In the given circuit the internal resistance of the 18V 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 5V, then the value of $R_{2}$ will be

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$300 Ω$

$230 Ω$

$450 Ω$

$550 Ω$

answer is D.

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Detailed Solution

$R_{1} = 400 Ω$

$R_{5} = R_{3} + R_{4}$

= 100 + 500 = 600

Voltage across $R_{4}, V_{4} = I_{1} R_{4}$

Or $I_{1} = \frac{5}{500} = \frac{1}{100} A$

Voltage Across $R_{3}, V_{3} = I_{1} R_{3} = \frac{1}{100} \times 100 = 1 V$

$V_{2} = V_{3} + V_{4} = 6 V$ or $I_{2} R_{2} = 6$ …. (i)

$V_{1} = 18 - 6$ or $I R_{1} = 12$ or $I = \frac{12}{400} = \frac{3}{100} A$

$I_{2} = I - I_{1} = \frac{2}{100} A$

From equation (i)

$R_{2} = \frac{6}{I_{2}} = 300 Ω$

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