An alternating voltage given as, E = 100 2 (100t) volt is applied to a capacitor of 1\ . The current reading of the ammeter will be equal to

Correct answer: B. 10\ mA Given alternating voltage E = 100 2 (100t) V. Peak voltage E_ 0 = 100 2 V. RMS voltage E_ rms = E_ 0 2 = 100 V. Angular frequency ω = 100 rad/s. Capacitance C = 1\ = 10^ -6 F. Capacitive reactance X_ C = (1)/(ω C) = 1 100 × 10^ -6 = 10^ 4 \ . The ammeter reads the RMS value of the current. I_ rms = E_ rms X_ C = 100 10^ 4 = 10^ -2 A = 10 mA. Answer: 10\ mA

MHT CET 202624 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 24 April
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

MHT CET 2026 · 24 April · Evening Shift · Physics · Alternating Current

An alternating voltage given as, E = 100 2 (100t) volt is applied...

An alternating voltage given as,

E = 100\sqrt{2}\sin(100t)

volt is applied to a capacitor of

1\ \mu\text{F}

. The current reading of the ammeter will be equal to

Options

A $20\ \text{mA}$
B $10\ \text{mA}$
C $40\ \text{mA}$
D $80\ \text{mA}$

Correct answer

B. $10\ \text{mA}$

Step-by-step solution

Given alternating voltage

E = 100\sqrt{2}\sin(100t)

V.

Peak voltage

E_{0} = 100\sqrt{2}

V.

RMS voltage

E_{\text{rms}} = \dfrac{E_{0}}{\sqrt{2}} = 100

V.

Angular frequency

\omega = 100

rad/s.

Capacitance

C = 1\ \mu\text{F} = 10^{-6}

F.

Capacitive reactance

X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 10^{-6}} = 10^{4}\ \Omega

.

The ammeter reads the RMS value of the current.

I_{\text{rms}} = \dfrac{E_{\text{rms}}}{X_{C}} = \dfrac{100}{10^{4}} = 10^{-2}

= 10

mA.

Answer:

10\ \text{mA}

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