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Alternating Current — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsAlternating Current5 Marks
Question
A capacitor of capacitance $10\mu\text{F}$ is connected to an oscillator with output voltage $\in=(10\text{V})\sin\omega\text{t}.$ Find the peak currents in the circuit for $\omega=10\text{s}^{-1},100\text{s}^{-1},500\text{s}^{-1},1000\text{s}^{-1}.$
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Answer

$\text{C}=10\mu\text{F}=10\times10^{-6}\text{F}=10^{-5}\text{F}$$\text{E}=(10\text{V})=\sin\omega\text{t}$
  1. $\text{I}=\frac{\text{E}_0}{\text{X}_\text{C}}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$
$=\frac{10}{\Big(\frac{1}{10\times10^{-5}}\Big)}=1\times10^{-3}\text{A}$
  1. $\omega=100\text{s}^{-1}$
$\text{I}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$
$=\frac{10}{\Big(\frac{1}{100\times10^{-5}}\Big)}=1\times10^{-2}\text{A}=0.01\text{A}$
  1. $\omega=500\text{s}^{-1}$
$\text{I}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$
$=\frac{10}{\Big(\frac{1}{500\times10^{-5}}\Big)}=5\times10^{-2}\text{A}=0.05\text{A}$
  1. $\omega=1000\text{s}^{-1}$
$\text{I}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$
$=\frac{10}{\Big(\frac{1}{1000\times10^{-5}}\Big)}=1\times10^{-1}\text{A}=0.1\text{A}$

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