Download Chapter 02-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 2, Problem 1. The voltage across a 5-kΩ resistor is 16 V. Find the current through the resistor. Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = 3.2 mA Chapter 2, Problem 2. Find the hot resistance of a lightbulb rated 60 W, 120 V. Chapter 2, Solution 2 p = v2/R → R = v2/p = 14400/60 = 240 ohms Chapter 2, Problem 3. A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 240 Ω at room temperature, what is the cross-sectional radius of the bar? Chapter 2, Solution 3 For silicon, 26.4 10xρ = Ω-m. 2A rπ= . Hence, 2 2 2 2 6.4 10 4 10 0.033953 240 L L L x x xR r A r R x ρ ρ ρ π π π − = = ⎯⎯→ = = = r = 0.1843 m Chapter 2, Problem 4. (a) Calculate current i in Fig. 2.68 when the switch is in position 1. (b) Find the current when the switch is in position 2. Chapter 2, Solution 4 (a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA docsity.com Chapter 2, Problem 5. For the network graph in Fig. 2.69, find the number of nodes, branches, and loops. Chapter 2, Solution 5 n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Problem 6. In the network graph shown in Fig. 2.70, determine the number of branches and nodes. Chapter 2, Solution 6 n = 12; l = 8; b = n + l –1 = 19 Chapter 2, Problem 7. Determine the number of branches and nodes in the circuit of Fig. 2.71. Figure 2.71 For Prob. 2.7. 1 Ω 4 Ω + _ 8 Ω 5 Ω 12 V 2 A Chapter 2, Solution 7 6 branches and 4 nodes. docsity.com Chapter 2, Problem 12. In the circuit in Fig. 2.76, obtain v1, v2, and v3. Chapter 2, Solution 12 + V1 - + V2 - + V3 - – 25V + 10V - + 15V - + 20V - LOOP LOOP LOOP For loop 1, -20 -25 +10 + v1 = 0 v1 = 35v For loop 2, -10 +15 -v2 = 0 v2 = 5v For loop 3, -V1 + V2 + V3 = 0 v3 = 30v docsity.com Chapter 2, Problem 13. For the circuit in Fig. 2.77, use KCL to find the branch currents I1 to I4. I1 I2 I4 I3 7 A 2 A 4 A3 A Figure 2.77 Chapter 2, Solution 13 2A I2 7A I4 1 2 3 4 4A I1 3A I3 At node 2, 3 7 0 102 2+ + = ⎯ →⎯ = −I I A 12 2 A 2 5 A At node 1, I I I I A1 2 1 22 2+ = ⎯ →⎯ = − = At node 4, 2 4 2 44 4= + ⎯ →⎯ = − = −I I At node 3, 7 7 4 3 3+ = ⎯ →⎯ = − =I I I Hence, I A I A I A I1 2 3 412 10 5 2= = − = A= −, , , docsity.com Chapter 2, Problem 14. Given the circuit in Fig. 2.78, use KVL to find the branch voltages V1 to V4. V2 V4 V1 V3 3 V 4 V 5 V + – – – + + – – 2 V + + ++ – – + – Figure 2.78 Chapter 2, Solution 14 + + - 3V V1 I4 V2 - I3 - + 2V - + - + V3 - + + 4V I2 - I1 + - V4 5V For mesh 1, − + + = ⎯ →⎯ =V V4 42 5 0 7 V 11 8 For mesh 2, + + + = ⎯ →⎯ = − − = −4 0 4 73 4 3V V V V For mesh 3, − + − = ⎯ →⎯ = + = −3 0 31 3 1 3V V V V V For mesh 4, − − − = ⎯ →⎯ = − − =V V V V V1 2 2 12 0 2 6 Thus, V V V V V V V1 2 3 48 6 11= − = = − V7=, , , docsity.com
