Question

A single-phase full-bridge diode rectifier feeds a resistive load of 50 \(\Omega\) from a 200 V, 50 Hz single-phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _________. (round off to nearest integer)

Hint:

For a resistive load, the active power drawn from the AC supply can be calculated using the formula \( P = \frac{V_{rms}^2}{R} \).

Answer 1

Correct Answer: 795

The active power drawn by a resistive load from an AC supply is given by: \[ P = \frac{V_{rms}^2}{R} \] where: - \( V_{rms} = 200 \, \text{V} \) (rms voltage of the AC supply),
- \( R = 50 \, \Omega \) (resistive load).
Substituting the values: \[ P = \frac{200^2}{50} = \frac{40000}{50} = 800 \, \text{W} \] Thus, the active power drawn by the load is \( 800 \, \text{W} \).