- #1

Jupiter_10

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## Homework Statement

A Transformer with a stepdown ratio of 10:1 is supplied with a 230 RMS voltage at 50hz connected to a four diode full wave bridge rectifier and smoothing circuit. If the smoothing capacitor value is 1000uF and the load resistance R

_{L}is 25 ohms, find:

the secondary winding

**rms**and

**peak**voltage,

the

**peak**voltage across the capacitor

the

**peak to**and

**peak****rms**ripple voltages

the load

**dc**voltage and current

## Homework Equations

## The Attempt at a Solution

Ok the peak secondary winding voltage is 23V +1.4V(assuming 2 diodes at 0.7v each)=24.4V

The secondary rms is 24.4/sqrt2= 17.253V

Peak capacitor voltage V

_{CP}is secondary rms - 1.4V = 17.253-1.4= 15.853V

Peak to Peak ripple voltage is V

_{RPP}= (1/(2fCR

_{L}))(V

_{CP}-(V

_{RPP}/2). Solving for V

_{RPP}, V

_{RPP}=0.4(15.853-V

_{RPP}/2),

V

_{RPP}=5.28V

RMS ripple voltage V

_{R(RMS)}is 5.28/(2*3sqrt)= 1.52V

DC load voltage is V

_{CP}- (V

_{RPP}/2) = 15.853 - (5.28/2)= 13.213V

Load Current I

_{L}is V

_{L}/R

_{L}= 13.213/25= 0.528A

Can someone verify the methods I have used.