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 2.3 A thin-film 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, R = l l = 2 mm, A = 1 mm 0.2 mm = m 2 σa = 4,

3 Problem 2.17 Determine currents 1 to 4 in the circuit of Fig. P Ω 6 A 2 Ω Figure P2.17: Circuit for Problem Solution: The same voltage exists across all four resistors. Hence, 2 1 = 4 2 = 2 3 = 4 4. Also, KCL mandates that = 6 t follows that 1 = 2 A, 2 = 1 A, 3 = 2 A, and 4 = 1 A.

4 Problem 2.18 Determine the amount of power dissipated in the 3-kΩ 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: n the left loop, V 0 = = 20 V. The dependent current source is 0 = 10 3 V 0 = 20 ma. The power dissipated in the 3-kΩ resistor is p = 2 0R = ( ) = 1.2 W.

5 Problem 2.19 Determine x and y in the circuit of Fig. P V 2 Ω x 6 Ω y 4 x Figure P2.19: Circuit for Problem Solution: Application of KVL to the two loops gives 102 x 4 = y 4 x = 0. Additionally, = x y. Solution of the three equations yields x = 3.57 A, y = 2.86 A.

6 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 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 = 0, or Node voltages (relative to V 4 ): 1 = 31 = 2 A. V 1 = 6 1 = 6 V, V 2 = V = 6 3( 2) = 0, V 3 = 6 1 = 6 V.

7 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.

8 Problem 2.49 Determine current in the circuit of Fig. P2.49. Solution: Resistance combining leads to 10 Ω 40 Ω 2 30 Ω 60 Ω 50 V 10 Ω 10 Ω 40 Ω = 2 50 V 6 30 Ω 50 V 50 V = Ω = Fig. P2.49 (a) 50 = 1.97 A

9 Problem 2.50 Determine the equivalent resistance R eq at terminals (a,b) in the circuit of Fig. P2.50. Solution: a R eq 6 Ω b a 10 Ω b a R = 4 4 (5 5 10) = 10 Ω b Fig. P2.50 (a) R = 44(5 5 10) = 10 Ω.

10 Problem 2.52 Determine voltage V a in the circuit of Fig. P2.52. Solution: 2 A V a 2 Ω 2 A 2.5 A 2 Ω 5 A 2 A V a 2 Ω 4 V 2 Ω 10 V 10 V 2 A V a 8 Ω 16 V 2 A 2 A V a 8 Ω 4 A V a 8 Ω Fig. P2.52 (a) By current division, = = 2.67 A

11 and V = 4 = V.

12 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 x. 4 A V 3 Ω 2 6 Ω x V 1 V 3 2 Ω 2 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, x = V 2 V 3 6 = 30 12/5 6 = 4.6 A.

13 Problem 3.29 Apply mesh analysis to find in the circuit of Fig. P V 1 Ω 1 Ω 1 1 Ω 1 Ω 1 Ω 2 3 Figure P3.29: Circuit for Problem Solution: Mesh 1: 16 1 ( 1 2 ) = 0 Mesh 2: ( 2 1 ) 2 ( 2 3 ) = 0 Mesh 3: ( 3 2 ) 3 = 0 Solution is: 1 = 10 A, 2 = 4 A, 3 = 2 A. = ( 1 2 ) = 10 4 = 6 A.