A full-wave rectifier is designed using a center-tapped transformer and two diodes. The primary winding of the transformer, with N₁ turns, is connected to a 240V (RMS) AC supply, and the secondary winding has N₂ turns. What is the approximate rectified DC output voltage (V) if N₁/N₂ = 15 : 1?

Question

Question

This question was previously asked in

RRB JE CBT 2 (Electronics Engineering) ReExam Official Paper (Held On: 04 Jun, 2025)

Answer 1

Concept:

For a full-wave rectifier with a center-tap transformer, the average DC output voltage is given by:

\( V_{DC} = \frac{2V_{m}}{\pi} \)

where \( V_{m} \) is the maximum secondary voltage.

Calculation:

Given:

Primary RMS voltage = 240 V, turns ratio \( N_1/N_2 = 15:1 \)

Secondary RMS voltage: \( V_{2(RMS)} = \frac{240}{15} = 16~V \)

Secondary peak voltage: \( V_{m} = V_{2(RMS)} \times \sqrt{2} = 16\sqrt{2}~V \)

DC output voltage of full-wave rectifier:

\( V_{DC} = \frac{2V_{m}}{\pi} = \frac{2 \times 16\sqrt{2}}{\pi} = \frac{16\sqrt{2}}{\pi}~V \)

Answer:

The approximate rectified DC output voltage is:\(\frac{16\sqrt{2}}{\pi}~V \)