Question

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

• A. $20\Omega$
• B. $15\Omega$
• C. $10\Omega$
• D. $30\Omega$

Answer 1

Answer: (C)

$P=\frac{V^2}{R_{eq}}$
$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5R}$
$R_{eq}=\left(\frac{5R}{5+R}\right)P=30W$
Substituting the values in equation (i)
$30=\frac{(10{)}^2}{\left(\frac{5R}{5+R}\right)}$
$\Rightarrow R=10\Omega$