Question:
The resistive network shown below is connected to a D.C. source of $16 \mathrm{~V}$. The power consumed by the network is 4 Watt. The value of $R$ is :
Correct Option: , 2
Solution:
(2) Equivalent resistance,
$R_{\mathrm{eq}}=\frac{4 R \times 4 R}{4 R+4 R}+R+\frac{6 R \times 12 R}{6 R+12 R}+R$
$=2 R+R+4 R+R$
$=8 R$
Using, $P=\frac{V^{2}}{R_{\mathrm{eq}}}$
$4=\frac{16^{2}}{8 R}$
$\therefore R=\frac{16^{2}}{4 \times 8}=8 \Omega$