## anonymous one year ago 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?

1. anonymous

the answer should be 5.0 V

2. anonymous

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

3. KenLJW

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

4. anonymous

$V_r = \frac{V_m}{2fRC}$ 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.

5. KenLJW

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

6. anonymous

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