To determine Find the total power developed in the circuit. Explanation Given data: Refer to the given circuit and table in the problem. Calculation: The given circuit is redrawn as Figure 1. The positive sign should be used in power equation, if the reference direction of the current is in the direction of the reference voltage drop across the element. The negative sign should be used in power equation, if the reference direction of the current is in the direction of the reference voltage rise across the element. Such sign convention is called as Passive Sign Convention. For element “a”: The direction of the current is in the direction of voltage rise across the element. Therefore, the power equation should have the negative sign. Write the expression for power in element “a”. p a = − v a i a (1) Here, v a is the voltage across the element “a”, and i a is the current through the element “a”. Substitute 40 V for v a and − 4 mA for i a in equation (1) to find p a . p a = − ( 40 V ) ( − 4 mA ) = − ( 40 ) ( − 4 × 10 − 3 ) V ⋅ A { ∵ 1 m = 10 − 3 } = ( 160 × 10 − 3 ) W { ∵ 1 W = 1 V ⋅ 1 A } = 160 mW { ∵ 1 m = 10 − 3 } For element “b”: The direction of the current is in the direction of voltage drop across the element. Therefore, the power equation should have the positive sign. Write the expression for power in element “b”. p b = v b i b (2) Here, v b is the voltage across the element “b”, and i b is the current through the element “b”. Substitute − 24 V for v b and − 4 mA for i b in equation (2) to find p b . p b = ( − 24 V ) ( − 4 mA ) = ( − 24 ) ( − 4 × 10 − 3 ) V ⋅ A { ∵ 1 m = 10 − 3 } = ( 96 × 10 − 3 ) W { ∵ 1 W = 1 V ⋅ 1 A } = 96 mW { ∵ 1 m = 10 − 3 } For element “c”: The direction of the current is in the direction of voltage rise across the element. Therefore, the power equation should have the negative sign. Write the expression for power in element “c”. p c = − v c i c (3) Here, v c is the voltage across the element “c”, and i c is the current through the element “c”. Substitute − 16 V for v c and 4 mA for i c in equation (3) to find p c . p c = − ( − 16 V ) ( 4 mA ) = ( 16 ) ( 4 × 10 − 3 ) V ⋅ A { ∵ 1 m = 10 − 3 } = ( 64 × 10 − 3 ) W { ∵ 1 W = 1 V ⋅ 1 A } = 64 mW { ∵ 1 m = 10 − 3 } For element “d”: The direction of the current is in the direction of voltage rise across the element. Therefore, the power equation should have the negative sign. Write the expression for power in element “d”. p d = − v d i d (4) Here, v d is the voltage across the element “d”, and i d is the current through the element “d”. Substitute − 80 V for v d and − 1.5 mA for i d in equation (4) to find p d . p d = − ( − 80 V ) ( − 1.5 mA ) = − ( 80 ) ( 1.5 × 10 − 3 ) V ⋅ A { ∵ 1 m = 10 − 3 } = − ( 120 × 10 − 3 ) W { ∵ 1 W = 1 V ⋅ 1 A } = − 120 mW { ∵ 1 m = 10 − 3 } For element “e”: The direction of the current is in the direction of voltage drop across the element

Electrical Circuits and Modified MasteringEngineering - With Access

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ISBN: 9780133992793

Author: NILSSON

Publisher: PEARSON

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Chapter 1, Problem 29P

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