(a) Explanation Given data: Refer to Figure P4.60 in the textbook. Calculation: In the given circuit of Figure P4.60, the voltage source 50 V with series resistance 10 kilo ohms is converted into current source with parallel resistance using source transformation method. That is, I 1 = 50 V 10 k Ω = 50 V 10 × 10 3 Ω { ∵ 1 k = 10 3 } = 5 × 10 − 3 A { ∵ 1 A = 1 V 1 Ω } = 5 mA { ∵ 1 m = 10 − 3 } Figure 1 shows the modified circuit diagram. In Figure 1, 10 kilo ohms and 15 kilo ohms are connected in parallel. Therefore, the equivalent resistance for parallel connection is calculated as follows. 10 k Ω | | 15 k Ω = 10 k Ω × 15 k Ω 10 k Ω + 15 k Ω = 6 k Ω Figure 2 shows the reduced circuit diagram. In Figure 2, the current source (left side) 5 mA with parallel resistance 6 kilo ohms is converted into voltage source with series resistance using source transformation method. That is, V 1 = ( 5 mA ) ( 6 k Ω ) = ( 5 × 10 − 3 A ) ( 6 × 10 3 Ω ) { ∵ 1 m = 10 − 3 , 1 k = 10 3 } = 30 V { ∵ 1 V = 1 A ⋅ 1 Ω } Figure 3 shows the modified circuit diagram. In Figure 3, the 6 kilo ohms, 3 kilo ohms, and 6 kilo ohms are connected in series. Therefore, the equivalent resistance for series connection is calculated as follows. 6 k Ω + 3 k Ω + 6 k Ω = 15 k Ω Figure 4 shows the reduced circuit diagram. In Figure 4, the voltage source 30 V with series resistance 15 kiloohms is converted into current source with parallel resistance using source transformation method. That is, I 2 = 30 V 15 k Ω = 30 V 15 × 10 3 Ω { ∵ 1 k = 10 3 } = 2 × 10 − 3 A { ∵ 1 A = 1 V 1 Ω } = 2 mA { ∵ 1 m = 10 − 3 } Figure 5 shows the modified circuit diagram. In Figure 5, the current source 2 mA and 1 mA are in parallel connection with opposite direction. Therefore, the equivalent current source is calculated as follows. 2 mA − 1 mA = 1 mA Since, the positive sign of 1 mA current source shows the current direction in upward direction which is shown in Figure 6. Figure 6 shows the reduced circuit diagram. In Figure 6, the current source 1 mA with parallel resistance 15 kilo ohms is converted into voltage source with series resistance using source transformation method. That is, V 2 = ( 1 mA ) ( 15 k Ω ) = ( 1 × 10 − 3 A ) ( 15 × 10 3 Ω ) { ∵ 1 m = 10 − 3 , 1 k = 10 3 } = 15 V { ∵ 1 V = 1 A ⋅ 1 Ω } Figure 7 shows the modified circuit diagram (b) To determine Calculate the power developed by the 50 V source in the given circuit using PSPICE.

10th Edition

ISBN: 9780133760033

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 4, Problem 60P

(a)

To determine

Calculate the current io in the given circuit of Figure P4.60 by making a succession of source transformation method using PSPICE.

(b)

To determine

Calculate the power developed by the 50 V source in the given circuit using PSPICE.

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Chapter 4, Problem 59P

Chapter 4, Problem 61P

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Calculate the current i o in the given circuit of Figure P4.60 by making a succession of source transformation method using PSPICE.