The power dissipated in the circuit shown in figure is 36 W. The value of R is:

12 Ω

The correct answer is Option (2) → 12 Ω

As resistor are connected in parallel,

$\frac{1}{R_{net}}=\frac{1}{6}+\frac{1}{R}=\frac{R+6}{6R}$

$⇒R_{net}=\frac{6R}{R+6}$

electron into the conduction band.

Power dissipated = $\frac{V^2}{R}=\frac{12×12×(R+6)}{6R}=36$

$⇒\frac{R+6}{R}=\frac{36×6}{12×12}$

$⇒\frac{R+6}{R}=\frac{3×1}{2}$

$⇒2R+12=3R$

$⇒R=12Ω$