Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1632 Upvotes
Verified Answer
The correct answer is:
$3.8 \mathrm{~W}$
$$
\text { } R=10 \Omega L=6 \mu \mathrm{H}, C=10 \mu \mathrm{H}
$$

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.
\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.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.