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A series $L-C-R$ circuit is connected to an $\mathrm{AC}$ source of $220 \mathrm{~V}$ and $50 \mathrm{~Hz}$ shown in figure. If the readings of the three voltmeters $V_1, V_2$ and $V_3$ are $65 \mathrm{~V}, 415 \mathrm{~V}$ and $204 \mathrm{~V}$ respectively, the value of inductance and capacitance will be
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Verified Answer
The correct answer is:
$1.0 \mathrm{H}, 5 \mu \mathrm{F}$
As $\quad V=I_{\mathrm{rms}} R$
$$
\begin{aligned}
I_{\mathrm{rms}} & =\frac{V_R}{R}=\frac{65}{100}=0.65 \mathrm{~A} \\
V_L & =I_{\mathrm{rms}} \times X_L \\
X_L & =\frac{V_L}{I_{\mathrm{rms}}}=\frac{204}{0.65}=313.85 \Omega \\
X_L & =\omega L=2 \pi f L \\
L & =\frac{X_L}{2 \pi f} \\
L & =\frac{313.85}{2 \times \pi \times 50}=1.0 \mathrm{H} \\
X_C & =\frac{V_C}{I_{\mathrm{rms}}}=\frac{415}{0.65}=638.46 \Omega \\
X_C & =\frac{1}{\omega C}=\frac{1}{2 \pi f C} \\
C & =\frac{1}{2 \times \pi \times 50 \times 638.46} \\
& =5 \times 10^{-6}=5 \mu \mathrm{F}
\end{aligned}
$$
$$
\begin{aligned}
I_{\mathrm{rms}} & =\frac{V_R}{R}=\frac{65}{100}=0.65 \mathrm{~A} \\
V_L & =I_{\mathrm{rms}} \times X_L \\
X_L & =\frac{V_L}{I_{\mathrm{rms}}}=\frac{204}{0.65}=313.85 \Omega \\
X_L & =\omega L=2 \pi f L \\
L & =\frac{X_L}{2 \pi f} \\
L & =\frac{313.85}{2 \times \pi \times 50}=1.0 \mathrm{H} \\
X_C & =\frac{V_C}{I_{\mathrm{rms}}}=\frac{415}{0.65}=638.46 \Omega \\
X_C & =\frac{1}{\omega C}=\frac{1}{2 \pi f C} \\
C & =\frac{1}{2 \times \pi \times 50 \times 638.46} \\
& =5 \times 10^{-6}=5 \mu \mathrm{F}
\end{aligned}
$$
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