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

The average current will be "10.12 mA".

Step-by-step explanation:

In the transformer:

⇒
`(V1)/(V2)=(N1)/(N2)`

where,

V1 = 120

N2 = 1

N1 = 10

So that,

⇒
`V2=V1* (N2)/(N1)`

On putting the values,

`=120* (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√(2)`

`=15.9V`

⇒
`Vaverage=2Vmax \ \pi`

`=10.12 \ V`

The amount conducted by each diode is 50%.

⇒
`Iaverage=(Vaverage)/(R)`

`=(10.12)/(1)`

`=10.12 \ mA`