SOURCE TRANSFORMATIONS TECHNIQUE Lecture 11 SOURCE TRANSFORMATIONS TECHNIQUE

Learning Outcomes After completing this module, you will be able to: convert a voltage source into an equivalent current source, convert a current source into an equivalent voltage source, Use source transformation to simplify and analyze circuits.

Equivalent Practical Sources Two sources are equivalent if they produce identical values of vL and iL when they are connected to identical values of RL, no matter what the value of RL may be. Since RL =∞ and RL = 0 are two such values, equivalent sources provide the same open-circuit voltage and short-circuit current. Cct 1 RL v(t) i(t) i(t) v(t) Vo/c Is/c Cct 2 RL v(t) i(t)

Thevenin's Theorem for Linear Resistive Circuits Any combination of batteries and resistances with two terminals can be replaced by a single voltage source VTH and a single series resistance RTH. Linear Circuit A B VTH RTH

The value of VTH is the open circuit voltage at the terminals. In practice, the Thevenin voltage, VTH, of the network is obtained by measuring the open-circuit voltage VAB. By definition, VTH = VAB Note In circuit analysis, VTH is found by computing the voltage drop across the terminals A and B. A Linear Circuit Voltmeter B

Measure the open-circuit voltage VAB. The value of RTH is obtained by dividing VAB by the short circuit current ISC. In practice, the Thevenin resistance, RTH, of the network is obtained as follows: Measure the open-circuit voltage VAB. Short the output terminals with an ammeter and read off the short circuit current displayed by the ammeter. Let the short-circuit current reading displayed by the ammeter be ISC (say). Divide the open-circuit voltage by the short circuit current. The ratio obtained is the Thevenin resistance, RTH; that is, A Linear Circuit Isc B

Worked Example Find the Thevenin equivalent of the circuit in the figure below. A B 3 A 3 Ω Solution By inspection, we see that the Thevenin voltage VTH that we want to find is given by the voltage drop across the 3 Ω resistor; that is VTH = VAB = V3Ω Therefore,

Finding the Thevenin resistance RTH. One way to find the Thevenin resistance is to first find the short circuit current through A and B when the two terminals are short-circuited by a wire, as shown in the following figure. A B 3 A 3 Ω ISC Thus, Therefore,

Alternative method for finding the Thevenin resistance RTH. An alternative method for finding the Thevenin resistance of the given circuit is as follows: Step 1. Replace the current source with an open circuit. Step 2. Find the equivalent resistance of the circuit, looking into the circuit from the terminals A and B. The resistance obtained is the Thevenin resistance we are looking for. Thus, replacing the current source in the previous figure with open circuit leads us to the following circuit. A B 3 Ω RTH Thus, from inspecting the figure on the left, we see that

A B 3 A 3 Ω 3 Ω A 9 V B

Find the Thevenin equivalent of the circuit in the following figure. Worked Example Find the Thevenin equivalent of the circuit in the following figure. A B 2 A 5 Ω Solution By inspection, we see that the Thevenin voltage VTH that we want to find is given by the voltage drop across the 5 Ω resistor; that is VTH = VAB = V5Ω Therefore,

Finding the Thevenin resistance RTH. One way to find the Thevenin resistance is to first find the short circuit current through A and B when the two terminals are short-circuited by a wire, as shown in the following figure. A B 2A 5Ω ISC Thus, Therefore,

Alternative method for finding the Thevenin resistance RTH. An alternative method for finding the Thevenin resistance of the given circuit is as follows: Step 1. Replace the current source with an open circuit. Step 2. Find the equivalent resistance of the circuit, looking into the circuit from the terminals A and B. The resistance obtained is the Thevenin resistance we are looking for. Thus, replacing the current source in Figure x with open circuit leads us to the following circuit. Thus, from inspecting the figure shown on the left, we see that A B 5 Ω RTH

A B 2 A 5 Ω 5 Ω A -10 V B

Find the Thevenin equivalent of the circuit in the following figure. Worked Example Find the Thevenin equivalent of the circuit in the following figure. A B 2 Ω 3 V Solution By inspection, we see that the Thevenin voltage VTH that we want to find is given by the voltage drop across the 2 Ω resistor; that is VTH = VAB = V2Ω Therefore,

Finding the Thevenin resistance RTH. One method for finding the Thevenin resistance of the given circuit is as follows: Step 1. Replace the voltage source with a short circuit. Step 2. Find the equivalent resistance of the circuit, looking into the circuit from the terminals A and B. The resistance obtained is the Thevenin resistance we are looking for. Thus, replacing the voltage source in the previous figure with a short circuit leads us to the following circuit. Thus, from inspecting the figure shown on the left, we see that A B 2 Ω RTH

A B 2 Ω 3 V 0 Ω A 3 V B

Norton's Theorem for Linear Resistive Circuits Any collection of batteries and resistances with two terminals is electrically equivalent to an ideal current source IN in parallel with a single resistor RN. B IN RN A Linear Circuit A B

The value of IN is the short circuit current at the terminals. In practice, the Norton current, IN, of the network is obtained by measuring the short circuit IAB. By definition, IN= IAB Note In circuit analysis, IN is found by computing the short circuit current flowing through the shorting wire connecting the terminals A and B. Linear Circuit A B Ammeter

Measure the open-circuit voltage VAB. The value of RN is obtained by dividing VAB by the short-circuit current, ISC. Like the Thevenin resistance, RTH, the Norton resistance of the network is obtained as follows: Measure the open-circuit voltage VAB. Short the output terminals with an ammeter and read off the short circuit current displayed by the ammeter. Let the short-circuit current reading displayed by the ammeter be ISC (say). Divide the open-circuit voltage by the short circuit current. The ratio obtained is the Norton resistance, RN; that is, Linear Circuit A B Isc

Find the Norton equivalent of the circuit in the figure shown below. Worked Example Find the Norton equivalent of the circuit in the figure shown below. 5 Ω A B 7 V Solution By inspection, we see that the Norton current IN that we want to find is given by the short current flowing through terminals A and B when they are shorted by a wire, as shown in the following figure.

5 Ω A B 7 V ISC The Norton resistance RN is obtained by finding the resistance looking into the A-B terminals when the voltage source is replaced by a short circuit. 5 Ω A B RTH

A 7/5 A 5 Ω B Both Thevenin’s and Norton’s Theorems can also be used in ac circuit analysis.

Source Transformations Technique for Linear AC Circuits Source transformations is a technique of analysing circuits which involve transforming part of a circuit into its Thevenin equivalent or into its Norton equivalent, while retaining the terminal characteristics of the original circuit. By transforming a subcircuit into its Thevenin or Norton equivalent, impedances or sources which then appear as being connected in parallel or in series can thus be combined and simplified. By doing repeated source transformations, we can eventually simplify a circuit to its simplest form for solution. The following examples show how we can use source transformations technique to solve a circuit.

Worked Example Find the steady-state expression for io(t) if ig(t) = 125cos(500t) mA R1 = 50 Ω R2 = 250 Ω L = 1 H C = 20 F R1 R2 C L

Solution Obtain the phasor equivalent circuit. ZR1 = R1 = 50 Ω ZL = jL) = j500 Ω ZC = 1/(jC) = - j100Ω Ig = 0.1250 A ZR1 ZR2 ZC ZL Ig Io

Replace the subcircuit on the left with its Thevenin equivalent. ZAB A B A B VAB

Applying KVL to the circuit leads us to the equation Calculate Io. Applying KVL to the circuit leads us to the equation ZAB giving VAB Hence Therefore,

Worked Example Find the steady-state expression for vo(t) if vg(t) = 64cos(8000t)

Solution Obtain the phasor equivalent circuit. ZR = R = 2000 Ω ZC = 1/(jC) = - j4000 Ω ZL = jL = j4000 Ω Vg = 640 V ZR ZC ZL Vg = 640 V Vo

Replace the subcircuit on the left with its Thevenin equivalent. ZAB A VAB B

Applying voltage divider rule to the circuit leads us to the equation Calculate Vo. Applying voltage divider rule to the circuit leads us to the equation ZL Hence,

Exercise If the current source iS outputs a current (1cos2t) A, find the voltage drop vo across the 2-Ω resistor. iS

Worked Example Use source transformations to solve for the steady-state part of vo. The sinusoidal voltages are: v1(t) = 240cos(4000t + 53.13) V v2(t) = 96sin(4000t) V

Solution Obtain the phasor equivalent circuit. ZR2 = R2 = 20 Ω ZC = 1/(jC) = 1/(j4000 x 25 x 10-6) = - j10 Ω ZL = jL = j4000 x 15 x 10-3 = j60 Ω V1 = 24053.13 V ZR2 = R2 = 20 Ω V2 = 96-90 V

Phasor equivalent circuit: ZR1 ZR2 ZL ZC V1 V2 V0 ZR2 = R2 = 20 Ω ZR1 30 Ω ; ZL = j60 Ω ; ZC = - j10 Ω V2 = 96-90 V V1 = 24053.13 V

Replace the subcircuit on the left of ZC with its Norton equivalent. IAB ZAB A B IAB B

Replace the subcircuit on the right of ZC with its Norton equivalent. ZAB IAB IAB ;

Draw the simplified circuit by replacing the left and right subcircuits with the Norton equivalents. Z1 Z2 I2 where

Combine the two current sources into one equivalent source Combine the two current sources into one equivalent source. Combine the three impedances to one equivalent impedance. IEQ = I1 + (– I2) = 4-36.87 - (j4.8) = 436.87 A 1 1 1 1 ZEQ = 6.845-56.88 Ω ZC = + + ZEQ Z1 Z2 IEQ ZEQ Therefore, and

Worked Example Find currents Ib, Ic and Id.

Solution Method I Replace the subcircuit to the right of the j5Ω impedance with its Thevenin equivalent.

Therefore, we get KVL gives Solving the above equation for Ib, we obtain

Replace the subcircuit to the right of the -j5Ω impedance with its Thevenin equivalent. VAB ZAB

Therefore, we get KVL gives Solving the above equation for Ic, we obtain

Calculate Id using KCL. Therefore,

Method II Temporarily remove the j5 Ω inductive impedance and replace it with an open circuit. From the resulting circuit obtained, find the open circuit voltage VAB and the terminal impedance ZAB.

VAB V5Ω Voltage drop across the 5 Ω resistance is

Therefore, voltage drop across terminals A-B is Thevenin impedance ZAB is equal to the impedance seen when looking into terminals A-B with the voltage sources replaced with short-circuits. ZAB

Connect the j5 Ω inductive impedance (which was removed earlier from the original circuit) across the A-B terminals of the Thevenin equivalent circuit and compute the current Ib.

Temporarily remove the -j5 Ω capacitive impedance from the original circuit and replace it with an open circuit. From the resulting circuit obtained, find the open circuit voltage VAB and the terminal impedance ZAB.

VAB V5Ω Voltage drop across the 5 Ω resistance is

Therefore, voltage drop across terminals A-B is Thevenin impedance ZAB is equal to the impedance seen when looking into terminals A-B with the voltage sources replaced with short-circuits. ZAB

Connect the -j5 Ω capacitive impedance (which was removed earlier from the original circuit) across the A-B terminals of the Thevenin equivalent circuit and compute the current Ic. Ic

Calculate Id using KCL. Therefore,

Worked Example Find the phasor voltage Vo.

Solution Temporarily remove the –j3 Ω capacitive impedance and replace it with an open circuit. From the resulting circuit obtained, find the open circuit voltage VAB and the terminal impedance ZAB. A B

A B Vj3Ω VA VB Voltage drop across the j3 Ω inductive impedance is

Potential of terminal A w.r.t. ground, Potential of terminal B w.r.t. ground,

Therefore, Thevenin impedance ZAB is equal to the impedance seen when looking into terminals A-B with the voltage source replaced with a short-circuit and the current source replaced with an open-circuit.

From the circuit on the right, we get A B

The Thevenin equivalent circuit between terminals A and B is as shown below. Applying voltage divider rule to the circuit above, we obtain Vo.

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