half-wave rectifier with filter capacitor

Discussion in 'Homework Help' started by notoriusjt2, Oct 3, 2010.

Feb 4, 2010
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0

book states

thats basically 6540 micro farads, but that answer was deemed incorrect when I submitted my assignment. so my question is where did i go wrong?

2. t_n_k AAC Fanatic!

Mar 6, 2009
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783
Your value for R of 288Ω is incorrect because you based this on the RMS source voltage - rather than the estimated DC load voltage.

The approximate DC load voltage will be Vm-Vr/2, where Vr is the peak-to-peak ripple.

Feb 4, 2010
209
0
so the value for Rload would be

= ((169.7-1.5)/2)^2/50
=141.46ohms

is this correct?

4. t_n_k AAC Fanatic!

Mar 6, 2009
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Not quite - I wrote Vm-Vr/2 not (Vm-Vr)/2

So Vdc would be 120*√2-0.75=168.96V

Feb 4, 2010
209
0

so

=(120(1.414))/(50)(570.95)(1.5)
=169.7/42821.25
=.00396297
=3962microF

thats really not even close to any answers on there

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
This was my approach ....

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