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 when a voltage of $100V$ at $50Hz$ is applied.

$X_{L_1}=\omega L_1=(2\mathrm{\pi }\times 50)(0.02)=6.28\mathrm{\Omega }$

$\begin{array}{rcl}\therefore Z_1& =& \sqrt{R_1^2+X_{L_1}^2}=\sqrt{{\left(5\right)}^2+{(6.28)}^2}=8.0\Omega \end{array}$

$P_1={\left(I_{rms}\right)}_1{V}_{rms} \cos {\varphi }_1=\left(\frac{V_{rms}}{Z_1}\right)V_{rms}\left(\frac{R_1}{Z_1}\right)=\left(\frac{100}{8}\right)\left(100\right)\left(\frac{5}{8}\right)=781.25W$

Again,

$X_{L_2}=\omega L_2=(2\mathrm{\pi }\times 50)(0.08)=25.13\mathrm{\Omega }$

$\begin{array}{rcl}\therefore Z_2& =& \sqrt{R_2^2+X_{L_2}^2}=\sqrt{{\left(1\right)}^2+{(25.13)}^2}=25.15\Omega \end{array}$

$P_2={\left(I_{rms}\right)}_2{V}_{rms} \cos {\varphi }_2=\left(\frac{V_{rms}}{Z_2}\right)V_{rms}\left(\frac{R_2}{Z_2}\right)=\left(\frac{100}{25.15}\right)\left(100\right)\left(\frac{1}{25.15}\right)=15.8W$

$\therefore P_{Total}=P_1+P_2=797W$.