A capacitor of capacitance 2 μ F is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz . If the current flowing in the circuit is 2 ~mA , the voltage across the capacitor will be

Correct answer: C. 79.5 ~V 50. aligned & X_C=(1)/(ω C)=(1)/(2 π f C) & = 1 2 × 3.14 × 1 × 10^3 × 2 × 10^ -6 =(10^3)/(4 × 3.14) aligned Voltage, V=i × X_C=2 × 10^ -3 × 10^ -3 4 × 3.14 =0.16 ~V

AIIMS 2003

Exam: AIIMS
Year: 2003
Subject: Physics
Chapter: Alternating Current

AIIMS 2003 · Physics · Alternating Current

A capacitor of capacitance 2 μ F is connected in the tank circuit...

A capacitor of capacitance

2 \mu \mathrm{F}

is connected in the tank circuit of an oscillator oscillating with a frequency of

1 \mathrm{kHz}

. If the current flowing in the circuit is

2 \mathrm{~mA}

, the voltage across the capacitor will be

Options

A $0.16 \mathrm{~V}$
B $0.32 \mathrm{~V}$
C $79.5 \mathrm{~V}$
D $159 \mathrm{~V}$ .

Correct answer

C. $79.5 \mathrm{~V}$

Step-by-step solution

50.

\begin{aligned}& X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\\ & =\frac{1}{2 \times 3.14 \times 1 \times 10^3 \times 2 \times 10^{-6}}=\frac{10^3}{4 \times 3.14} \end{aligned}

Voltage,

V=i \times X_C=2 \times 10^{-3} \times \frac{10^{-3}}{4 \times 3.14}=0.16 \mathrm{~V}

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