Question

a half-wave rectifier, the peak value of the

ac voltage across the secondary of the

transformer is 20√2 V. If no filter circuit is used,

the maximum dc voltage across the load will be​

Answer 1

Answer:

In a half-wave rectifier, the peak value of the ac voltage across the secondary of the transformer is 20 V. If no filter circuit is used, the maximum dc voltage across the load will be—

28.28 V

20 V

14.14 V

9 V

Correct Option: C

Vrms =

Vm

√2

Vrms =

20 √2

= 20 V

√2

Answer 2

Answer:

Therefore, the maximum dc voltage across the load will be​ $V_{dc}$ = 9 V

Explanation:

Before solving this question we have to understand what is half-wave rectifier.

A half wave rectifier is a type of rectifier that allows only one half-cycle of an AC voltage waveform to pass through while blocking the other half. Half-wave rectifiers are used to convert AC electricity to DC voltage and are made up of only one diode.

The dc value of a rectified sinusoidal half-wave is $\frac{V_{m} }{\pi }$, where $V_{m}$ is the sinusoidal voltage's peak value.

Calculation:

Given, $V_{m}$ = 20√2 V

So, for a half-wave rectified wave

$V_{dc}$ = $\frac{V_{m} }{\pi }$

$V_{dc}$ = $\frac{20\sqrt{2} }{\pi }$

$V_{dc}$ = $\frac{20\sqrt{2} }{3.14}$

$V_{dc}$ = 9.0077 ≈ 9 V

So,  the maximum dc voltage across the load will be​ $V_{dc}$ = 9 V

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