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In the following circuit an $\mathrm{AC}$ input $V_i(t)=(20 \mathrm{mV}) \sin \left(10^5 t\right)$ is applied at the left end. The amplitude of the output voltage $V_0$ at the right end across the capacitor will be
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Verified Answer
The correct answer is:
14.14 mV
Given that, input AC voltage,
Comparing Eqs. (i) and (ii), we get $\omega=10^5 \mathrm{rad} / \mathrm{s}, V_{\max }=20 \mathrm{mV}=20 \times 10^{-3} \mathrm{~V}$
The $V_i$ is distributed across $R$ and $C$. by $V_R$ and $V_C$, as phasor diagram.
Reactance of capacitor,
$$
X_C=\frac{1}{\omega C}=\frac{1}{10^5 \times 10^{-8}} \Omega
$$
$$
=10^3 \Omega=1000 \Omega
$$
$\therefore$ Now, impedance of $R$ - $C$ series circuit.
$$
Z=\sqrt{R^2+X_C^2}=\sqrt{(1000)^2+(1000)^2}=1000 \sqrt{2} \Omega
$$
Now, voltage across capacitor,
$$
\begin{aligned}
V_C & =V_0=\frac{X_C}{Z} V_{\max }=\frac{1000}{1000 \sqrt{2}} \times 20 \mathrm{mV} \\
& =\sqrt{2} \times 10 \mathrm{mV} \\
& =14.14 \mathrm{mV}
\end{aligned}
$$
Comparing Eqs. (i) and (ii), we get $\omega=10^5 \mathrm{rad} / \mathrm{s}, V_{\max }=20 \mathrm{mV}=20 \times 10^{-3} \mathrm{~V}$
The $V_i$ is distributed across $R$ and $C$. by $V_R$ and $V_C$, as phasor diagram.
Reactance of capacitor,
$$
X_C=\frac{1}{\omega C}=\frac{1}{10^5 \times 10^{-8}} \Omega
$$
$$
=10^3 \Omega=1000 \Omega
$$
$\therefore$ Now, impedance of $R$ - $C$ series circuit.
$$
Z=\sqrt{R^2+X_C^2}=\sqrt{(1000)^2+(1000)^2}=1000 \sqrt{2} \Omega
$$
Now, voltage across capacitor,
$$
\begin{aligned}
V_C & =V_0=\frac{X_C}{Z} V_{\max }=\frac{1000}{1000 \sqrt{2}} \times 20 \mathrm{mV} \\
& =\sqrt{2} \times 10 \mathrm{mV} \\
& =14.14 \mathrm{mV}
\end{aligned}
$$
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