Question

A coil having a resistance of $5\Omega$ and an inductance of $0.02H$ is arranged in parallel with another coil having a resistance of $1\Omega$ and an inductance of $0.08H$. Calculate the power absorbed (in W) when a voltage of $100V$ at $50Hz$ is applied.

Answer 1

$X_L=\omega L_1=(2\pi \times 50)\left(0.02\right)=6.28\Omega$
$\therefore Z_1=\sqrt{R_1^2+X^2{L}_1}$
$=\sqrt{(5{)}^2+(6.28{)}^2}=8.0\Omega$
$P_1=(I_{rms}{)}_1{V}_{rms}\cos {\varphi }_1$ $=\left(\frac{100}{8}\right)$ $\left(100\right)$ $\left(\frac{5}{8}\right)$
$=781.25W$
$|X_{L_2}=\omega L_2=(2\pi \times 50)\left(0.08\right)=25.13\Omega$
$\therefore Z_2=\sqrt{R_2^2+X_{L_2}^2}$
$25.15\Omega$
$\therefore P_2=(i_{rms}{)}_2{V}_{rms}\cos {\varphi }_2$
$=\left(\frac{100}{25.15}\right)\left(100\right)\left(\frac{1}{25.15}\right)$
$=15.8W$
$\therefore P_{Total}=P_1+P_2=797W$