A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω 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)

The correct answer is 800.

Step-by-Step Explanation:

1. Identify the given parameters from the problem statement:
- Type of rectifier: Single-phase full-bridge diode rectifier
- Load: Purely resistive load, $R=50\Omega$
- AC Supply Voltage (RMS value): $V_s=200\text{V}$
- Supply Frequency: $f=50\text{Hz}$
- Diode characteristics: Ideal (zero forward voltage drop and zero resistance)

2. Determine the RMS output voltage across the load:
For a single-phase full-bridge diode rectifier operating with a resistive load, the output voltage waveform consists of rectified half-cycles of the input sine wave. Since the diodes are ideal, the absolute value of the input voltage is transferred to the load. Therefore, the RMS value of the output voltage ($V_{0\left(\text{rms}\right)}$) is exactly equal to the RMS value of the input supply voltage ($V_s$):
$V_{0\left(\text{rms}\right)}=V_s=200\text{V}$

3. Calculate the active power drawn by the load:
The active power ($P$) dissipated in a purely resistive load is determined by the RMS value of the voltage across it and the resistance:
$P=\frac{V_{0\left(\text{rms}\right)}^2}{R}$

Substituting the values:
$P=\frac{{200}^2}{50}$
$P=\frac{40000}{50}=800\text{W}$

Thus, the active power drawn by the load is 800 W.