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A parallel plate capacitor in series with a resistance of $100 \Omega$, an inductor of $20 \mathrm{mH}$ and an AC voltage source of variable frequency shows resonance at a frequency of $\frac{1250}{\pi}$ Hz If this capacitor is charged by a DC voltage source to a voltage 25 v, what amount of charge will be stored in cach plate of the capacitor?
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Verified Answer
The correct answer is:
$0.2 \mathrm{mC}$
In given, Series $R-L-C$ circuit Resistance $R=100 \Omega$
Inductance of Inductor, $L=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}$
Resonance frequency $f=\frac{1250}{\pi} \mathrm{Hz}$
Source voltage $V_{D C}=25 \mathrm{V}$
According to resonant frequency.
$$
\omega_{0}=2 \pi f_{0}=\frac{1}{\sqrt{L C}}
$$
or $\left(2 \pi f_{0}\right)^{2}=\frac{1}{L C}$
or $\quad 4 \pi^{2} \frac{1250 \times 1250}{\pi \times \pi}=\frac{1}{L C}$
Or
$$
\begin{aligned}
C=& \frac{1000}{1250 \times 1250 \times 4 \times 20} \\
&\left(\because \mathrm{By} \text { substituting } L=20 \times 10^{-3}\right) \\
C=& 8 \times 10^{-6} \mathrm{F}
\end{aligned}
$$
or
We know that, Charge $Q_{s}=C V$
$$
Q_{s}=8 \times 10^{-6} \times 25=0.2 \mathrm{mC}
$$
So, the amount of charge stored in each plate of capacitor is $0.2 \mathrm{mC}$.
Inductance of Inductor, $L=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}$
Resonance frequency $f=\frac{1250}{\pi} \mathrm{Hz}$
Source voltage $V_{D C}=25 \mathrm{V}$
According to resonant frequency.
$$
\omega_{0}=2 \pi f_{0}=\frac{1}{\sqrt{L C}}
$$
or $\left(2 \pi f_{0}\right)^{2}=\frac{1}{L C}$
or $\quad 4 \pi^{2} \frac{1250 \times 1250}{\pi \times \pi}=\frac{1}{L C}$
Or
$$
\begin{aligned}
C=& \frac{1000}{1250 \times 1250 \times 4 \times 20} \\
&\left(\because \mathrm{By} \text { substituting } L=20 \times 10^{-3}\right) \\
C=& 8 \times 10^{-6} \mathrm{F}
\end{aligned}
$$
or
We know that, Charge $Q_{s}=C V$
$$
Q_{s}=8 \times 10^{-6} \times 25=0.2 \mathrm{mC}
$$
So, the amount of charge stored in each plate of capacitor is $0.2 \mathrm{mC}$.
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