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JEE Main 2019 (12 Apr, Shift 1) · Combination of Resistors

Combination of Resistors question from JEE Main 2019 (12 Apr, Shift 1)

Q64JEE Main 2019 · 12 APR · SHIFT 1Medium

Combination of Resistors

The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is:

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Answer: (c)

$4R \parallel 4R = 2R$ and $6R \parallel 12R = 4R$ .
$R_{eq} = 2R + R + 4R + R = 8R$
$P = \dfrac{\varepsilon^2}{R_{eq}} \Rightarrow 4 = \dfrac{256}{8R} \Rightarrow R = 8\ \Omega$
So the answer is (c) $8\ \Omega$ .

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