A full wave rectifier uses a load resistor of 1500 ohms. Assume the diodes have Rf=10 Ohms and Rr= infinite. The voltage applied to the diodes is 30V with a frequency of 50Hz. Calculate the AC power input.
Correct answer: B. 275.2 mW
- A. 358.98 mW
- B. 275.2 mW
- C. 145.76 mW
- D. 456.78 mW
Explanation
The correct AC power input is calculated using the formula: A.c power input = Irms^2 (Rf + Rr). First, determine the RMS current: Irms = Imax / √2 = Vmax / (Rf + Rr) √2. Substituting the values: Irms = 30 / (1500 + 10) * 1.414 = 13.5 mA. Now, calculate the power input: Power input = (13.5 * 10^-3)^2 * (1500 + 10) = 275.2 mW. The other options are incorrect because they either overestimate or underestimate the AC power input based on the given circuit parameters.
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A p-n junction forms when p-type and n-type semiconductor regions meet, creating a depletion layer, barrier potential and rectifying action. The work covers forward and reverse bias, junction current, diode characteristics, and how a diode converts alternating current into pulsating direct current through half-wave and full-wave rectification.
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