Figure shows a typical circuit for a low-pass filter. An AC input…

Question

Figure shows a typical circuit for a low$-$pass filter. An $AC$ input $V_{i }= 10mV$ is applied at the left end and the output $V_0$ is received at the right end. Find the output voltage for$ ν = 10\ kHz, 1.0\ MHz$ and $10.0\ MHz.$ Note that as the frequency is increased the output decreases and, hence, the name low$-$pass filter.

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Answer

$\text{V}_1=10\times10^{-3}\text{V}$ $\text{R}=1\times10^3\Omega$ $\text{C}=10\times10^{-9}\text{F}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10\times10^3\times10\times10^{-9}}$
$=\frac{1}{2\pi\times10^{-4}}$
$=\frac{10^4}{2\pi}=\frac{5000}{\pi}$
$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$
$=\sqrt{\big(1\times10^3\big)^2+\Big(\frac{5000}{\pi}\Big)^2}$
$=\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}$
$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$
$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10^5\times10\times10^{-9}}$
$=\frac{1}{2\pi\times10^{-3}}=\frac{10^3}{2\pi}=\frac{500}{\pi}$
$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$
$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{500}{\pi}\Big)^2}$
$=\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}$
$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$
$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}$
$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}\times\frac{500}{\pi}$
$=1.6124\text{V}\approx1.6\text{mV}$

$\text{f}=1\text{MHz}=10^6\text{Hz}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$
$=\frac{1}{2\pi\times10^6\times10\times10^{-9}}$
$=\frac{1}{2\pi\times10^{-2}}=\frac{10^2}{2\pi}=\frac{50}{\pi}$
$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$
$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{50}{\pi}\Big)^2}$
$=\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}$
$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$
$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}$
$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}\times\frac{50}{\pi}$
$\approx1.16\mu\text{V}$

$\text{f}=10\text{MHz}=10^7\text{Hz}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$
$=\frac{1}{2\pi\times10^7\times10\times10^{-9}}$
$=\frac{1}{2\pi\times10^{-1}}=\frac{10}{2\pi}=\frac{5}{\pi}$
$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$
$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{5}{\pi}\Big)^2}$
$=\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}$
$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$
$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}$
$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}\times\frac{5}{\pi}$
$\approx16\mu\text{V}$