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

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Given reading of an ideal voltmeter for resistance $R_{4}$ is $V_{4} = 5 V$ , then current (let $i_{1}$ )is given branch (as resistance $R_{3}$ in series with resistance $R_{4}$ ) is
$i_{1} = \frac{V}{R_{4}} = \frac{5}{500} = 0.01 A$
Then for resistance $R_{3}$ , the potential difference will be

$V_{3} = i_{1} R_{3} = 0.01 \times 100 = 1 V$

So, for given branch of resistance potential difference is

$V_{3} + V_{4} = 1 + 5 = 6 V$

From figure for resistance $R_{1}, R_{3}$ and $R_{4}$ a total applied potential difference is $18 V$ then,

$V_{1} + V_{3} + V_{4} = 18 ∴ V_{1} = 18 - (V_{3} + V_{4}) = 18 - 6 = 12 V$

So, the total current drawn from the source is

$∴$ Current through $R_{2}$

$i_{2} = 0 \cdot 03 - 0 \cdot 01 = 0 \cdot 02 A$
and potential difference for $R_{2}$ is

$V_{2} = V_{3} + V_{4} = 6 V$ ( $R_{2}$ is in parallel arrangement with the resistances $R_{3}$ and $R_{4}$ )

Now the value of resistance $R_{2}$ is :
$V_{2} = i_{2} R_{2} \Rightarrow R_{2} = \frac{V_{2}}{i_{2}} = \frac{6}{0 \cdot 02} = 300 Ω$

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