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In the circuit shown in the figure, the $\mathrm{AC}$ source gives a voltage $V=20 \cos (2000 t)$ neglecting source resistance, the voltmeter and ammeter reading will be
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Verified Answer
The correct answer is:
$282 \mathrm{~V}, 1.41 \mathrm{~A}$
$282 \mathrm{~V}, 1.41 \mathrm{~A}$
Given, $R_1=8 \Omega$
$\begin{aligned} R_2 & =2 \Omega \\ L & =5 \mathrm{mH} \\ C & =50 \mu \mathrm{F}\end{aligned}$
and $\quad V_0=20 \cos (2000 t)$
Impedance, $Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
As, here, $X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega$
Similarly, $X_L=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega$ $\because \quad X_L=X_C$, Hence, $i_{\max }=\frac{V_0}{Z}=\frac{20}{(8+2)}=2 \mathrm{~A}$
Hence, $\quad i_{\operatorname{rms}}=\frac{i_{\max }}{\sqrt{2}}=\frac{2}{\sqrt{2}}=1.41 \mathrm{~A}$ and
$$
\begin{aligned}
V & =R_2 i_{\mathrm{rms}}=1.41 \times 2 \\
& =282 \mathrm{~V}
\end{aligned}
$$
$\begin{aligned} R_2 & =2 \Omega \\ L & =5 \mathrm{mH} \\ C & =50 \mu \mathrm{F}\end{aligned}$
and $\quad V_0=20 \cos (2000 t)$
Impedance, $Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
As, here, $X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega$
Similarly, $X_L=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega$ $\because \quad X_L=X_C$, Hence, $i_{\max }=\frac{V_0}{Z}=\frac{20}{(8+2)}=2 \mathrm{~A}$
Hence, $\quad i_{\operatorname{rms}}=\frac{i_{\max }}{\sqrt{2}}=\frac{2}{\sqrt{2}}=1.41 \mathrm{~A}$ and
$$
\begin{aligned}
V & =R_2 i_{\mathrm{rms}}=1.41 \times 2 \\
& =282 \mathrm{~V}
\end{aligned}
$$
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