# Bridge Full Wave Rectifier Peak Voltages and Capacitcane Value

eximius
Note: This is not a homework or coursework, simply revision for an exam. Thanks

## Homework Statement

[PLAIN]http://img194.imageshack.us/img194/223/question2z.jpg [Broken]

## Homework Equations

Peak voltage = RMS voltage x root2
r = T/(2RlC)
T = 1/f

## The Attempt at a Solution

part d)

Peak Supply Voltage = 240 x root2 = 339.41V
Peak Va = 339.41 x 0.1 = 33.941V

part e)

r=10%=0.1 T=1/50 Rl=200ohms

.:.

r = T/(2RlC)

0.1 = 0.02/400C
40C = 0.02
C = 5x10-4F

## The Attempt at a Solution

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d) You got 0.7V voltage drop over each diode. How many diodes do the current go trough per half period (one pulse)?

eximius
d) You got 0.7V voltage drop over each diode. How many diodes do the current go trough per half period (one pulse)?

Ahhh, so because it goes through 2 diodes there's a 1.4 V drop?

You tell me...

Draw a simple diagram for one half period ( (?) diodes, eqv. voltage source, and load). 'Kirchhoff voltage law' will tell you the answer.

eximius
During one half period there is current flow through 2 of the diodes. In the positive half cycle there's current flow through D2 and D3. But looking at current flow diagrams has made me think that the Voltage drop at Vb is just 0.7 V. Because from Va to Vb there's only one diode.

I'm just uncertain. And I have no clue when it comes to KVL, haven't studied it yet.