Physics Circle home Student Login

Home›Chapters›Alternating Current PYQs›Pure R, L and C Circuits›Question 28

NEET 2024 · Pure R, L and C Circuits

Pure R, L and C Circuits question from NEET 2024

Q28NEET 2024Easy

Pure R, L and C Circuits

A 10 μF10\ \mu F capacitor is connected to a 210 V,50 Hz210\text{ V}, 50\text{ Hz} source as shown in figure. The peak current in the circuit is nearly (π=3.14)(\pi = 3.14):

Figure

Show answer and solution

Answer: (b)

  1. XC=12πfC=12×3.14×50×10×10−6≈318.5 ΩX_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \approx 318.5\ \Omega
  2. 210 V is the rms value: Irms=210318.5≈0.66I_{rms} = \frac{210}{318.5} \approx 0.66 A.
  3. I0=2 Irms≈1.414×0.66≈0.93I_0 = \sqrt{2}\, I_{rms} \approx 1.414 \times 0.66 \approx 0.93 A
  4. So the answer is (b) 0.93 A.
Report a mistake

More questions on Pure R, L and C Circuits