Chapter 29, Problem 27P

(a)

To determine

The average current in a half wave rectifier.

Answer to Problem 27P

  $\text{2}\text{.4}\text{mA}$ .

Explanation of Solution

Given:

RMS voltage, $V_{rms}=120V$

Resistance, $R=25k\Omega$

Formula used:

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

Where,

  $I$ is current.

  $V$ is voltage

  $R$ is resistance

Calculation:

For a half-wave rectifier without a capacitor, the current is zero for half the time.

So, the average current is

  ${\text{I}}_{\text{av}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{V}}_{\text{rms}}}{\text{R}}\text{=}\frac{\text{1}}{\text{2}}\frac{\left(\text{120}\text{V}\right)}{\text{25}\text{kΩ}}\text{=2}\text{.4}\text{mA}$

(b)

To determine

The average current in a full wave rectifier.

Answer to Problem 27P

  $\text{4}\text{.8}\text{mA}$ .

Explanation of Solution

Given:

RMS voltage, $V_{rms}=120V$

Resistance, $R=25k\Omega$

Formula used:

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

Where,

  $I$ is current.

  $V$ is voltage

  $R$ is resistance

Calculation:

For a full-wave rectifier without a capacitor, the current is positive all the time.

So, the average current is

  ${\text{I}}_{\text{av}}\text{=}\frac{{\text{V}}_{\text{rms}}}{\text{R}}\text{=}\frac{\left(\text{120}\text{V}\right)}{\text{25}\text{kΩ}}\text{=4}\text{.8}\text{mA}$