A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is `1000 Omega`. The forward resistance `R_(f)` ideal diode is `10 Omega`. Calculate.
(i) Peak, average and rms values of load current
(ii) d.c power output
(ii) a.c power input
(iv) % Rectifier efficiency
(v) Ripple factor.

Explanation

To solve the problem step by step, we will calculate the peak, average, and RMS values of the load current, followed by the DC power output, AC power input, percentage rectifier efficiency, and the ripple factor.

### Step 1: Calculate the Peak Current (I_peak)
The peak current (I_peak) can be calculated using the formula:
\[ I_{peak} = \frac{V_{m}}{R_{eq}} \]
Where:
- \( V_{m} = 25 \, \text{V} \) (peak voltage)
- \( R_{eq} = R_{f} + R_{L} = 10 \, \Omega + 1000 \, \Omega = 1010 \, \Omega \)
...