SSC JE Electrical Engineering 2014 Set 1 Paper

Question : 196 of 200

Marks: +1, -0

An AC supply of 230 V is applied to half-wave rectifier through transformer of turns ratio 10 : 1 as shown in figure. Determine the peak inverse voltage across the diode.

37.6 V
32.5 V
23.0 V
14.54 V

Validate

Solution:

V_2 = V_1, N_2 / N_1

= 230×1/10= 23V

\PIV =√ 2 ×23

= 32.52 Volts.

Go to Question:

Prev Question

Next Question