22823 A12Vfig. 5wwww+1

Answered: 85 3A 12V 2 fig. 5 +1

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Step 1

Given circuit:

Step 2

Solution:

Let the Voltage across the current source is V_x.

so modified circuit is,

$From the circuit, V_{o} = 2 (i_{1} - i_{2})$

$At the current source, i_{3} - i_{2} = 3 . . . . . . (1)$

On applying KVL in loop 1,

$\begin{array}{rcl} - 12 + 2 (i_{1} - i_{2}) + 4 (i_{1} - i_{3}) & = & 0 \\ 6 i_{1} - 2 i_{2} - 4 i_{3} & = & 12 \\ 3 i_{1} - i_{2} - 2 i_{3} & = & 6 . . . . . . . . . . . . (2) \end{array}$

On applying KVL in loop 2,

$\begin{array}{rcl} 2 (i_{2} - i_{1}) + 8 i_{2} - V_{x} & = & 0 \\ - 2 i_{1} + 10 i_{2} - V_{x} & = & 0 . . . . . . . . . . . . . . . . (3) \end{array}$

On applying KVL in loop 3,

$4 (i_{3} - i_{2}) + V_{x} + 2 V_{o} = 0 plugging the value of V_{o} in the above equation, 4 (i_{3} - i_{2}) + V_{x} + 2 \times 2 (i_{1} - i_{2}) = 0 4 i_{1} - 8 i_{2} + 4 i_{3} + V_{x} = 0 . . . . . . . . . . . . . . . . (4)$

Adding equation (3) and (4),

$2 i_{1} + 2 i_{2} + 4 i_{3} = 0 i_{1} + i_{2} + 2 i_{3} = 0 . . . . . . . . . (5)$

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