A 10$\Omega$ resistor, 10mH inductor and 100$\mu F$ capacitor are connected in series to a 50V (rms) source having variable frequency. The energy that is delivered to the circuit during one period if the operating frequency is twice the resonance frequency, is $\frac{\pi }{x}J$. The value of $x$ is

Since at resonance the angular frequency ${\omega }_0$ is given by${\omega }_0=\frac{1}{\sqrt{L}C}$

Now, when $\omega =2{\omega }_0=\frac{2}{\sqrt{LC}}$, then

$X_L=L\omega =\frac{2L}{\sqrt{LC}}=2\sqrt{\frac{L}{C}}and$$X_C=\frac{1}{C\omega }=\frac{\sqrt{LC}}{2C}=\frac{1}{2}\sqrt{\frac{L}{C}}$

Since by definition, we have

$Z^2={(X_L-X_C)}^2+R^2$

$\Rightarrow Z^2=R^2+2.25\left(\frac{L}{C}\right)$

Further, we know that

$P=\left(\frac{V^2}{Z}\right)\cos \varphi =\left(\frac{V^2}{Z}\right)\left(\frac{R}{Z}\right)=\frac{V^{2}R}{Z^2}$

$\Rightarrow P=\frac{V^{2}R}{R^2+2.25\left(\frac{L}{C}\right)}$

So, energy delivered to the circuit in one cycle is

$E=PT=P\left(\frac{2\pi }{\omega }\right)=\left(\frac{2\pi RC{V}^2}{R^{2}C+2.25L}\right)\frac{\sqrt{LC}}{2}$

$\Rightarrow E=\frac{4\pi RC{V}^2\sqrt{LC}}{4R^{2}C+9L}$

Substituting values, we get

$E=\frac{\pi }{13}J$