Question

A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 step-down transformer having a single secondary winding. It uses four diodes, each of which can be modeled to have a 0.7-V drop for any current. What is the peak value of the rectified voltage across the load? For what fraction of a cycle does each diode conduct? What is the average voltage across the load? What is the average current through the load?

Answer 1

Answer:

The average current will be "10.12 mA".

Explanation:

In the transformer:

⇒  $\frac{V1}{V2}=\frac{N1}{N2}$

where,

V1 = 120

N2 = 1

N1 = 10

So that,

⇒  $V2=V1\times \frac{N2}{N1}$

On putting the values,

         $=120\times \frac{1}{10}$

         $=12V \ rms$

The full wave rectifier would conducts during positive and negative half.

⇒  $Vmax=12-0.7$

               $=11.3V \ rms$

⇒  $Vpeak=11.3\sqrt{2}$

               $=15.9V$

⇒  $Vaverage=2Vmax \ \pi$

                     $=10.12 \ V$

The amount conducted by each diode is 50%.

⇒  $Iaverage=\frac{Vaverage}{R}$

                    $=\frac{10.12}{1}$

                    $=10.12 \ mA$