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A $100 \mu \mathrm{F}$ capacitor is connected to a $100 \mathrm{~V}$, $50 \mathrm{~Hz}$ AC supply. The rms value of the current is
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Verified Answer
The correct answer is:
$3.14 \mathrm{~A}$
Given, $V_{\mathrm{rms}}=100 \mathrm{~V}, f=50 \mathrm{~Hz}$
$C_1=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}$
Now, reactance of capacitor,
$X_C=\frac{1}{C \omega}=\frac{1}{C \times 2 \pi f}$
$\Rightarrow \quad X_C=\frac{1}{100 \times 10^{-6} \times 2 \pi \times 50}=\frac{100}{\pi} \Omega$
r. m. s value of current in circuit will be
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}$
$=\frac{100}{\frac{100}{\pi}}=\pi \mathrm{A}=3.14 \mathrm{~A}$
$C_1=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}$
Now, reactance of capacitor,
$X_C=\frac{1}{C \omega}=\frac{1}{C \times 2 \pi f}$
$\Rightarrow \quad X_C=\frac{1}{100 \times 10^{-6} \times 2 \pi \times 50}=\frac{100}{\pi} \Omega$
r. m. s value of current in circuit will be
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}$
$=\frac{100}{\frac{100}{\pi}}=\pi \mathrm{A}=3.14 \mathrm{~A}$
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