pls help me in this.
A 60 V peak full-wave rectified voltage is applied to a capacitor-input filter.?
If f= 120Hz, RL = 10Kohms, and C= 10µF. what is the ripple voltage?

Question

pls help me in this.
A 60 V peak full-wave rectified voltage is applied to a capacitor-input filter.?
If f= 120Hz, RL = 10Kohms, and C= 10µF. what is the ripple voltage?

anonymous · Answer

the answer should be 5.0 V

anonymous · Answer

i've tried to solve but i cant get the right answer

kenljw · Answer

T = 1/f  :  TC = RL C
ripple = (T/TC) Vp
This gives you 5 V ripple, leave you to derive theory, please show

anonymous · Answer

$$V_r = \frac{V_m}{2fRC} $$
substitution of the given values leaves me with 2.5 V.
if it was half wave it would have been 5 V. I don't understand why it is 5.

kenljw · Answer

I suspect the f in your equations is the frequency unretified, 60 HZ
The 2 signifies fullwave.

anonymous · Answer

It is 5 v. RF=1/4*1.7328fcr and Vdc=Vrms/rf