Q 12-03-279JEE MainJEE Main 2019 (12 Apr, Shift 1)Easy
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:
Answer: (B) $8\ \Omega$
$4R \parallel 4R = 2R$ and $6R \parallel 12R = 4R$, so the total is $2R + R + 4R + R = 8R$.
$$P = \frac{V^2}{8R} = \frac{256}{8R} = 4 \Rightarrow R = 8\ \Omega$$
Solution by Sreeraj P, M.Sc Physics