Fundamentals of Electrical Engineering

Published by McGraw-Hill Education
ISBN 10: 0073380377
ISBN 13: 978-0-07338-037-7

Chapter 3 - Resistive Network Analysis - Part 1 Circuits - Homework Problems - Page 116: 3.8



Work Step by Step

This question references the same circuit as that in the previous question (3.7). We are now expected to use mesh current analysis to obtain $i_{1}$ and $i_{2}$. By KCL, $\sum{i_{n}}=0$. Taking KCL at the node immediately to the left of $R_{2}$, $1A$ of current enters the node and $i_{1}A$ exits it, leaving $0+1-i_{1}=(1-i_{1})A$ of current to pass through $R_{2}$. Taking KCL at the node above $R_{3}$, $$(1-i_{1})-i_{2}-2=0$$ $$\Rightarrow{i_{1}}+i_{2}=-1$$ By considering current flow in the leftmost and rightmost loops of the circuit, we can deduce that $3i_{1}=1$ and $6i_{2}=2$. (This is known as the loop current method). By solving the above two equations and taking ratios in $i_{1}$ and $i_{2}$, we can deduce that $i_{1}=i_{2}$. Substituting $i_{1}=i_{2}$ into the equation we derived earlier, $$i_{2}+i_{2}=2i_{2}=-1\Rightarrow{i_{2}}=-0.5A$$ $$\therefore{i_{1}}=i_{2}=-0.5A$$
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