AC alternating emf mathrm{E}=110 √{2} sin 100 mathrm{t} volt is applied to a capacitor of 2 μ mat

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7.Alternating Current

hard

$AC$ alternating emf $\mathrm{E}=110 \sqrt{2} \sin 100 \mathrm{t}$ volt is applied to a capacitor of $2 \mu \mathrm{F}$, the rms value of current in the circuit is.. . . . . .$\mathrm{mA}$.

A

$22$

B

$20$

C

$25$

D

$30$

(JEE MAIN-2024)

Solution

$C=2 \mu f ; \quad E=110 \sqrt{2} \sin (100 \mathrm{t})$

$\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega c}=\frac{1}{100 \times 2 \times 10^6}$

$=\frac{10000}{2}=5000 \Omega$

$\mathrm{i}_0=\frac{110 \sqrt{2}}{5000}$

$\mathrm{i}_{\mathrm{rms}}=\frac{110 \sqrt{2}}{5000 \sqrt{2}}$

$=\frac{110}{5} \mathrm{~mA}$

$=22 \mathrm{~mA}$

Standard 12

Physics

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