Chapter 29, Problem 28P

(a)

To determine

The average current in a half wave rectifier.

Answer to Problem 28P

  $\text{38}\text{mA}$ .

Explanation of Solution

Given:

The ac voltage varies from, $0.6V-12.0Vrms$

Resistance, $R=150\Omega =0.150k\Omega$

Formula used:

It is known that $\text{V=IR}$

Where,

  $I$ is current.

  $V$ is voltage

  $R$ is resistance

Calculation:

According to the question, $V_{rms}$ across the resistor is $12.0V-0.6V=11.4V$

So, the average current (for a half wave) is

  ${\text{I}}_{\text{av}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{V}}_{\text{rms}}}{\text{R}}\text{=}\frac{\text{1}}{\text{2}}\frac{\left(\text{11}\text{.4}\text{V}\right)}{\left(0.150\text{kΩ}\right)}\text{=}\text{38}\text{mA}$

(b)

To determine

The average current in a full wave rectifier.

Answer to Problem 28P

  $\text{76}\text{mA}$ .

Explanation of Solution

Given:

The ac voltage varies from, $0.6V-12.0Vrms$

Resistance, $R=150\Omega =0.150k\Omega$

Formula used:

It is known that $\text{V=IR}$

Where,

  $I$ is current.

  $V$ is voltage

  $R$ is resistance

Calculation:

According to the question, $V_{rms}$ across the resistor is $12.0V-0.6V=11.4V$

So, the average current (for a full wave) is

  ${\text{I}}_{\text{av}}\text{=}\frac{{\text{V}}_{\text{rms}}}{\text{R}}\text{=}\frac{\left(\text{11}\text{.4}\text{V}\right)}{\left(0.150\text{kΩ}\right)}\text{=}\text{76}\text{mA}$