Explanation Concept used: Write the expression for current. i = E S Z ......... (1) Here, i is the current, E S is the source voltage and Z is the value of impedance. Calculation: The circuit diagram is redrawn as shown in Figure 1. Voltage source in series with impedance can be converted to current source by using same value of impedance in parallel with current source. For voltage source connected at node 1 Value of impedance is Z = ( 0.1 + j 0.5 ) Ω Convert in polar form. Z = ( 0.1 ) 2 + ( 0.5 ) 2 ∠ tan − 1 ( 0.5 0.1 ) Ω = 0.01 + 0.25 ∠ tan − 1 ( 5 ) Ω = 0.26 ∠ tan − 1 ( tan 78.69 ° ) Ω = 0.51 ∠ 78.69 ° Ω Substitute 1 ∠ 30 ° V for E S , i 1 for i and 0.51 ∠ 78.69 ° Ω for Z in equation (1). i 1 = 1 ∠ 30 ° V 0.51 ∠ 78.69 ° Ω = 1.96 ∠ ( 30 ° − 78.69 ° ) A = 1.96 ∠ − 48.69 ° A For voltage source connected at node 2 Value of impedance is Z = ( 0.1 + j 0.5 ) Ω Convert in polar form. Z = ( 0.1 ) 2 + ( 0.5 ) 2 ∠ tan − 1 ( 0.5 0.1 ) Ω = 0.01 + 0.25 ∠ tan − 1 ( 5 ) Ω = 0.26 ∠ tan − 1 ( tan 78.69 ° ) Ω = 0.51 ∠ 78.69 ° Ω Substitute 1 ∠ 0 ° V for E S , i 2 for i and 0.51 ∠ 78.69 ° Ω for Z in equation (1). i 2 = 1 ∠ 0 ° V 0.51 ∠ 78.69 ° Ω = 1.96 ∠ ( 0 ° − 78.69 ° ) A = 1.96 ∠ − 78.69 ° A The circuit diagram after source conversion is redrawn as shown in Figure 2. V 1 is voltage at node 1 and V 2 is the voltage at node 2 . Write the KCL equation at node 1 . − 1.96 ∠ − 48.69 ° A+ V 1 0.1 + j 0.5 + V 1 − j 0.1 + ( V 1 − V 2 ) 0.02 + j 0.2 = 0 Simplify further. V 1 ( 0.1 − j 0.5 ) 0.1 + j 0.5 ( 0.1 − j 0.5 ) − j 10 V 1 j 2 1 + ( V 1 − V 2 ) ( 0.02 − j 0.2 ) 0.02 + j 0.2 ( 0.02 − j 0.2 ) = 1.96 ∠ − 48.69 ° A V 1 ( 0.1 − j 0.5 ) 0

6th Edition

ISBN: 9780357700907

Author: Glover

Publisher: INTER CENG

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Question

Chapter 2, Problem 2.35P

To determine

Nodal equation in matrix form after convert the voltage source to equivalent current source.

Expert Solution & Answer

Chapter 2 Solutions

POWER SYS. ANALYSIS+DESIGN

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