JEE PHYSICS MCQ: ALTERNATING CURRENT

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Analyzing +4 | -1

An AC source producing emf

e=e_0\left[\cos \left(100 \pi \mathrm{s}^{-1}\right) t+\cos \left(500 \pi \mathrm{s}^{-1}\right) t\right]

is connected in series with a capacitor and resistor. The steady state current in the circuit is found to be

I=I_1 \cos \left[\left(100 \pi \mathrm{s}^{-1}\right) t+\phi_1\right]+I_2 \cos \left[\left(500 \pi \mathrm{s}^{-1}\right) t+\phi_2\right]

View answer and explanation

Correct option: C

I_1 < I_2

Option C (

I_1 < I_2

) is correct because the impedance is lower at higher frequencies in a capacitor circuit, leading to a larger current.

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R=300 \, \Omega

C=25 \, \mu \mathrm{F}, \varepsilon_0=50 \mathrm{~V}

v=50 / \pi \, \mathrm{Hz}

R=300 \Omega, L=0.9 \mathrm{~H}, C=2.0

\mu F

\omega=1000 \mathrm{~rad} / \mathrm{sec}

100 \Omega

200 \mathrm{~V}

300 \, \mathrm{rad} / \mathrm{s}

60^{\circ}

60^{\circ}

L-C \cdot R, \mathrm{AC}

L-C-R, \mathrm{AC}

L-C-R

\mathrm{AC}

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\frac{\pi}{4}

C

\frac{\pi}{4}