Question
Find the current through the $10\Omega$ resistor shown in figure.
✓
Answer
In the circuit ADCBA,
$3\text{i}+6\text{i}_1-4.5=0$
In the circuit GEFCG,
$3\text{i}+6\text{i}_1=4.5$
$=10\text{i}-10\text{i}_1-6\text{i}_1=-3$
$\Rightarrow[10\text{i}+16\text{i}_1=-3]3\ ...(1)$
$[3\text{i}+6\text{i}_1=4.5]10\ ...(2)$
From (1) and (2)
$-108\text{i}_1=-54$
$\Rightarrow\text{i}_1=\frac{54}{108}=\frac{1}{2}=0.5$
$3\text{i}+6\times\frac{1}{2}-4.5=0$
$3\text{i}-1.5=0\Rightarrow\text{i}=0.5$
Current through $10\Omega$ resistor = 0A.
$3\text{i}+6\text{i}_1-4.5=0$
In the circuit GEFCG,
$3\text{i}+6\text{i}_1=4.5$
$=10\text{i}-10\text{i}_1-6\text{i}_1=-3$
$\Rightarrow[10\text{i}+16\text{i}_1=-3]3\ ...(1)$
$[3\text{i}+6\text{i}_1=4.5]10\ ...(2)$
From (1) and (2)
$-108\text{i}_1=-54$
$\Rightarrow\text{i}_1=\frac{54}{108}=\frac{1}{2}=0.5$
$3\text{i}+6\times\frac{1}{2}-4.5=0$
$3\text{i}-1.5=0\Rightarrow\text{i}=0.5$
Current through $10\Omega$ resistor = 0A.
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