- #1

JC2000

- 186

- 16

- Homework Statement
- In a full wave bridge rectifier circuit having 230V, 50Hz at the primary input of a transformer, the peak value of the output voltage without filter is found to be 25 Volts. Calculate the (a) DC value of the output voltage with filter and (b) Value of capacitance needed for a ripple factor of 0.004 at IL= 0.5 mA

- Relevant Equations
- Without filter:

##V_{dc} = \frac{2*V_p}{ pi }##

##V_{rms} = \frac{V_p}{\sqrt{2}}##

With filter :

##V_{dc} = \frac{4fCR_L*V_p}{1 + 4fCR_L} ##

I need help with part (a)... I know that the root-mean-square voltage is the dc-equivalent voltage for an AC waveform and what my book labels "##V_{dc}## is actually the average voltage. Hence I am assuming the question is asking me to find ##V_{rms}## ...

Is my assumption that the root-mean-square voltage and average voltage of a full wave rectifier with filter are the same correct?! If so, solving the problem becomes easy since we can find ##V_{rms}## without filter using the peak voltage, which will remain the same when a filter is introduced to the circuit...

For part (b), since we now know ##V_{rms}## and the load current, we can apply Ohm's law and find out the load resistance. Then use the following equation to find the capacitance :

##Ripple Factor = \frac{1}{4\sqrt{3}fCR_L}##

Is my assumption that the root-mean-square voltage and average voltage of a full wave rectifier with filter are the same correct?! If so, solving the problem becomes easy since we can find ##V_{rms}## without filter using the peak voltage, which will remain the same when a filter is introduced to the circuit...

For part (b), since we now know ##V_{rms}## and the load current, we can apply Ohm's law and find out the load resistance. Then use the following equation to find the capacitance :

##Ripple Factor = \frac{1}{4\sqrt{3}fCR_L}##