Answer:
Option C
Explanation:
The power dissipated in the circuit
$P = \frac{V^{2}}{R_{eq}}$ ....(1)
v = 10 volt
$\frac{1}{R_{eq}} = \frac{1}{R}+\frac{1}{5}$ = $\frac{5+R}{5R}$
$R_{eq}$ = $\frac{5R}{5+R}$
P = 30W
Substituting the values in equation (1)
$30 = \frac{10^{2}}{\frac{5R}{5+R}}$
$\frac{15R}{5+R}$ = 10 → 15R = 50 + 10R
5R=50 → R = 10Ω
