Solution: Based on the slope of q(t): 20 A for 0 t 1 s dt = 0 for 3 t 4 s. 20 A for 4 t 5 s 0 for t 5 s 20 C. t (s) 20 C. i (A) Fig. P1.


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1 Problem 1.24 The plot in Fig. P1.24 displays the cumulative charge q(t) that has entered a certain device up to time t. Sketch a plot of the corresponding current i(t). q 20 C t (s) 20 C Figure P1.24: q(t) for Problem Solution: Based on the slope of q(t): 20 A for 0 t 1 s i(t) = dq 20 A for 1 t 3 s dt = 0 for 3 t 4 s 20 A for 4 t 5 s 0 for t 5 s 20 i (A) t (s) 20 Fig. P1.24
2 Problem 1.34 The voltage across a device and the current through it are shown graphically in Fig. P1.34. Sketch the corresponding power delivered to the device and calculate the energy absorbed by it. i(t) 10 A 0 υ(t) 1 s 2 s t 5 V 0 1 s 2 s t Figure P1.34: i(t) and υ(t) of the device in Problem Solution: For 0 t 1 s, p(t) = υi = (5t)(10t) = 50t 2 For 1 s t 2 s, υ = 5(2 t) i = 10(2 t) p(t) = 50(2 t) 2 p(t) 50 W 1 s 2 s t 2 w = = = 33.3 J. p(t) dt 2 50t 2 dt 50(2 t) 2 dt 1
3 Problem 2.3 A thinfilm resistor made of germanium is 2 mm in length and its rectangular cross section is 0.2 mm 1 mm, as shown in Fig. P2.3. Determine the resistance that an ohmmeter would measure if connected across its: (a) Top and bottom surfaces (b) Front and back surfaces (c) Right and left surfaces z 2 mm y 0.2 mm 1 mm x Figure P2.3: Film resistor of Problem 2.3. Solution: (a) (b) (c) R = l l = 0.22 mm, A = 1 mm 2 mm = m 2 σa = 47 Ω R = l l = 1 mm, A = 2 mm 0.2 mm = m 2 σa 10 3 = 1,174 Ω R = l l = 2 mm, A = 1 mm 0.2 mm = m 2 σa = 4,695 Ω
4 Problem 2.17 Determine currents I 1 to I 4 in the circuit of Fig. P2.17. I 1 I 2 I 3 I 4 2 Ω 4 Ω 6 A 2 Ω 4 Ω Figure P2.17: Circuit for Problem Solution: The same voltage exists across all four resistors. Hence, 2I 1 = 4I 2 = 2I 3 = 4I 4. Also, KCL mandates that I 1 I 2 I 3 I 4 = 6 It follows that I 1 = 2 A, I 2 = 1 A, I 3 = 2 A, and I 4 = 1 A.
5 Problem 2.18 Determine the amount of power dissipated in the 3kΩ resistor in the circuit of Fig. P ma V 0 2 kω 3 kω 10 3 V 0 Figure P2.42: Circuit for Problem Solution: In the left loop, V 0 = = 20 V. The dependent current source is I 0 = 10 3 V 0 = 20 ma. The power dissipated in the 3kΩ resistor is p = I 2 0R = ( ) = 1.2 W.
6 Problem 2.19 Determine I x and I y in the circuit of Fig. P V 2 Ω I x 6 Ω I 4 Ω I y 4I x Figure P2.19: Circuit for Problem Solution: Application of KVL to the two loops gives 102I x 4I = 0 4I 6I y 4I x = 0. Additionally, I = I x I y. Solution of the three equations yields I x = 3.57 A, I y = 2.86 A.
7 Problem 2.25 After assigning node V 4 in the circuit of Fig. P2.25 as the ground node, determine node voltages V 1, V 2, and V 3. 3 A 12 V 3 Ω V 3 Ω V 2 1 V 3 6 Ω 6 Ω 6 Ω 1 A V 4 1 A Figure P2.25: Circuit of Problem Solution: 3 A 12 V 3 Ω I 1 V 3 Ω V 2 1 V 3 6 Ω 6 Ω 6 Ω 1 A V 4 Fig. P2.25 (a) 1 A From KCL at node V 1, the sum of currents leaving the node is 3I 1 1 = 0, or Node voltages (relative to V 4 ): I 1 = 31 = 2 A. V 1 = 6 1 = 6 V, V 2 = V 1 3I 1 = 6 3( 2) = 0, V 3 = 6 1 = 6 V.
8 Problem 2.36 Use resistance reduction and source transformation to find V x in the circuit of Fig. P2.36. All resistance values are in ohms. Solution: Figure P2.36: Circuit for Problem V x A V x A V x 6 10 A A V x V V x V x = = 8 V.
9 Problem 2.43 Apply voltage and current division to determine V 0 in the circuit of Fig. P2.43 given that V out = 0.2 V. Solution: I 5 V 0 V 5 8 I 3 I 4 V 3 4 I 1 V I V 1 V Figure P2.43: Circuit for Problem V out = 0.2 V I 1 = = 0.2 A I 2 = V 2 2 = I 1 (21) = 0.3 A 2 I 3 = I 1 I 2 = 0.5 A I 4 = V 4 4 = V 3 V 2 = 4I 3 2I = 0.65 A I 5 = I 3 I 4 = 1.15 A V 0 = V 4 V 5 = 4I 4 8I 5 = 11.8 V.
10 Problem 2.44 Apply source transformations and resistance reductions to simplify the circuit to the left of nodes (a,b) in Fig. P2.44 into a single voltage source and a resistor. Then, determine I. 3 A 10 Ω a 5 A 2 Ω 12 Ω I 4 Ω b Figure P2.44: Circuit of Problem Solution:
11 3 A 10 Ω a 5 A 2 Ω 12 Ω b 30 V 2 Ω 10 Ω a 10 V 12 Ω b 12 Ω a 40 V 12 Ω b a Ω 12 Ω Ω b a b 20 V 6 Ω a I 4 Ω Fig. P2.44 (a) I = = 2 A. b
12 Problem 2.47 Determine currents I 1 to I 4 in the circuit of Fig. P Ω I 1 I 3 3 Ω 6 Ω I 2 I 4 6 Ω 12 V Figure P2.47: Circuit of Problems 2.47 and Solution: I 1 = = 1 A, I 2 = 12 6 = 2 A, I 3 = 12 3 = 4 A, I 4 = 12 6 = 2 A.
13 Problem 3.9 Apply nodal analysis to find node voltages V 1 to V 3 in the circuit of Fig. P3.9 and then determine I x. 4 A V I 3 Ω 2 6 Ω x V 1 V 3 2 Ω 2 Ω 4 Ω 48 V Figure P3.9: Circuit for Problem 3.9. Solution: At nodes V 1, V 2, and V 3 : Node 1: Node 2: Node 3: V 1 Simplification of the three equations leads to: Simultaneous solution of Eqs. (4) (6) leads to: 2 V 1 V 2 4 = 0 3 (1) V 2 V 1 V 2 48 V 2 V 3 = (2) V 3 V 2 V = 0 (3) 5V 1 2V 2 = 24 (4) 2V 1 6V 2 V 3 = 144 (5) 2V 2 5V 3 = 48 (6) V 1 = 84 5 V, V 2 = 30 V, V 3 = 12 5 V. Hence, I x = V 2 V 3 6 = 30 12/5 6 = 4.6 A.
14 Problem 3.29 Apply mesh analysis to find I in the circuit of Fig. P V 1 Ω 1 Ω I I 1 1 Ω 1 Ω 1 Ω I 2 I 3 Figure P3.29: Circuit for Problem Solution: Mesh 1: 16I 1 (I 1 I 2 ) = 0 Mesh 2: (I 2 I 1 )I 2 (I 2 I 3 ) = 0 Mesh 3: (I 3 I 2 )I 3 = 0 Solution is: I 1 = 10 A, I 2 = 4 A, I 3 = 2 A. I = (I 1 I 2 ) = 10 4 = 6 A.
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