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MHT CET 202624 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
24 April
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

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

An alternating voltage is given as, E=100 2 100t This voltage E is...

An alternating voltage is given as, E=1002sin⁡100tE=100\sqrt{2}\sin 100t
This voltage E is applied to a capacitor of 1 μF1\ \mu\text{F}. The current reading in an a.c. ammeter in mA is

Options

  1. A1010
  2. B2020
  3. C4040
  4. D8080

Correct answer

A. 1010

Step-by-step solution

Given alternating voltage is E=1002sin⁡(100t)E = 100\sqrt{2}\sin(100t)

Comparing with the standard equation E=E0sin⁡(ωt)E = E_0\sin(\omega t), we get:

Peak voltage, E0=1002E_0 = 100\sqrt{2} V

Angular frequency, ω=100\omega = 100 rad/s

The rms voltage is Erms=E02=10022=100E_{rms} = \dfrac{E_0}{\sqrt{2}} = \dfrac{100\sqrt{2}}{\sqrt{2}} = 100 V

Capacitive reactance, XC=1ωC=1100×10−6=104 ΩX_C = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 10^{-6}} = 10^4\ \Omega

The reading of an a.c. ammeter is the rms current, which is given by:

Irms=ErmsXC=100104=10−2I_{rms} = \dfrac{E_{rms}}{X_C} = \dfrac{100}{10^4} = 10^{-2} A

Converting to milliamperes:

Irms=10−2×1000I_{rms} = 10^{-2} \times 1000 mA =10= 10 mA

Answer: 1010
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