Question

If in the circuit, power dissipation is $150\text{W}$, then $R$ is

• A. $2\Omega$
• B. $6\Omega$
• C. $5\Omega$
• D. $4\Omega$

Answer 1

The correct option is B $6\Omega$

Given:
$P_{net}=150\text{W};R_1=R;R_2=2\Omega :V=15\text{V}$

Since both resistances are in parallel, so net potential will be

$P_{net}=P_1+P_2$

$P_{net}=\frac{V^2}{R_1}+\frac{V^2}{R_2}$

putting the given value in equation

$150=\frac{{\left(15\right)}^2}{R}+\frac{{\left(15\right)}^2}{2}$

$\Rightarrow \frac{1}{R}+\frac{1}{2}=\frac{150}{{15}^2}=\frac{2}{3}\Rightarrow \frac{1}{R}=\frac{1}{6}$

$\therefore R=6\Omega$

Hence, option $\left(b\right)$ is correct.