Page 1 :
age |1, , V.D.S.T.C BOYS' HIGHSCHOOL GADAG’, PHYSICS NUMERICAL PROBLEMS, , 10° Physics Electricity Numerical Problems and Solutions, , 1. Calculate the number of electrons constituting one coulomb of charge., The value of the charge of an electron is 1.6 107°C., 1C=nx16x 107”, , 1, = = 18, = Fpy aps = 625 x 10, , Therefore, the number of electrons constituting one coulomb of charge is 6. 25 « 10'8., , 2. How much energy is given to each coulomb of charge passing through a 6 V, battery?, , Potential difference, V=6V, Charge , Q =1C, we know that,, V=WQ, W=VxQ, =6x1, =6], , 3. Redraw the circuit of Question 1, putting in an ammeter to measure the current, through the resistors and a voltmeter to measure the potential difference across, the 12 Q resistor. What would be the readings in the ammeter and the voltmeter?, An ammeter should always be connected in series with resistors while the voltmeter, , should be connected in parallel to the resistor to measure the potential difference, The total resistance of the circuit is 5 Q +8 Q+12Q=25Q., , We know that the potential difference of the circuit is 6 V, hence the current flowing, through the circuit or the resistors can be calculated as follows:, , I= V/R = 6/25 = 0.24A, , Let the potential difference across the 12 Q resistor be V)., , From the obtained current V; can be calculated as follows:, , V; =0.24A «12 Q2= 2.88 V, , Therefore, the ammeter reading will be 0.24 A and the voltmeter reading be 2.88 V., , 4. Judge the equivalent resistance when the following are connected in parallel — (a), 1Q and 10° Q, (b) 1 Q, 10° Q, and 10° Q.
Page 2 :
Page |2, , , , a), riila, R 1° 108, pu 1 _ ag, ~ 14108 108, b), 11,11, R 1 103° 108, Solving, we get, 10°+10%+1 1000000, R=———_ = ——_= 0.999 2, , 10° ~ 1000001, , 5. An electric lamp of 100 Q, a toaster of resistance 50 Q, and a water filter of, resistance 500 Q 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 three appliances, and what is the current through it?, , The electric lamp, the toaster and the water filter connected in parallel to a 220 V, source can be shown as using a circuit diagram as follows:, , , , , , R,=1000, ANA. + ~, [ R,=500, AAA. +, R,=5009, AK. +, , , , , , , , ai, V=220V, , The equivalent resistance of the resistors can be calculated as follows:, , , , 11 11, === 4 =, R R1- R2_ RB, 11 1%, R 100 50 500, 1 _ 541041, , R 500
Page 3 :
Page |3, , , , , , _ 500, 16, v, I =a, R, 220, 1= 00, 16, 220 x 16, I ——, 500, 1=7.04A, v, R=, 1, _220, 7.04, , The resistance of the electric iron box is 31.25 Q., , 6. How can three resistors of resistances 2 Q, 3 Q, and 6 Q be connected to give a, total resistance of (a) 4 Q, (b) 1 Q?, (a) The circuit diagram below shows the connection of three resistors, , 6a, , 20, , From the circuit above, it is understood that 3 Q and 6 Q are connected in parallel. The, equivalent resistance is given by, , 6n, Vv, 20, , 2 1, “ate, , p 1, cur, , 241, , Hl, |, , I, wlan IW, , DW RIP RIP RIF RIE
Page 4 :
Page |4, , R=20, Total resistance, , R=20Q+2Q0=40, , (b) The circuit diagram below, shows the connection of three resistors., , Vv $10 62, , , , From the circuit, it is understood that all the resistors are connected in parallel. Therefore,, their equivalent resistance can be calculated as follows:, , , , v..2 , tf, ===] 4 +=, R R1- R2_ RB, a_4@, 1.4, Se5.1..cuS, R273 "6, 1_ 34241, , R 6, , 1_6, , R 6, , R=19, , The total resistance of the circuit is 1 Q., , 7. What is (a) the highest, (b) the lowest total resistance that can be secured by, combinations of four coils of resistance 4 Q, 8 Q, 12 Q, 24 Q?, (a) The total equivalent resistance of the resistors connected in series will be, 4248 Q4+12Q+240Q=48Q., , (b) If the resistors are connected in parallel, then their equivalent resistances will be the, , lowest., , Their equivalent resistance connected in parallel is, 1,1,1,1, ie afi, , pa, R1 R2 R3 R4, 4, 4, , PIR mle, , =<+, , ole, sl, vl
Page 5 :
Page |5, , , , Hence, the lowest total resistance is 2 Q., , 8. Compute the heat generated while transferring 96000 coulomb of charge in one, hour through a potential difference of 50 V., The heat generated can be calculated by Joule’s law of heating., , we know that., , H=I?Rt, H=VIt ( V=IR), H=VIt, V=50V, Q= 9600C, T=,1h = 3600s, p=2, t, a eee, 3600, I= 26.66A, Amount of heat liberated, H =VIt, = 50 x 26.66 x 3600, =48x 108 J, , 9. An electric iron of resistance 20 Q takes a current of 5 A. Calculate the heat, , developed in 30 s., The amount of heat generated can be calculated by using the Joule’s law of heating., , H=PRt, H=VIt ( V=IR, H= 100 x 5 x 30=1.5 x 104 J, The amount of heat developed by the electric iron in 30s is 1.5 = 10*J.
