Thevenin Equivalent Circuits

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1 hevenin Equivalent Circuits Introduction In each of these problems, we are shown a circuit and its hevenin or Norton equivalent circuit. he hevenin and Norton equivalent circuits are described using three parameters: V oc, the open circuit voltage of the circuit, I sc, the short circuit of the circuit and R th, the hevenin resistance of the circuit. Each problem, asks us to determine the value of asked to determine the value of V oc, I sc or R th. hevenin equivalent circuits are discussed in Section 5.5 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda. Norton equivalent circuits are discussed in Section 5.6. Worked Examples Example 1: he circuit shown in Figure 1b is the hevenin equivalent circuit of the circuit shown in Figure 1a. Find the value of the open circuit voltage, V oc and hevenin resistance, R th. Figure 1 he circuit considered in Example 1. Solution: he circuit from Figure 1a can be reduced to its hevenin equivalent circuit in four steps shown in Figure 2a, b, c and d. A source transformation transforms the series voltage source and 20 Ω resistor in Figure 1a into the parallel current source and 20 Ω resistor in Figure 2a. he current source current is calculated from the voltage source voltage and resistance as 20 V = 1 A. After the source 20 Ω transformation, the 20 Ω resistor is parallel to the 80 Ω resistor. Replacing these parallel resistors with the equivalent 16 Ω resistor produces the circuit shown in Figure 2b. 2

2 A second source transformation transforms the parallel current source and 16 Ω resistor in Figure 2b into the series voltage source and 16 Ω resistor in Figure 2c. he voltage source 1 A 16 Ω = 16 V. After voltage is calculated from the current source current and resistance as ( )( ) the source transformation, the two16 Ω resistors are in series. Replacing these series resistors with the equivalent 32 Ω resistor produces the circuit shown in Figure 2d. Comparing Figure 2d to Figure 1b shows that the hevenin resistance is R th = 32 Ω and the open circuit voltage, V oc = -16 V. (a) (b) (c) (d) Figure 2 he circuit from Figure 1a can be reduced to its hevenin equivalent circuit in four steps shown here as (a), (b), (c), and (d). 3

3 Example 2: he circuit shown in Figure 3b is the hevenin equivalent circuit of the circuit shown in Figure 1a. Find the value of the open circuit voltage, V oc and hevenin resistance, R th. Figure 3 he circuit considered in Example 2. Solution: he circuit from Figure 3a can be reduced to its hevenin equivalent circuit in five steps shown in Figure 4a, b, c, d and e. A source transformation transforms the parallel current source and 3 Ω resistor in Figure 3a into the series voltage source and 3 Ω resistor in Figure 4a. he voltage source voltage is 2 A 3 Ω = 6 V. After the source calculated from the current source current and resistance as ( )( ) transformation, the 3 Ω and 6 Ω resistors are in series. Also, the 6V and 3 V voltage sources are in series. Replacing the series resistors with the equivalent 9 Ω resistor and the series voltage sources with the equivalent 8 V source produces the circuit shown in Figure 4b. A second source transformation transforms the series 8 V voltage source and 9 Ω resistor in Figure 4b into the parallel current source and 9 Ω resistor in Figure 4c. he current source current is calculated from the voltage source voltage and resistance as 8 V = 0.89 A. After the 9 Ω source transformation, the 9 Ω resistor is parallel to the 6 Ω resistor. Replacing these parallel resistors with the equivalent 3.6 Ω resistor produces the circuit shown in Figure 4d. A third source transformation transforms the parallel 0.89 A current source and 3.6 Ω resistor in Figure 4d into the series voltage source and 3.6 Ω resistor in Figure 4e. he voltage source voltage is calculated from the current source current and resistance as 0.89 A 3.6 Ω = 3.2 V. ( )( ) Comparing Figure 4e to Figure 3b shows that hevenin resistance is R th = 3.6 Ω and that the open circuit voltage, V oc = -3.2 V. 4

4 (a) (b) (c) (d) Figure 4 he circuit from Figure 3a can be reduced to its hevenin equivalent circuit in five steps shown here as (a), (b), (c), (d) and (e). (e) 5

5 Example 3: he circuit shown in Figure 5b is the hevenin equivalent circuit of the circuit shown in Figure 5a. Find the value of the open circuit voltage, V oc and hevenin resistance, R th. Figure 5 he circuit considered in Example 3. Solution: he circuit from Figure 5a can be reduced to its hevenin equivalent circuit in four steps shown in Figure 6a, b, c and d. A source transformation transforms the series 10 V voltage source and 5 Ω resistor in Figure 5a into the parallel current source and 5 Ω resistor in Figure 6a. he current source current is calculated from the voltage source voltage and resistance as 10 V = 2 A. After the 5 Ω source transformation, the 5 Ω resistor is parallel to the 20 Ω resistor. Also, the 2 A current source is parallel to the 1 A current source. Replacing these parallel resistors with the equivalent 4 Ω resistor and replacing the parallel current sources with the equivalent 1 A current source produces the circuit shown in Figure 6b. A second source transformation transforms the parallel 1 A current source and 4 Ω resistor in Figure 6b into the series voltage source and 4 Ω resistor in Figure 6c. he voltage 1 A 4 Ω = 4 V source voltage is calculated from the current source current and resistance as ( )( ). After the source transformation, the two 4 Ω resistors are in series. Replacing the series resistors with the equivalent 8 Ω produces the circuit shown in Figure 6d. Comparing Figure 6d to Figure 5b shows that hevenin resistance is R th = 8 Ω and that the open circuit voltage, V oc = 4 V. 6

6 (a) (b) (c) (d) Figure 6 he circuit from Figure 5a can be reduced to its hevenin equivalent circuit in four steps shown here as (a), (b), (c), and (d). Example 4: he circuit shown in Figure 7b is the hevenin equivalent circuit of the circuit shown in Figure 7a. Find the value of the open circuit voltage, V oc and hevenin resistance, R th. Also, determine the value of the short circuit current, I sc. Figure 7 he circuit considered in Example 4. 7

7 Solution: o determine the value of the open circuit voltage, V oc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure 8 shows the circuit from Figure 7a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i 1 and i 2 denote the mesh currents in meshes 1 and 2, respectively. In Figure 8, mesh current i 2 is equal to the current in the open circuit. Consequently, i 2 = 0 A. he controlling current of the CCVS is expressed in terms of the mesh currents as ia = i1 i2 = i1 0 = i 1 Apply KVL to mesh 1 to get ( ) ( ) ( ) ( ) 3i 2 i i + 6 i i 10= 0 3i 2 i i 0 10= Apply KVL to mesh 2 to get 10 i1 = = 1.43 A 7 ( ) oc ( ) ( ) 5 i + V 6 i i = 0 V = 6 i = = 8.58 V 2 oc Next, to determine the value of the short circuit current, I sc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure 9 shows the circuit from Figure 7a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i 1 and i 2 denote the mesh currents in meshes 1 and 2, respectively. In Figure 9, mesh current i 2 is equal to the current in the short circuit. Consequently, i = I. he controlling current of the CCVS is expressed in terms of the mesh currents as 2 sc ia = i1 i2 = i1 Is c Figure 8 Calculating the open circuit voltage, V oc, using mesh equations. Apply KVL to mesh 1 to get 8

8 ( ) ( ) 3i 2 i i + 6 i i 10= 0 7i 4i = 10 (1) Apply KVL to mesh 2 to get 11 5i2 6( i1 i2) = 0 6i1+ 11i2 = 0 i1 = i 2 6 Substituting into equation 1 gives 11 7 i2 4 i2 = 10 i2 = 1.13 A I sc = 1.13 A 6 Figure 9 Calculating the short circuit current, I sc, using mesh equations. Figure 10 Calculating the hevenin resistance, R th v i =, using mesh equations. o determine the value of the hevenin resistance, R th, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure 10. he hevenin resistance will be calculated from the current and voltage of the current source as 9

9 R th v = i In Figure 10, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i 1 and i 2 denote the mesh currents in meshes 1 and 2, respectively. In Figure 10, mesh current i 2 is equal to the negative of the current source current. Consequently, i = i. he controlling current of the CCVS is expressed in terms of the mesh currents as 2 Apply KVL to mesh 1 to get Apply KVL to mesh 2 to get i = i i = i + i a i1 2( i1 i2) + 6( i1 i2) = 0 7i1 4i2 = 0 i1 = i 2 (2) 7 ( ) Substituting for i 1 using equation 2 gives 5i + v 6 i i = 0 6i + 11i = v Finally, 4 6 i2 + 11i2 = v 7.57i2 = v 7 R th v v v = = = = 7.57 Ω i i i 2 As a check, notice that ( )( ) R I = = 8.55 V th sc oc 10

10 Example 5: he circuit shown in Figure 11b is the hevenin equivalent circuit of the circuit shown in Figure 11a. Find the value of the open circuit voltage, V oc and hevenin resistance, R th. Also, determine the value of the short circuit current, I sc. Figure 11 he circuit considered in Example 5. Solution: o determine the value of the open circuit voltage, V oc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure 12 shows the circuit from Figure 11a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v 1, v 2 and v 3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure 12, node voltage v 1 is equal to the negative of the voltage source voltage. Consequently, v 1 = 24 V. he controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2. he voltage at node 3 is equal to the open circuit voltage, i.e. v3 = Voc. Apply KCL at node 2 to get v1 v2 v2 v3 = 2v1+ v3 = 3v2 48+ Voc =3 v 3 6 a Figure 12 Calculating the open circuit voltage, V oc, using node equations. 11

11 Apply KCL at node 3 to get v v 4 + v2 = 0 9v2 v3 = 0 9va = V o c Combining these equations gives ( ) V = 9v = V V = 72 V oc a oc oc Next, to determine the value of the short circuit current, I sc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure 13 shows the circuit from Figure 7a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v 1, v 2 and v 3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure 13, node voltage v 1 is equal to the negative of the voltage source voltage. Consequently, v 1 = 24 V. he voltage at node 3 is equal to the voltage across a short, v 3 = 0. he controlling voltage of the VCCS, v a, is equal to the node voltage at node 2, i.e. v a = v 2. he voltage at node 3 is equal to the voltage across a short, i.e. v 3 = 0. Apply KCL at node 2 to get Apply KCL at node 3 to get v1 v2 v2 v3 = 2v1+ v3 = 3v2 48= 3va va = 16 V 3 6 v v v2 = Isc va = Isc Isc = ( 16) = 24 A Figure 13 Calculating the short circuit current, I sc, using mesh equations. 12

12 Figure 14 Calculating the hevenin resistance, R th v i =, using mesh equations. o determine the value of the hevenin resistance, R th, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure 14. he hevenin resistance will be calculated from the current and voltage of the current source as v Rth = i Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v 1, v 2 and v 3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure 14, node voltage v 1 is equal to the across a short circuit, i.e. v 1 = 0. he controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. v a = v 2. he voltage at node 3 is equal to the voltage across the current source, i.e. v3 = v. Apply KCL at node 2 to get Apply KCL at node 3 to get Finally, v 2 3 As a check, notice that v1 v2 v2 v3 = 2v1+ v3 = 3v2 v =3 v 3 6 v 4 + v2 + i = 0 9v2 v3+ 6i = v v + 6i = 0 a 3v v + 6i = 0 2v = 6i R th v = = 3 Ω i a 13

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