Q22. An a.c. source of 200 V is connected through a diode 'D' to a capacitor as shown in figure. Fin

Q: What is the correct answer for CUET UG Physics Q22?

The correct answer is c. The AC source of 200 V refers to the RMS (Root Mean Square) voltage (V_rms). The diode 'D' acts as a half-wave rectifier when connected in series with a capacitor. During the positive half-cycle, the diode is forward-biased, and the capacitor charges up to the peak voltage (V_peak) of the AC source. Once charged, the capacitor holds this peak voltage. The relationship between peak voltage and RMS

Q: Q22. An a.c. source of 200 V is connected through a diode 'D' to a capacitor as shown in figure. Find the maximum potential difference across the capacitor: (a) 100 V (b) 200 V (c) 283 V (d) 310 V

The correct answer is c. The AC source of 200 V refers to the RMS (Root Mean Square) voltage (V_rms). The diode 'D' acts as a half-wave rectifier when connected in series with a capacitor. During the positive half-cycle, the diode is forward-biased, and the capacitor charges up to the peak voltage (V_peak) of the AC source. Once charged, the capacitor holds this peak voltage. The relationship between peak voltage and RMS

Q: What are the options for this CUET UG Physics question?

A. 100 V · B. 200 V · C. 283 V · D. 310 V

Question

Q22. An a.c. source of 200 V is connected through a diode 'D' to a capacitor as shown in figure. Find the maximum potential difference across the capacitor:
(a) 100 V
(b) 200 V
(c) 283 V
(d) 310 V

FirstInTest · Accepted Answer

The AC source of 200 V refers to the RMS (Root Mean Square) voltage (V_rms). The diode 'D' acts as a half-wave rectifier when connected in series with a capacitor. During the positive half-cycle, the diode is forward-biased, and the capacitor charges up to the peak voltage (V_peak) of the AC source. Once charged, the capacitor holds this peak voltage. The relationship between peak voltage and RMS voltage for a sinusoidal AC source is V_peak = V_rms * √2.Given V_rms = 200 V.V_peak = 200 V * √2 ≈ 200 V * 1.414 = 282.8 V.Rounding to the nearest whole number, the maximum potential difference across the capacitor will be 283 V.

Answer

100 V

Answer

200 V

Answer

283 V

Answer

310 V

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