × t = V × Q = 50 × 96000 C =, , 50 × 96000, kJ = 4800 kJ, 1000

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2., , 3., , , 4., , R = 20 Ω, I = 5 A, t = 30 s, H = I2 × R × t = 5 × 5 × 20 × 30 = 15000 J = 15 kJ, I = 5 A, V = 220 V, P = ?, E = ?, t = 2 hours, P = V × I = 220 V × 5 A = 1100 W, E = P × t = 1100 W × 2 h = 2200 W h = 2.2 kW h, Total power = N × P where ‘N’ is the number of bulbs, ‘P’ power of each bulb., , , I=, 5. P =, , N×P, V, , V2, R, , ⇒ 5=, , ⇒ R=, , N × 10, 220, , ⇒ N=, , 5 × 220, = 110, 10, , 250 × 250, V2, =, = 625 Ω, 100, P, , 6. The commercial unit of electrical energy is kW h. (Kilowatt hour), 1 kW h =1 kW × 1 hour = 1000 W × 3600 s = 3.6 × 106 J, P 1.1 × 1000 1100, =, =, =5A, V, 220, 220, The rating of fuse that can be used is more than 5 A., It is also called Board of Trade Unit (BOTU), V. 1. R = 8 Ω, I = 15 A, t = 2 hours, H I2 × R × t, =, = I2 × R, Rate of heat developed =, t, t, 7. I =, , , = 152 × 8 = 1800 J s–1 = 1800 Watt = 1.8 kW, V 4, 2. R =, = = 1.33 Ω, I, 3, 4, H = I2 × R × t = 3 × 3 × × 10 = 120 J, 3, 3. P = V × I ⇒ 460 = 230 × I ⇒ I = 2 Ampere, A 3A fuse should be fitted (It should be slightly more than 2A)., 4. E = P × t = 1000 W × 8 × 30 = 240000 W h = 240 kW h, Cost of electric energy = 240 × 4.70 = ` 1128., 5. Water provides conducting path for a current to flow through the human body in damp conditions like, bathroom. We get electric shock if we are bare footed. Wet body has low resistance, high current can, easily pass through, leading to electric shock., 6. (i) Rs = R1 + R2 = 1 Ω + 2 Ω = 3 Ω, V = 6 V, I = V = 6 = 2 A, R 3, , P = I2 × R (Since same current flows through each resistor) = 2 × 2 × 2 = 8 W, , (Q R = 2 Ω), , 1, 1, 1 1+6, 7, 1, 1, 1, 12, =, + =, =, =, +, , (ii), ⇒, Ω, ⇒ Rp =, R p 12 2, 12, 12, R p R1 R 2, 7, , Resistance will change, voltage remains the same., , V2 4 × 4, =, =8W, 2, R, Power used are in the ratio 1 : 1., , Power across 2 W resistor, P =, , 7. R100 W =, I =, , V 2 220 × 220, V 2 220 × 220, =, ; R60 W =, =, P, P, 100, 60, , V, 220 × 100 100, V, 220 × 60, 6, =, =, = 0.45 A ; I =, =, =, = 0.27 A, R100 220 × 220 220, R 60 220 × 220 22, , , \ Total current = 0.45 A + 0.27 A = 0.72 A, 8. (i) The current whose direction gets reversed after every half cycle is called alternating current or A.C., There is no change in the direction of D.C. D.C. is a uni directional current.

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The most important advantage of using A.C. over D.C. is that in the A.C. mode electric power can, be transmitted over long distances at a high voltage and low current with very little loss of power., , (ii) Here P = 2 kW = 2000 W, V = 220 Volt, P 2000, = 9.09 A, , P = VI, so the current, I =, =, V, 220, As the current is 9.09 A below the rating of fuse, therefore the fuse will withstand, i.e. it will not, blow off when A.C. is on., 9. It is in opposite direction offered to the flow of electric current., , (i) No effect on resistance, low current, hence no appreciable rise in temperature so there no change, in resistance., , (ii) Heavy current for 30 seconds may increase the temperature so resistance will increase., 10. (a) P = 2 kW, t = 2 h, E =P × t = 2 × 2 h = 4 kW h, , (b) Total energy consumed per month = 4 kW h × 30 = 120 kW h, Total cost = 120 × 3 = ` 360., 11. R =, P =, 12., , V2, 220 V × 220 V, = 484 Ω, =, P, 100 W, V2, 110 × 110, =, = 25 W, R, 484, , (i) P = 750 W, V = 220 V, , P = V × I ⇒ 750 = 220 × I ⇒ I =, , (ii) P =, , V2, R, , ⇒, , R=, , V 2 220 × 220, =, P, 750, , 750, = 3.409 A, 220, ⇒ R = 64.53 Ω, , 13., , , , Maximum current through 4 W resistor =, , , , Maximum current through each 8 W resistor =, , 14. R1 =, , P, 16 W, =, =, R, 8Ω, , 2 = 1.414 A, , V 2 220  220, V 2 220  220, ,  242 , ,  484  ; R2 =, P2, 200, P1, 100, In series: Rs= R1 + R2 = 484 + 242 = 726 W, , , , In parallel:, , , P, 16 W, =2A, =, R, 4Ω, , ∴ Is =, , V 220 10, =, = A = 0.30A, R s 726 33, , 484, 1, 1, 1, 1, 1, Ω, , , =, ⇒ Rp =, +, 3, R p R1 R 2, 484 242, V 220  3 30, , , A  1.36A, Ip =, RP, 484, 22

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N × 40 W, 220 V × 5 A 110, N×P, ⇒ 5A =, ⇒N=, =, = 27 lamps, 220, V, 40 W, 4, V, VI. 1. (a) Rs = R1 + R2 = 20 Ω + 4 Ω = 24 Ω, 15. I =, , V, 6V, , (b) I = R = 24 Ω = 0.25A, , (c) (i) V = I × R = 0.25 A × 20 Ω = 5 V, , (ii) V = I × R = 0.25 A × 4 = 1V, 2, , 1,  1, × 20 = 1.25 W, , (d) (i) P = I2 × R =   × 20 =,  4, 16, 2. (i) It has high resistance and high melting point. So it does not melt when current is passed through, it., , (ii) Alloys have more resistivity and higher melting point., , (iii) It increases the resistance and current decreases. Also if one component fails to work, others will, also not work, , (iv) Resistance is inversely proportional to the area of cross section., , (v) It is because Cu and Al have low resistivity and it allows current to flow., 3. (a) Electric power is the rate of doing work by an energy source or the rate, at which the electrical energy dissipated or consumed per unit time in the, electric circuit., So,, , , , Power (P) =, , Work done (W), Time (t), , Electrical energy dissipated, =, Time (t), ⇒ VI or P =, , 3, V (Volt) 2, 1, , V2, R, , 1, 0.5, 1.5, I (Ampere), , , (b) It means the maximum current will flow through it is only 2 A. Fuse wire will melt if the current, flowing through it exceeds 2 A value., , (c) Given:, P = 1 kW = 1000 W, V = 220 V, P, 1000, 50, =, =, = 4.54 A, V, 220, 11, To run electric iron of 1 kW efficiently, rated fuse of 5 A should be used., , Current drawn, I =, , , 4. (a) Difference between electric energy and electric power:, Electric Energy, , Electric Power, , (i) The work done or energy supplied by the (i) The time rate at which electric energy is, consumed or dissipated by an electrical device, source in maintaining the flow of electric, is called electric power and is given by:, current is called electrical energy. It, appears in the form of heat and is given, V2, = I2 R, P = VI =, , by:, R, V2t, H = VIt =, = I 2 Rt, , R, (ii) It is also equal to the product of power and (ii) It is also equal to the rate of doing work by an, time taken:, energy source and is given by:, , E=P×t, W, P=, , t, (iii) Its SI unit is Joule (J), (iii) Its SI unit is Watt (W), , 1J=1W×1s,       1 W = 1 J s–1      (Any two)

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1, ,, R, i.e. smaller the power of electrical device, higher is its electrical resistance., So a 40 W lamp has higher electrical resistance than a 60 W lamp., , (c) Kilowatt hour – Commercial unit of electrical energy., 1 kW h = 1000 W h = 1000 J/s × 3600 s = 3600000 J = 3.6 × 106 J, 5. The resistivity of nichrome is more than that of copper, so its resistance is also high. Therefore large, amount of heat is produced in the nichrome wire for the same current passed as compared to that of, the copper wire. Accordingly greater change in temperature is observed in case of nichrome wire. This, can be explained by Joule’s law of heating., Joule’s law of heating states that the amount of heat produced in a conductor is:, , (i) Directly proportional to the square of current flowing through it, i.e., , H ∝ I2, , (ii) Directly proportional to the resistance offered by the conductor to the flow current, i.e., H ∝ R, , (iii) Directly proportional to the time for which current is flowing through the conductor, i.e., H ∝ t, Combining these, we get, H ∝ I2Rt, or H = KI2Rt, where K is a proportionality constant and in SI units, it is equal to unity., , (b) For the same applied voltage, P ∝, , 6. (a) Linear equation, hence a straight line passing through the origin, is obtained as follows:, Ohm’s law states that potential difference across the given metallic, wire in an electric circuit is directly proportional to the electric, current flowing through it, if the temperature remains constant., Mathematically, V ∝ I., or, or, , V, = Constant, I, V = IR (where R is a constant known as the, constant of proportionality)., , , (b) Energy consumed per day = 1 kW h., , , P = V × I, E = P × t  E = V × I × t = 220 × 2.27 × 2 = 1 kWh, 500 , P, 500, , =, = 2.27 A, ∵P = V × I ∴I =, , , V, 220, 220 , Cost of electric energy for 30 days = 1 × 5 × 30 = ` 150.00, 7. (i) The lamps are in parallel., , (ii) Advantages:, If one lamp is faulty, it will not affect the working of the other lamps., They will also be using the full potential of the battery as they are connected in parallel., , (iii) The lamp with the highest power will glow the brightest., , P = VI, In this case, all the bulbs have the same voltage. But lamp C has the highest current., Hence, for lamp C, P = 5 × 60 Watt = 300 W. (the maximum)., , (iv) The total current in the circuit = (3 + 4 + 5 + 3) A = 15A, , The Voltage = 60V, , V = IR and hence R = V/I, , R = 60/15 W = 4 W, 8. (a) Fuse wire protects the electrical circuits from over voltage and high current., , (b) 2.2 A flows through the circuit, fuse should be rated 3 A., , (c) 1 kW h = 3.6 × 106 J, 9. (i) Resistivity will not change as it depends on the nature of the material of the conductor., , (ii) The length of each part becomes L/4. While ρ, A remain constant.  R = ρL/A., Resistance of each part = Rpart = (ρL/4)/A = R/4., So, I =

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(a) In parallel connection,, , 1, 1, 1, 1, 1, 4, 16, R, =, +, +, +, =, =, ⇒ R eqv =, Ω, R eqv R part R part R part R part R part R, 16, , , (b) In series connection, Reqv =, , R R R R, + + +, = R Ω (ohms), 4 4 4 4, , , (iii) P = V2/R., If Reqv is less, power consumed will be more., In the given case, Reqv is lesser in the parallel connection and the power consumed will be more., , C ase study questions, , 1. Ohm’s law gives the relationship between current flowing, , through a conductor with potential difference across it, provided the physical conditions and temperature remains, constant. The electric current flowing in a circuit can be, measured by an ammeter., Potential difference is measured by voltmeter connected, in parallel to the battery or cell. Resistances can reduce, current in the circuit. A variable resistor or rheostat is used, to vary the current in the circuit., , (i) What type of conductor is represented by the following, graph?, , Current, , I, , V, Voltage, , , (a) Non-ohmic conductor like thermistor, (b) Non-ohmic conductor like metal filament, , (c) Ohmic conductor like copper, (d) None of these, , (ii) What type of conductors are represented by the following graph?, , Current, , I, , V, Voltage, I, , Current, , , (a) Non-ohmic conductor like thermistor, , (b) Non-ohmic conductor like metal filament, , (c) Ohmic conductor like copper, , (d) None of these, , (iii) Which type of conductor is represented by the graph given alongside?, , (a) Non-ohmic conductor like thermistor, , (b) Non-ohmic conductor like metal filament, , (c) Ohmic conductor like copper, , (d) None of these, , V, Voltage

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(iv) What is the slope of graph in (i) equal to?, , (a) V, , (b) I, , (c) R, , (d) VI, , , (v) Which of the following is the factor on which resistance of a conductor does not depend?, , (a) Length, (b) Area, (c) Temperature, (d) Pressure, Ans. (i) (c)  (ii) (b)  (iii) (a)  (iv) (c)  (v) (d), 2. Study this table related to material and their resistivity and answer the questions that follow., Material, , Resistivity (W m), , Conductors, , Silver, Copper, Sluminium, Tungsten, Nickel, Iron, Chromium, Mercury, Manganese, , 1.60 × 10–8, 1.62 × 10–8, 2.63 × 10–8, 5.20 × 10–8, 6.84 × 10–8, 10.0 × 10–8, 12.9 × 10–8, 94.0 × 10–8, 1.84 × 10–6, , Alloys, , Constantan (alloy of Cu and Ni), Manganin (alloy of Cu, Mn and Ni), Nichrome (alloy of Ni, Cr, Mn and Fe), , 49 × 10–6, 44 × 10–6, 100 × 10–6, , Insulators, , Glass, Hard rubber, Ebonite, Diamond, Paper (dry), , 1010 – 1014, 1013 – 1016, 1015 – 1017, 1012 – 1013, 1012, , , (i) What is the range of resistivity in metals, good conductors of electricity?, (b) 10–6 to 10–4 Ωm (c), (a) 10–8 to 10–6 Ωm , (d) 1012 to 1014 Ωm, 1010 to 1014 Ωm , , (ii) Which property of the alloy makes it useful in heating devices like electric iron, toasters,, immersion rods, etc.?, (a) Higher resistivity, (b) Do not oxidise at low temperature, (c) Do not reduce at high temperature, (d) Oxidise at high temperature, , (iii) Which of the following is used in transmission wires?, (a) Cr (b) Al, (c) Zn, (d) Fe, , (iv) Which is the best conducting metal?, (a) Cu (b) Ag, (c) Au, (d) Hg, , (v) Which of the following is used as a filament in electric bulbs?, (a) Nichrome (b) Tungsten (c) Manganese, (d) Silver, Ans. (i) (a)  (ii) (a)  (iii) (b)  (iv) (b)  (v) (b), , Quick revision notes, , • Resistance is the property of a conductor that resists or opposes the flow of current through it. It controls the, magnitude of current passing through the substance., • Higher resistance reduce the current in a circuit., • The S.I. unit of resistance is Ohm (W).

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• Rheostat is used to provide variable resistance, e.g. speed of fan can be decreased by increasing the resistance, and can be increased by decreasing the resistance., • Ohm’s law gives the relationship between potential difference (V) and current (i):, V ∝ I or V = RI, where R is a constant called resistance of the conductor., • The resistance of a wire is directly proportional to the length of wire and is inversely proportional to its area, of cross section., • Thinner the wire, more will be resistance, that is why a thin wire is used in electric fuse which can resist, high voltage only to some extent and prevent electric circuit from damage., • Longer the wire, more will be resistance., • Resistance also depends upon the material of conductor, e.g. copper is a good conductor of electricity due to, its low resistance., • Temperature also affects the resistance, e.g. resistance of metals increases with increase in temperature and, decreases with decrease in temperature., • That is why some metals become superconductors at very low temperature because practically they offer no, resistance in the path of flow of current., • The equivalent resistance of many resistors connected in series is equal to the sum of the resistances of, individual resistors: RS = R1 + R2 + R3 + …, • The equivalent resistance of resistors connected in parallel is given by:, 1, 1, 1, 1, =, +, +, +…, RP, R1, R2, R3, • A cell can supply energy but not charge. Its function is to drive the free electrons in a circuit to flow., • The electrical energy dissipated (lost) in a resistor is given by:, W=V×I×t, where ‘V’ is voltage, ‘I’ is current, ‘t’ is time for which current flows., • Electric kettles, cookers, electric iron and heaters make use of electric energy which is converted into heat, energy. The heating element is made up of a material having high resistance., • When electrons pass through the metal wire, they give some of their energy to the particles of wire and make, them vibrate more randomly, and that is why wire gets hotter., • For the same current flowing in the circuit, greater the resistance, the hotter it becomes. That is why electric, iron, immersion rod becomes red hot but electric transmission wires remain cool due to low resistance., • Energy is measured in Joules., • Electric power is the rate of using electrical energy. It is measured in joules per second J s –1 or, in a unit called Watt (W)., • One watt power is consumed when 1 Ampere (a) current flows at a potential difference of 1 Volt., • The commercial unit of electrical energy is kilowatt hour (kW h)., , 1 W = 1 J s–1,  , 1 kW = 1000 W, , 1 hour = 3600 seconds, , \ , 1 kW h = 1 × 1000 × 60 × 60 J = 36 × 105 = 3.6 × 106 J, • Power (P) = V × I, where ‘V’ is voltage (potential difference) and I is current in Ampere., V, V, \ I=,      , Ohm’s law, R =, I, R, P=V × I =V×, ,             , , V, V2, =, R, R, , ⇒, , • Electric energy (E) = V × I × t, E=, , V2t, R, , E = I2 Rt, , V, , , ∵ I = R by Ohm’s law , , [∵, , V = I × R by Ohm’s law ], , P=, , V2, or P = I 2 R, R

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important Formulae, 1. Ohm’s law: V = IR , where R is resistance., l, 2. R = ρ, where, ‘l’ is length of conductor, ‘A’ is, A, area of cross section, rho (r) is resistivity., 3. R S = R1 + R 2 + R3, , where,, , RS, , is, , total, , resistance connected in series combination, In series combination, current across each, resistor remains same, 1, 1, 1, 1, =, +, +, 4., where RP is equivalent, R P R1 R 2 R 3, resistance in parallel combination; R1, R2, R3, are resistors connected in parallel., , In parallel combination potential difference across, each resistor remains same., 5. H = VIt ; H = P × t, ‘H’ is heat produced in a conductor, I is current, t, is time, ‘V’ is voltage (Potential difference), ‘P’ is, power., 6. Joule’s Law: H = I 2Rt, 7. P = I2 × R, , 13. P =, , V2, R, , 6, 8. 1 kW h = 3.6 × 10 J, , Common Errors, Errors, ●, ●, , ●, , ●, , ●, , ●, , ●, , Students do not write the correct SI units., , Corrections, , ☞ Always write the correct SI unit., , Students do not show arrowhead in circuit ☞ Direction of current should be shown by arrowhead, diagrams., in circuit diagram., Students connect Ammeter in parallel and ☞ Ammeter is always connected is series, Voltmeter, Voltmeter in series., in parallel., Students do not write formulae in numericals., , ☞ Formula must be written before starting a, numerical., , Students interchange or write incorrect formulae ☞ Rs = R1 + R2 + R3: for series combination;, for series and parallel combination of resistances., For parallel combination:, 1, 1, 1, 1, and not, =, +, +, RP, R1, R2, R3, 1, 1, 1, +, +, RP =, R1, R2, R3, Students get confused between resistance and ☞ ‘R’ (resistance) depends upon length, area,, resistivity., temperature; while resistivity only depends upon, temperature and nature of material., Circuit diagrams and other diagrams are not ☞ Circuit diagrams and other diagrams must be, drawn and properly labeled., made with pencil with proper labeling.

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Assignment, I. Multiple Choice Questions, , Total Marks : 20, (1 Mark), , Choose the correct answer from the given options., 1. Which of the following does not represent electric power?, (a) I2R, (b) IR2, (c) VI, (d) V2/R, [CBSE 2020], 1, 2. If a person has five resistors each of value Ω , then the maximum resistance he can obtain by, 5, connecting them is, (a) 1 W, (b) 5 W, (c) 10 W, (d) 25 W , [CBSE 2020], II. Assertion-Reason Type Questions, (1 Mark), , Note: Use instructions as given in topical exercises of the chapter., 1. Assertion: The graph between V and I is a straight line., , Reason: V is directly proportional to ‘R’ and the slope gives value of ‘R’., 2. Assertion: At high temperature, metal wires have greater chances of short circuiting., , Reason: Both resistance and resistivity of metallic wires varies with temperature., , III. Very Short Answer Type Questions, , (1 Mark), , 1. What is the function of rheostat?, 2. List two factors on which resistance depends., , IV. Short Answer Type Questions-I, , (2 Marks), , 1. Which has more resistance, ammeter or voltmeter? Give reason., 2. How are voltmeter and ammeter connected in an electric circuit?, 3. If you connect three resistors having values 2 W, 6 W and 8 W in parallel, calculate the equivalent, resistance., , V. Short Answer Type Question-II, , (3 Marks), , 1. (a) Write mathematical expression for Joule’s Law of heating., , (b) Compare the heat generated while transferring 96500 coloumb of charge in two hours through a, potential difference of 40 V., [CBSE 2020], , VI. Long Answer Type Question, , (5 Marks), , 1. (a) Two lamps rated 100 W, 200 V and 10 W, 220 V are connected in parallel to 220 V supply. Calculate, the total current through the circuit., , (b) Two resistors X and Y of resistance 2 W and 3 W respectively are first joined in parallel and then in, series. In each case the voltage is 5 V., (i) Draw circuit diagram to show the combination of resistors in each case., (ii) Calculate the voltage across the 3 W resistor in the series combination of resistors., [CBSE 2020]