Question

The total power dissipated in watt in the circuit shown here is

• A. 40 W
• B. 54 W
• C. 4 W
• D. 16 W

Answer 1

Answer: (B)

The resistance of $6\Omega$ and $3\Omega$ are in parallel in the given circuit, their equivalent resistance is
$\frac{1}{R_1}=\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6}=\frac{1}{2}$ or $R_1=2\Omega$
Again, $R_1$ is in series with $4\Omega$ resistance, hence ltbgt
$R=R_1+4=2+4=6\Omega$
Thus, the total power dissipatesin the circuit
$P=\frac{V^2}{R}$
Here, $V=18V,R=6\Omega$
Thus, $P=\frac{(18{)}^2}{6}=54W$