Discussion Forum

1)

The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is

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