An oscillating circuit consisting of a capacitor with capacitance $C=10 \mu \mathrm{F}$, a coil with inductance $L=6.0 \mu \mathrm{H}$ and active resistance $R=10 \Omega$. The mean power that should be fed to the circuit to maintain undamped harmonic oscillations with an external driving power with $50 \mathrm{~Hz}$ and a $V_m$ of $280 \mathrm{~V}$ is

• Correct answer
  A

  $3.8 \mathrm{~W}$
• B

  $48 \mathrm{~W}$
• C

  $3 \mathrm{~mW}$
• D

  $48 \mathrm{~mW}$

Solution

$$ \text { } R=10 \Omega L=6 \mu \mathrm{H}, C=10 \mu \mathrm{H} $$
Impedence of the circuit, $$ Z=\sqrt{R^2+\left(X_C-X_L\right)^2} $$ Capacitive reactance, $X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$ $$ X_C=\frac{1}{2 \times 314 \times 50 \times 10 \times 10^{-6}} $$ $$ =318.47 \Omega $$ Inductive reactance $X_L=\omega L=2 \pi f L$ $$ \begin{aligned} X_L & =2 \times 314 \times 50 \times 6 \times 10^{-6} \\ & =18.84 \times 10^{-4} \Omega \end{aligned} $$ Resistance, $R=10 \Omega$ $$ \begin{gathered} Z=\sqrt{10^2+\left(318.4-18.84 \times 10^{-4}\right)^2} \\ Z=\sqrt{100+(318.39)^2}=\sqrt{101477.36} \\ Z=318.55 \Omega \\ \text { rms current, } i_{\text {rms }}=\frac{V_{\text {rms }}}{Z}=\frac{V_0 / \sqrt{2}}{Z} \\ i_{\text {rms }}=\frac{280}{\sqrt{2} \times 318.55}=0.621 \mathrm{~A} \end{gathered} $$ As energy is only dissipated in resistance as heat and no power dissipates in capacitor and inductor, so average power dissipates as heat $$ P=i_{\mathrm{rms}}^2 \cdot R $$ $$ \begin{aligned} & P=(0.621)^2 \times 10 \\ & P=3.8 \mathrm{~W} \end{aligned} $$ $3.8 \mathrm{~W}$ power should be fed to maintain an undamped oscillation.

Hence, the correct answer is option: 'A'.