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Alternating Current — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current3 Marks
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Answer

$\text{V}_1=10\times10^{-3}\text{V}$

$\text{R}=1\times10^3\Omega$

$\text{C}=10\times10^{-9}\text{F}$

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

$=\frac{1}{2\pi\times10\times10^3\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-4}}$

$=\frac{10^4}{2\pi}=\frac{5000}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(1\times10^3\big)^2+\Big(\frac{5000}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}}$

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

$=\frac{1}{2\pi\times10^5\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-3}}=\frac{10^3}{2\pi}=\frac{500}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{500}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}\times\frac{500}{\pi}$

$=1.6124\text{V}\approx1.6\text{mV}$

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

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

$=\frac{1}{2\pi\times10^6\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-2}}=\frac{10^2}{2\pi}=\frac{50}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{50}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}\times\frac{50}{\pi}$

$\approx1.16\mu\text{V}$

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

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

$=\frac{1}{2\pi\times10^7\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-1}}=\frac{10}{2\pi}=\frac{5}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{5}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}\times\frac{5}{\pi}$

$\approx16\mu\text{V}$

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