Chapter 3 Nodal and Mesh Equations - Circuit Theorems 3.14 Exercises Multiple Choice 1. The voltage across the 2 Ω resistor in the circuit of Figure 3.67 is A. 6 V B. 16 V C. –8 V D. 32 V E. none of the above − + 6 V + 2 Ω − 8 A 8 A Figure 3.67. Circuit for Question 1 2. The current i in the circuit of Figure 3.68 is A. –2 A B. 5 A C. 3 A D. 4 A E. none of the above 4 V − + 2 Ω 2 Ω + − 2 Ω 2 Ω 10 V i Figure 3.68. Circuit for Question 2 3-52 Circuit Analysis I with MATLAB Applications Orchard Publications Exercises 3. The node voltages shown in the partial network of Figure 3.69 are relative to some reference node which is not shown. The current i is A. –4 A B. 83⁄ A C. –5 A D. –6 A E. none of the above 6 V 8 V − 4 V + 8 V 2 Ω − 12 V + 2 Ω i 8 V − 6 V + 13 V 2 Ω Figure 3.69. Circuit for Question 3 4. The value of the current i for the circuit of Figure 3.70 is A. –3 A B. –8 A C. –9 A D. 6 A E. none of the above 6 Ω 3 Ω i + 12 V 6 Ω 8 A 3 Ω − Figure 3.70. Circuit for Question 4 Circuit Analysis I with MATLAB Applications 3-53 Orchard Publications Chapter 3 Nodal and Mesh Equations - Circuit Theorems 5. The value of the voltage v for the circuit of Figure 3.71 is A. 4 V B. 6 V C. 8 V D. 12 V E. none of the above v + X − 2 Ω + + 2 A 2 Ω v − 2vX − Figure 3.71. Circuit for Question 5 6. For the circuit of Figure 3.72, the value of k is dimensionless. For that circuit, no solution is pos- sible if the value of k is A. 2 B. 1 C. ∞ D. 0 E. none of the above 4 Ω + + 2 A 4 Ω v − kv − Figure 3.72. Circuit for Question 6 3-54 Circuit Analysis I with MATLAB Applications Orchard Publications Exercises 7. For the network of Figure 3.73, the Thevenin equivalent resistance RTH to the right of terminals a and b is A. 1 B. 2 C. 5 D. 10 E. none of the above a 2 Ω 2 Ω 3 Ω RTH 4 Ω 2 Ω 2 Ω b 2 Ω Figure 3.73. Network for Question 7 8. For the network of Figure 3.74, the Thevenin equivalent voltage VTH across terminals a and b is A. –3 V B. –2 V C. 1 V D. 5 V E. none of the above − + a 2 V 2 Ω 2 Ω 2 A b Figure 3.74. Network for Question 8 Circuit Analysis I with MATLAB Applications 3-55 Orchard Publications Chapter 3 Nodal and Mesh Equations - Circuit Theorems 9. For the network of Figure 3.75, the Norton equivalent current source IN and equivalent parallel resistance RN across terminals a and b are A. 1 A2, Ω B. 1.5 A25, Ω C. 4 A2.5, Ω D. 0 A5, Ω E. none of the above a 5 Ω 5 Ω 2 A 2 A b Figure 3.75. Network for Question 9 10. In applying the superposition principle to the circuit of Figure 3.76, the current i4 due to the V source acting alone is A. 8 A B. –1 A C. 4 A D. –2 A E. none of the above i 2 Ω 2 Ω + 4 V 8 A 2 Ω − Figure 3.76. Network for Question 10 3-56 Circuit Analysis I with MATLAB Applications Orchard Publications Exercises Problems 1. Use nodal analysis to compute the voltage across the 18 A current source in the circuit of Figure 3.77. Answer: 1.12 V –1 4 Ω–1 5 Ω –1 8 Ω–1 10 Ω + –1 –1 4 Ω v18 A 6 Ω 18 A − 12 A 24 A Figure 3.77. Circuit for Problem 1 2. Use nodal analysis to compute the voltage v6 Ω in the circuit of Figure 3.78. Answer: 21.6 V 36 V + − 12 Ω 15 Ω + v 4 Ω 6 Ω 6Ω 18 A − 12 A 24 A Figure 3.78. Circuit for Problem 2 3. Use nodal analysis to compute the current through the 6 Ω resistor and the power supplied (or absorbed) by the dependent source shown in Figure 3.79. Answers: –3.9 A499.17, – w 4. Use mesh analysis to compute the voltage v36A in Figure 3.80. Answer: 86.34 V 5. Use mesh analysis to compute the current through the i6Ω resistor, and the power supplied (or absorbed) by the dependent source shown in Figure 3.81. Answers: –3.9 A499.33, – w 6. Use mesh analysis to compute the voltage v10Ω in Figure 3.82. Answer: 0.5 V Circuit Analysis I with MATLAB Applications 3-57 Orchard Publications Chapter 3 Nodal and Mesh Equations - Circuit Theorems 18 A 12 Ω 15 Ω iX 6 Ω + 5iX 4 Ω i6Ω − 12 A + 24 A − 36 V Figure 3.79. Circuit for Problem 3 120 V − 240 V + + − 4 Ω 3 Ω 8 Ω 12 Ω + 4 Ω v36A 36 A 6 Ω 12 A − 24 A Figure 3.80. Circuit for Problem 4 18 A 12 Ω 15 Ω iX 6 Ω + i 5iX 4 Ω 6Ω − + 12 A 24 A − 36 V Figure 3.81. Circuit for Problem 5 3-58 Circuit Analysis I with MATLAB Applications Orchard Publications Exercises 10iX 12 Ω 15 Ω 4 Ω + 8 Ω i v 6 Ω X 10Ω + − − 10 Ω − + 12 V 24 V Figure 3.82. Circuit for Problem 6 7. Compute the power absorbed by the 10 Ω resistor in the circuit of Figure 3.83 using any method. Answer: 1.32 w 2 Ω + 3 Ω 6 Ω 10 Ω − 12 V − + + − 24 V 36 V Figure 3.83. Circuit for Problem 7 8. Compute the power absorbed by the 20 Ω resistor in the circuit of Figure 3.84 using any method. Answer: 73.73 w + − 20 Ω 12 V 3 Ω 8 A 2 Ω 6 A Figure 3.84. Circuit for Problem 8 9. In the circuit of Figure 3.85: a. To what value should the load resistor RLOAD should be adjusted to so that it will absorb maximum power? Answer: 2.4 Ω Circuit Analysis I with MATLAB Applications 3-59 Orchard Publications Chapter 3 Nodal and Mesh Equations - Circuit Theorems b. What would then the power absorbed by RLOAD be? Answer: 135 w 36 V + − 12 Ω 15 Ω R 4 Ω 6 Ω LOAD 12 A 18 A Figure 3.85. Circuit for Problem 9 10. Replace the network shown in Figure 3.86 by its Norton equivalent. Answers: iN = 0R, N = 23.75 Ω a 15 Ω 4 Ω 5 Ω iX 5iX b Figure 3.86. Circuit for Problem 10 11. Use the superposition principle to compute the voltage v18A in the circuit of Figure 3.87. Answer: 1.12 V –1 4 Ω–1 5 Ω –1 8 Ω–1 10 Ω + –1 –1 4 Ω v18 A 6 Ω 18 A − 12 A 24 A Figure 3.87. Circuit for Problem 11 3-60 Circuit Analysis I with MATLAB Applications Orchard Publications Exercises 12. Use the superposition principle to compute voltage v6 Ω in the circuit of Figure 3.88. Answer: 21.6 V 36 V + − 12 Ω 15 Ω + v 4 Ω 6 Ω 6Ω 18 A − 12 A 24 A Figure 3.88. Circuit for Problem 12 13.In the circuit of Figure 3.89, vS1 and vS2 are adjustable voltage sources in the range –50≤≤ V 50 V, and RS1 and RS2 represent their internal resistances. Table 3.4 shows the results of several measurements. In Measurement 3 the load resistance is adjusted to the same value as Measurement 1, and in Measurement 4 the load resistance is adjusted to the same value as Mea- surement 2. For Measurements 5 and 6 the load resistance is adjusted to 1 Ω . Make the neces- sary computations to fill-in the blank cells of this table. TABLE 3.4 Table for Problem 13 Measurement Switch S1 Switch S2 vS1 (V)vS2 (V)iLOAD (A) 1ClosedOpen480 16 2OpenClosed0366 3ClosedOpen 0 −5 4OpenClosed0−42 5 Closed Closed 15 18 6 Closed Closed 24 0 Answers: –15 V , –7 A , 11 A24, – V Circuit Analysis I with MATLAB Applications 3-61 Orchard Publications Chapter 3 Nodal and Mesh Equations - Circuit Theorems + RS1 1 Ω i + R LOAD S2 1 Ω Adjustable − + Resistive vS1 vLOAD − Load vS2 S 1 S2 − Figure 3.89. Network for Problem 13 14. Compute the efficiency of the electrical system of Figure 3.90. Answer: 76.6% 0.8 Ω 0.5 Ω i i vS 1 100 A 2 80 A + 1st Floor 2nd Floor −− Load Load 480 V 0.5 Ω 0.8 Ω Figure 3.90. Electrical system for Problem 14 15. Compute the regulation for the 2st floor load of the electrical system of Figure 3.91. Answer: 36.4% 0.8 Ω 0.5 Ω i i2 VS 1 100 A 80 A + 1st Floor 2nd Floor −− Load Load 480 V 0.5 Ω 0.8 Ω Figure 3.91.