Physics Circle home Student Login

Home›Chapters›Alternating Current PYQs›Pure R, L and C Circuits›Question 8

MHT CET 2026 (15 May, Shift 1) · Pure R, L and C Circuits

Pure R, L and C Circuits question from MHT CET 2026 (15 May, Shift 1)

Q8MHT CET 2026 · 15 MAY · SHIFT 1Easy

Pure R, L and C Circuits

An alternating voltage V=2002sin⁡(100t)V = 200\sqrt{2} \sin(100t) volt is connected to 1 μF1\ \mu F capacitor through a.c. ammeter. The reading of ammeter is

Show answer and solution

Answer: (d)

  1. XC=1ωC=1100×10−6=104 ΩX_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = 10^4\ \Omega
  2. Vrms=200V_{rms} = 200 V
  3. Irms=200104=20I_{rms} = \frac{200}{10^4} = 20 mA
  4. So the answer is (d) 20 mA.
Report a mistake

More questions on Pure R, L and C Circuits