+91-8076753736 [email protected]

-->

Wave Optics

Question

Wired Faculty App

An a.c. supply of 230 V is applied to a half-wave rectifier circuit through a transformer of turn ratio 10: 1. Find the output d.c. voltage. Assume the diode to be ideal.

Solution

Given,

R.M.S. primary voltage, = 230 V
Primary to secondary turn ratio, n_p/n_s = 10
Therefore,
Maximum primary voltage,

V_{pm} = \sqrt{2} \times V_{rms} = \sqrt{2} \times 230 = 325.3 V

Maximum secondary voltage,

V_{s m} = V_{p m} \times \frac{n_{s}}{n_{p}} = 325.3 \times \frac{1}{10} = 32.53 V

Half-wave rectified current,

I_{d . c .} = \frac{I_{0}}{π}

∴

Output d.c. voltage =

I_{d . c} \times R_{L}

"

Some More Questions From Wave Optics Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation