Physics Circle home Student Login

Home›Chapters›Alternating Current PYQs›Pure R, L and C Circuits›Question 11

MHT CET 2026 (12 May, Shift 2) · Pure R, L and C Circuits

Pure R, L and C Circuits question from MHT CET 2026 (12 May, Shift 2)

Q11MHT CET 2026 · 12 MAY · SHIFT 2Easy

Pure R, L and C Circuits

A capacitor of 10 μF10\ \mu F has reactance 2×10−3 Ω2 \times 10^{-3}\ \Omega. The corresponding frequency is

Show answer and solution

Answer: (a)

  1. XC=12πfC⇒f=12πCXCX_C = \frac{1}{2\pi fC} \Rightarrow f = \frac{1}{2\pi CX_C}
  2. f=12π×10×10−6×2×10−3=14π×10−8=25×106πf = \frac{1}{2\pi \times 10\times10^{-6} \times 2\times10^{-3}} = \frac{1}{4\pi\times10^{-8}} = \frac{25\times10^{6}}{\pi} Hz
  3. So the answer is (a) 25π\frac{25}{\pi} MHz.
Report a mistake

More questions on Pure R, L and C Circuits