To determine Check whether the interconnection satisfy the power check Explanation Given data: Refer to Figure P1.31 and Table P1.31 in the textbook. 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 − 3 kV for v a and − 250 mA for i a in equation (1) to find p a . p a = − ( − 3 kV ) ( − 250 mA ) = − ( − 3 × 10 3 ) ( − 250 × 10 − 3 ) V ⋅ A { ∵ 1 k = 10 3 , 1 m = 10 − 3 } = − ( 750 ) W { ∵ 1 W = 1 V ⋅ 1 A } = − 750 W For element “b”: 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 “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 4 kV for v b and − 400 mA for i b in equation (2) to find p b . p b = − ( 4 kV ) ( − 400 mA ) = − ( 4 × 10 3 ) ( − 400 × 10 − 3 ) V ⋅ A { ∵ 1 k = 10 3 , 1 m = 10 − 3 } = − ( − 1600 ) W { ∵ 1 W = 1 V ⋅ 1 A } = 1600 W 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 1 kV for v c and 400 mA for i c in equation (3) to find p c . p c = − ( 1 kV ) ( 400 mA ) = − ( 1 × 10 3 ) ( 400 × 10 − 3 ) V ⋅ A { ∵ 1 k = 10 3 , 1 m = 10 − 3 } = − ( 400 ) W { ∵ 1 W = 1 V ⋅ 1 A } = − 400 W For element “d”: 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 “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 1 kV for v d and 150 mA for i d in equation (4) to find p d . p d = ( 1 kV ) ( 150 mA ) = ( 1 × 10 3 ) ( 150 × 10 − 3 ) V ⋅ A { ∵ 1 k = 10 3 , 1 m = 10 − 3 } = 150 W { ∵ 1 W = 1 V ⋅ 1 A } 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

10th Edition

ISBN: 9780133992793

Author: NILSSON

Publisher: PEARSON

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Fault Analysis

In an electrical engineering system, a fault is an abnormal electric current flowing through the system, and fault analysis is the study of these faults. For instance, a short-circuit is a fault where the live wire touches the ground or neutral wire. If…

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Fault Analysis in Power Systems part 1a; Author: GeneralPAC: Power System Tutorials;https://www.youtube.com/watch?v=g8itg4MOjok;License: Standard youtube license