In the given circuit the internal resistance of the $18 \mathrm{~V}$ cell is negligible. If $\mathrm{R}_{1}=400 \Omega, \mathrm{R}_{3}=100 \Omega$ and $\mathrm{R}_{4}=500 \Omega$ and the reading of an ideal voltmeter across $R_{4}$ is $5 V$, then the value of $R_{2}$ will be:
Correct Option: 1
(1)
Across $\mathrm{R}_{4}$ reading of voltmeter, $\mathrm{V}_{4}=5 \mathrm{~V}$
Current, $\mathrm{i}_{4}=\frac{\mathrm{V}_{4}}{\mathrm{R}_{4}}=0.01 \mathrm{~A}$
$\mathrm{V}_{3}=\mathrm{i}_{1} \mathrm{R}_{3}=1 \mathrm{~V}$
$\mathrm{~V}_{3}+\mathrm{V}_{4}=6 \mathrm{~V}=\mathrm{V}_{2}$
$\mathrm{~V}_{1}+\mathrm{V}_{3}+\mathrm{V}_{4}=18 \mathrm{~V}$
$\Rightarrow \mathrm{V}_{1}=12 \mathrm{~V}$
$\mathrm{i}=\frac{\mathrm{V}_{1}}{\mathrm{R}_{1}}=0.03 \mathrm{~A}$
$i=i_{1}+i_{2} \Rightarrow \mathrm{i}_{2}=i-i,=0.03-0.01 \mathrm{~A}=0.02 \mathrm{~A}$
$\therefore R_{2}=\frac{V_{2}}{i_{2}}=\frac{6}{0.02}=300 \Omega$
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