Question

The amount of power delivered by the AC source in the capacitor & resistance $R_1$ of the circuit given above is $\frac{X}{8}\times {10}^3$ watt. Then x is:

Answer 1

$i_{1_{rms}}=\frac{E_{rms}}{\sqrt{X_2^C+R_2^1}}=\frac{130}{13}=10A$
$i_{2_{rms}}=\frac{E_{rms}}{\sqrt{X_2^L+R_2^2}}=13A$
Power dissiputed $=i_2^{1rms}{R}_1={10}^2\times 5=500$ Watt
On compairing with the given form we can see that $X=4$