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Current Electricity

JEE Main level, single correct, medium

155 of 380

Question 155 of 380

In the given circuit the internal resistance of the $18V$ cell is negligible. If ${R}_{1}=400\Omega ,{R}_{3}=100\Omega$ and ${R}_{4}=500\Omega$ and the reading of an ideal voltmeter across ${R}_{4}$ is $5V,$ then the value of ${R}_{2}$ will be:

Show answer and solution

Answer: (B) $300\Omega$

Given reading of an ideal voltmeter for resistance

{R}_{4}

{V}_{4}=5V

, then current (let

{i}_{1}

)is given branch (as resistance

{R}_{3}

in series with resistance

{R}_{4}

) is

{i}_{1}=\dfrac{V}{{R}_{4}}=\dfrac{5}{500}=0.01A

Then for resistance

{R}_{3}

, the potential difference will be

{V}_{3}={i}_{1}{R}_{3}=0.01\times 100=1V

So, for given branch of resistance potential difference is

{V}_{3}+{V}_{4}=1+5=6V

From figure for resistance

{R}_{1},{R}_{3}

and

{R}_{4}

a total applied potential difference is

18V

then,

{V}_{1}+{V}_{3}+{V}_{4}=18 ∴{V}_{1}=18-\left({V}_{3}+{V}_{4}\right)=18-6=12V

So, the total current drawn from the source is

∴

Current through

{R}_{2}

{i}_{2}=0\cdot 03-0\cdot 01=0\cdot 02A

and potential difference for

{R}_{2}

{V}_{2}={V}_{3}+{V}_{4}=6V

(

{R}_{2}

is in parallel arrangement with the resistances

{R}_{3}

and

{R}_{4}

)
Now the value of resistance

{R}_{2}

is :

{V}_{2}={i}_{2}{R}_{2} \Rightarrow {R}_{2}=\dfrac{{V}_{2}}{{i}_{2}}=\dfrac{6}{0\cdot 02}=300\Omega

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