A 10 μ mathrm{F} capacitor is connected to a 210 mathrm{~V}, 50 mathrm{~Hz} source as shown in figu

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7.Alternating Current

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A $10 \mu \mathrm{F}$ capacitor is connected to a $210 \mathrm{~V}, 50 \mathrm{~Hz}$ source as shown in figure. The peak current in the circuit is nearly $(\pi=3.14)$ :

A

$0.93 \mathrm{~A}$

B

$1.20 \mathrm{~A}$

C

$0.35 \mathrm{~A}$

D

$0.58 \mathrm{~A}$

(NEET-2024)

Solution

$\text { Capacitive Reactance } X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}}$

$=\frac{1000}{3.14}$

$V_{\mathrm{ms}}=210 \mathrm{~V}$

$i_{\mathrm{ms}}=\frac{V_{\mathrm{ms}}}{X_C}=\frac{210}{X_C}$

$\text { Peak current }=\sqrt{2} i_{\text {mms }}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932$

$=0.93 \mathrm{~A}$

Standard 12

Physics

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