Page 1 : BEEE- BRANCH- (IT), Section A: Electrical Engineering, Unit-5 (DC Circuits), By, MEENAKSHI RASTOGI, LECTURER- ELECTRICAL ENGINEERING, Government Polytechnic Lucknow

Page 2 : Ohm’s Law, Ohm’s laws state that the current through any two points of the conductor is, directly proportional to the potential difference applied across the conductor,, provided physical conditions i.e. temperature, etc. do not change. It is measured in, (Ω) ohm., Mathematically it is expressed as, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 3 : Ohm’s Law, In other words, Ohm’s law can also be stated as;, The ratio of the potential difference across the end point of the conductor to the, current flowing between them is always constant, but the physical conditions of the, conductor i.e. temperature, etc. remain same., This constant is also called the resistance (R) of the conductor (or circuit), , It can be written as, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 4 : Ohm’s Law, In a circuit, when current flows through a resistor, the potential difference across, the resistor is known as voltage drops across it, i.e., V = IR., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 5 : Limitations of Ohm’s Law, 1. Ohm’s law is not applicable in unilateral networks. Unilateral networks allow the, current to flow in one direction. Such types of network consist elements like a, diode, transistor, etc., 2. It is not applicable for the non-linear network. In the nonlinear network, the, parameter of the network is varied with the voltage and current. Their parameter, likes resistance, inductance, capacitance and frequency, etc., not remain constant, with the times. So ohms law is not applicable to the nonlinear network., 3. Ohm’s law is used for finding the resistance of the circuit and also for knowing, the voltage and current of the circuit., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 9 : Effect of temperature on resistance, The electrical resistance changes with the change of temperature. The resistance, does not only increase with the rise in temperature but it also decreases in some, cases. In fact, for the different type of materials, the amount of change in, resistance due to change in temperature is different which are discussed as follow., Metal: The resistance of all pure metals increases linearly with increase in, temperature over a limited temperature range. At low temperature, the ions are, almost stationary. As the temperature increases, the ions inside the metal acquire, energy and start oscillating about their mean positions. These vibrating ions collide, with the electrons Hence resistance increases with increase in temperatures., The resistance of all metals such as tungsten, copper, aluminum etc. increases, linearly with increase in the temperature over a limited temperature range. For e.g., the resistance of copper is 100Ω at 0°c then it increases linearly upto 100°c. At a, temperature of -234.5°c the resistance of copper is almost zero as shown in the, figure., Meenakshi Rastogi-Lecturer-EE-GPL

Page 11 : Effect of temperature on resistance, The change in resistance of a material with the increase in temperature can be, expressed b means of the temperature coefficient of resistance. Consider a, conductor having resistance Ro at 0°c and Rt at t°c. From the above discussion, we, can conclude that the change in the resistance i.e (Rt – Ro) is, 1.Directly proportional to the initial resistance Ro, 2.Directly proportional to the rise in temperature t°c., 3.Depends on the nature of the material for conductor metals and alloy, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 14 : Heating effect of electric current, When a potential difference is applied across the ends of a conductor, an electric, field is set up across the conductor. The free electrons of the conductor are, accelerated ( during the intervals between collisions with the ions or atoms) in the, direction opposite to that of the electric field. As a result, the free electrons acquire, extra kinetic energy in addition to their own kinetic energy (due to their thermal, motion). Now, these electrons impart more kinetic energy to the atoms of the, lattice during collisions. As a result of the amplitude of thermal vibrations of the, ions and atoms of the lattice increases. So the number of collisions between free, electrons and ions in the lattice increases. Thus more work is done to carry free, electrons from one end of the conductor to the other end. This work done is, converted into heat energy of the conductor, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 16 : Uses- Heating effect of electric current, Examples- Electric lamp, Electric heater, OTG(Oven-Toaster-Griller), Electric Geyser,, Immersion rod(water-heater), Electric Welding, Arc -furnace, Electric iron, etc., , OTG(Oven-Toaster-Griller), Electric lamp, Electric Geyser, Meenakshi Rastogi-Lecturer-EE-GPL

Page 17 : Conversion of mechanical units into electrical, units, The Voltage Unit, , Where: voltage is in Volts, J is the work or energy in Joules and C is the charge in, Coulombs. Thus if J = 1 joule, C = 1 coulomb, then V will equal 1 volt., Electrical Energy Example No1, What is the terminal voltage of a battery that expends 135 joules of energy to, move 15 coulombs of charge around an electrical circuit., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 18 : Conversion of mechanical units into electrical, units, The Ampere Unit, Where: Q is the charge (in coulombs) and t is the interval in time (in seconds) that, the charge moves. In other words, electrical current has both a magnitude (the, amount of charge) and a specified direction associated with it., Electrical Energy Example No2, 1. How much current flows through a circuit if 900 coulombs of charge passes a, given point in 3 minutes., , ., Meenakshi Rastogi-Lecturer-EE-GPL

Page 19 : Conversion of mechanical units into electrical, units, 2. An electric current of 3 Amperes flows through a resistor. How many coulombs of, charge will flow through the resistor in 90 seconds, The Watt, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 20 : Conversion of mechanical units into electrical, units, So we can see that electrical power is also the rate at which work is performed, during one second. That is, one joule of energy dissipated in one second. As, electrical power is measured in Watts (W), therefore it must be also be measured, in Joules per Second. So we can correctly say that: 1 watt = 1 joule per second (J/s)., Electrical Power, 1 watt (W) = 1 joule/second (J/s), So if 1 watt = 1 joule per second, it therefore follows that: 1 Joule of energy = 1, watt over one unit of time, that is: Work equals Power multiplied by Time, (V*I*t, joules). So electrical energy (the work done) is obtained by multiplying power by, the time in seconds that the charge (in the form of a current) flows. Thus units of, electrical energy depend on the units used for electric power and time. So if we, measure electrical power in kilowatts (kW), and the time in hours (h), then the, electrical energy consumed equals kilowatts*hours (Wh) or simply: kilowatthours (kWh)., Meenakshi Rastogi-Lecturer-EE-GPL

Page 21 : Conversion of mechanical units into electrical, units, Electrical Energy Example No3, A 100 Watt light bulb is illuminated on for one hour only. How many joules of, electrical energy have been used by the lamp., , • Note that when dealing with the joule as a unit of electrical energy, it is more, convenient to present them in kilo-joules. Therefore the answer can be given as:, 360kJ., Meenakshi Rastogi-Lecturer-EE-GPL

Page 22 : Kirchhoff’s Current Law, • This is also called as the law of conservation of charge because charge or current, cannot be created or destroyed at the junction or node. It states that the, algebraic sum of currents at any node is zero. Thus the current entering at a node, must be equal to sum of current out of the node., I1 + I2 + (-I3) + (-I4) = 0, I1 + I2 = I3 + I4, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 23 : Example Problem of KCL, Consider the below figure where we have to determine the currents IAB and Ix by, using KCL ., By applying Kirchhoff’s Current Law at point A, we get, IAB = 0.5 – 0.3, IAB = 0.2 Amps, Similarly by applying KCL at point B, we get, IAB = 0.1 + Ix, 0.2 = 0.1 + Ix, Ix = 0.2 – 0.1 = 0.1 Amps, Meenakshi Rastogi-Lecturer-EE-GPL

Page 24 : Kirchhoff’s Voltage Law (KVL), Kirchhoff’s Voltage Law states that the algebraic, sum of voltages in a closed path is equal to zero, that is the sum of source voltages is equal to the, sum of voltage drops in a circuit. If the current, flows from higher potential to lower in an element,, then we consider it as a voltage drop., If the current flows from lower potential to higher, potential, then we consider it as a voltage rise., Thus, the energy dissipated by the current must be, equal to the energy given by the power supply in an, electric circuit., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 26 : Kirchhoff’s Voltage Law Example, 1. Let us consider the single loop circuit which is shown below and assume the, current flow direction as DEABCD closed path. In this circuit, by using KVL we have, to find the voltage V1., By applying KVL to this closed loop, we can write as, VED + VAE + VBA + VCB + VDC = 0, Where, Voltage of point E with respect to point D, VED = -50 V, Voltage of point D with respect to point C, VDC = -50 V, Voltage of point A with respect to point E. VAE = I * R, VAE = 500m* 200, VAE = 100 V, Meenakshi Rastogi-Lecturer-EE-GPL

Page 28 : Kirchhoff’s Voltage Law Example, 2. Consider the below typical two loop circuit where we have to find the currents I1, and I2 by applying the Kirchhoff’s laws., There are two loops inside the circuit and consider the loop paths as shown in, figure., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 30 : Thevenin’s Theorem, • Thevenin theorem is an analytical method used to change a complex circuit into a, simple equivalent circuit consisting of a single resistance in series with a source, voltage, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 31 : Thevenin’s Theorem, • Thevenin’s Theorem states that “Any linear circuit containing several voltages and, resistances can be replaced by just one single voltage in series with a single, resistance connected across the load“. In other words, it is possible to simplify, any electrical circuit, no matter how complex, to an equivalent two-terminal, circuit with just a single constant voltage source in series with a resistance (or, impedance) connected to a load as shown below., • Thevenin’s Theorem is especially useful in the circuit analysis of power or battery, systems and other interconnected resistive circuits where it will have an effect on, the adjoining part of the circuit., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 32 : Thevenin’s Theorem, • For example, consider the circuit from the previous tutorials., , • Firstly, to analyze the circuit we have to remove the center 40Ω load resistor, connected across the terminals A-B, and remove any internal resistance, associated with the voltage source(s). This is done by shorting out all the voltage, sources connected to the circuit, that is v = 0, or open circuit any connected, current sources making i = 0. The reason for this is that we want to have an ideal, voltage source or an ideal current source for the circuit analysis., Meenakshi Rastogi-Lecturer-EE-GPL

Page 33 : Thevenin’s Theorem, The value of the equivalent resistance, Rs is found by calculating the total, resistance looking back from the terminals A and B with all the voltage sources, shorted. We then get the following circuit., , Find the Equivalent Resistance (Rs), , Meenakshi Rastogi-Lecturer-EE-GPL

Page 34 : Thevenin’s Theorem, • The voltage Vs is defined as the total voltage across the terminals A and B when, there is an open circuit between them. That is without the load, resistor RL connected., • Find the Equivalent Voltage (Vs), , • We now need to reconnect the two voltages back into the circuit, and, as VS = VAB the current flowing around the loop is calculated as:, Meenakshi Rastogi-Lecturer-EE-GPL

Page 37 : Thevenin’s Theorem, The basic procedure for solving a circuit using Thevenin’s Theorem is as follows:, 1. Remove the load resistor RL or component concerned., 2. Find RS by shorting all voltage sources or by open circuiting all the current, sources., 3. Find VS by the usual circuit analysis methods., 4. Find the current flowing through the load resistor RL., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 38 : Norton’s Theorem, Norton’s Theorem states that “Any linear circuit containing several energy sources, and resistances can be replaced by a single Constant Current generator in parallel, with a Single Resistor“., As far as the load resistance, RL is concerned this single resistance, RS is the value of, the resistance looking back into the network with all the current sources open, circuited and IS is the short circuit current at the output terminals as shown below, , The value of this “constant current” is one which would flow if the two output, terminals where shorted together while the source resistance would be measured, looking back into the terminals, (the same as Thevenin)., Meenakshi Rastogi-Lecturer-EE-GPL

Page 39 : Norton’s Theorem, • For example, consider our now familiar circuit from the previous section., , • To find the Norton’s equivalent of the above circuit we firstly have to remove the, center 40Ω load resistor and short out the terminals A and B to give us the, following circuit., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 40 : Norton’s Theorem, , When the terminals A and B are shorted together the two resistors are connected, in parallel across their two respective voltage sources and the currents flowing, through each resistor as well as the total short circuit current can now be, calculated as:, with A-B Shorted Out, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 41 : Norton’s Theorem, • If we short-out the two voltage sources and open circuit terminals A and B, the, two resistors are now effectively connected together in parallel. The value of the, internal resistor Rs is found by calculating the total resistance at the, terminals A and B giving us the following circuit., , • Find the Equivalent Resistance (Rs), , Meenakshi Rastogi-Lecturer-EE-GPL

Page 42 : Norton’s Theorem, Having found both the short circuit current, Is and equivalent internal, resistance, Rs this then gives us the following Norton’s equivalent circuit., Norton’s equivalent circuit, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 43 : Norton’s Theorem, • Again, the two resistors are connected in, terminals A and B which gives us a total resistance of:, , parallel, , across, , the, , • The voltage across the terminals A and B with the load resistor connected is given, as:, • Then the current flowing in the 40Ω load resistor can be found as:, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 44 : Maximum Power Transfer Theorem, the maximum amount of power will be dissipated by a load resistance when that, load resistance is equal to the Thevenin/Norton resistance of the network, supplying the power. If the load resistance is lower or higher than the, Thevenin/Norton resistance of the source network, its dissipated power will be less, than the maximum., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 45 : Maximum Power Transfer Theorem, Taking our Thevenin equivalent example circuit, the Maximum Power Transfer, Theorem tells us that the load resistance resulting in greatest power dissipation is, equal in value to the Thevenin resistance (in this case, 0.8 Ω):, With this value of load resistance, the dissipated power will be 39.2 watts:, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 46 : Maximum Power Transfer Theorem, • If we were to try a lower value for the load resistance (0.5 Ω instead of 0.8 Ω, for, example), our power dissipated by the load resistance would decrease:, , • Power dissipation increased for both the Thevenin resistance and the total circuit,, but it decreased for the load resistor. Likewise, if we increase the load resistance, (1.1 Ω instead of 0.8 Ω, for example), power dissipation will also be less than it, was at 0.8 Ω exactly:, Meenakshi Rastogi-Lecturer-EE-GPL

Page 47 : Maximum Power Transfer Theorem, , If you were designing a circuit for maximum power dissipation at the load, resistance, this theorem would be very useful. Having reduced a network down to a, Thevenin voltage and resistance (or Norton current and resistance), you simply set, the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to, ensure maximum power dissipation at the load., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 49 : Steps for Solving network by Superposition, Theorem, • Considering the circuit diagram A, let us see the various steps to solve the, superposition theorem:, , Meenakshi Rastogi-Lecturer-EE-GPL

Page 51 : Steps for Solving network by Superposition, Theorem, • Step 1 – Take only one independent source of voltage or current and deactivate, the other sources., • Step 2 – In the circuit diagram B shown above, consider the source E1 and replace, the other source E2 by its internal resistance. If its internal resistance is not given,, then it is taken as zero and the source is short-circuited., • Step 3 – If there is a voltage source than short circuit it and if there is a current, source then just open circuit it., • Step 4 – Thus, by activating one source and deactivating the other source find the, current in each branch of the network. Taking the above example find the current, I1’, I2’and I3’., , Meenakshi Rastogi-Lecturer-EE-GPL

Page 52 : Steps for Solving network by Superposition, Theorem, • Step 5 – Now consider the other source E2 and replace the source E1 by its, internal resistance r1 as shown in the circuit diagram C., • Step 6 – Determine the current in various sections, I1’’, I2’’ and I3’’., • Step 7 – Now to determine the net branch current utilizing the superposition, theorem, add the currents obtained from each individual source for each branch., • Step 8 – If the current obtained by each branch is in the same direction then add, them and if it is in the opposite direction, subtract them to obtain the net current, in each branch., , Meenakshi Rastogi-Lecturer-EE-GPL