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

Question

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 (A) 0.16 V (B) 0.32 V (C) 79.5 V (D) 159 V

Tardigrade Admin · Accepted Answer

Reactance of capacitor or capacitive reactance is given by

X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}

where

f

is frequency of the alternating current.
Given,

f = 1 \times 1 0^{3} Hz, C = 2 \times 1 0^{- 6} F

\Rightarrow X_{C} = \frac{1}{2 \times 3.14 \times 1 0 ^{3} \times 2 \times 1 0 ^{- 6}}

= \frac{1 0 ^{3}}{4 \times 3.14}

Hence, voltage across the capacitor is

V = i X_{C} = 2 \times 1 0^{- 3} \times \frac{1 0 ^{3}}{4 \times 3.14} = 0.16 V

Q. A capacitor of capacitance $2 μ F$ is connected in the tank circuit of an oscillator oscillating with a frequency of $1 k Hz$ . If the current flowing in the circuit is $2 m A$ , the voltage across the capacitor will be

$0.16 V$

$0.32 V$

$79.5 V$

$159 V$

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

0.16V

0.32V

79.5V

159V

Reactance of capacitor or capacitive reactance is given by XC​=ωC1​=2πfC1​ where f is frequency of the alternating current. Given, f=1×103Hz,C=2×10−6F ⇒XC​=2×3.14×103×2×10−61​ =4×3.14103​ Hence, voltage across the capacitor is V=iXC​=2×10−3×4×3.14103​=0.16V

Q. A capacitor of capacitance $2 μ F$ is connected in the tank circuit of an oscillator oscillating with a frequency of $1 k Hz$ . If the current flowing in the circuit is $2 m A$ , the voltage across the capacitor will be

$0.16 V$

$0.32 V$

$79.5 V$

$159 V$