JEE Advanced 2017 Paper 2 · Physics · Electromagnetic Induction and Alternating Current · medium

A series circuit has a resistor R=30 Ω, an inductor L=0.80 H and a capacitor C=250 muF, connected to a source of rms voltage 200 V and angular…

One or more options correct

A series circuit has a resistor

R=30\ \Omega

, an inductor

L=0.80\ \mathrm{H}

and a capacitor

C=250\ \mu\mathrm{F}

, connected to a source of rms voltage

200\ \mathrm{V}

and angular frequency

100\ \mathrm{rad\,s^{-1}}

. Choose the correct statement(s).

A. The rms current is $4.0\ \mathrm{A}$ .
B. The power factor of the circuit is $0.8$ .
C. The rms voltage across the capacitor is $200\ \mathrm{V}$ .
D. If the frequency is varied to the resonant value with the source rms voltage unchanged, the rms current becomes $6.67\ \mathrm{A}$ .

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Answer: A, D

The reactances are

X_L=\omega L=80\ \Omega

and

X_C=1/(\omega C)=1/(100\times250\times10^{-6})=40\ \Omega

. The impedance is

Z=\sqrt{30^2+(80-40)^2}=50\ \Omega

. (A)

I=200/50=4.0\ \mathrm{A}

: true. (B)

\cos\phi=R/Z=30/50=0.6

, not

0.8

: false. (C)

V_C=IX_C=4\times40=160\ \mathrm{V}

, not

200\ \mathrm{V}

: false. (D) At resonance

X_L=X_C

, so

Z=R=30\ \Omega

and

I=200/30=6.67\ \mathrm{A}

: true. The average power in the original circuit is

I^2R=480\ \mathrm{W}

. The voltage across the inductor is

IX_L=320\ \mathrm{V}

, larger than the supply, because inductor and capacitor voltages are in antiphase and partly cancel. The phase angle is

\tan^{-1}(40/30)\approx53^\circ

, with voltage leading the current since

X_L>X_C

Original question written in the JEE Advanced 2017 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake