IIT-JEE 2012 · Paper 2 · Physics Q15

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PhysicsMultiple correct+4Question 15 of 60

In the given circuit, the AC source has $\omega=100$ rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

A 20 V AC source feeding two parallel branches: one a 100 μF capacitor in series with 100 Ω, the other a 0.5 H inductor in series with 50 Ω; I is the source current.

The current through the circuit,

I

is 0.3 A.

The current through the circuit,

I

0.3\sqrt2

The voltage across 100 Ω resistor

=10\sqrt2

The voltage across 50 Ω resistor = 10 V.

Answer and worked solution

Official answer: C

Capacitor branch. $X_C=\dfrac1{100\times100\times10^{-6}}=100\ \Omega$ , $Z_1=100\sqrt2\ \Omega$ :
$I_1=\frac{20}{100\sqrt2}=\frac{\sqrt2}{10}\text{ A},\qquad V_{100\,\Omega}=100I_1=10\sqrt2\text{ V},$
leading the source by 45°.

Inductor branch. $X_L=100\times0.5=50\ \Omega$ , $Z_2=50\sqrt2\ \Omega$ :
$I_2=\frac{20}{50\sqrt2}=\frac{2\sqrt2}{10}\text{ A},\qquad V_{50\,\Omega}=50I_2=10\sqrt2\text{ V},$
lagging by 45°.

Total current. The branch currents are 90° apart:
$I=\sqrt{I_1^2+I_2^2}=\sqrt{0.02+0.08}=\sqrt{0.1}\approx0.316\text{ A}.$

(A) $I\approx0.32$ A, only approximately 0.3 A.
(B) False.
(C) True.
(D) The 50 Ω resistor has $10\sqrt2$ V. False.

(C) (key: C or AC)

Topics:Reactance and impedance LCR circuits and resonance

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IIT-JEE 2012 · Paper 2, Physics question 15