Assignment task regarding the voltage change & wave form when changing a capasitor

[pdgarside]

Please can anyone help with the following task.

I need to know the following

1.Why the output voltage and D.C. voltage has changed as the value of the capacitor has been reduced?

2.The changes in the output waveform?

Here is the task.

Task 7 Enter the above circuit into Multisim, initially set the value of the capacitor at 10,000μF and connect a 16Vrms, 50Hz ac sine-wave generator to the input of the bridge rectifier.
a. Obtain a print-out of the output waveform and the D.C. voltage for a capacitor value of 10,000μF capacitor.

b. Replace the 10,000μF capacitor with a 4700μF capacitor and obtain a print-out of the output waveform and d.c. voltage.

c. Replace the 4700μF capacitor with a 1500μF capacitor and obtain a print-out of the output waveform and D.C. voltage.

d. Analyse the effect of varying the value of the capacitor on the output waveform and D.C. voltage obtained from the circuit. Explain why the output voltage and D.C. voltage has changed as the value of the capacitor has been reduced.

e. Repeat a. to c. above with R1 set to 150 Ω and explain the changes in the output waveform.

 

[crutschow]

You need to post the circuit.

 

[pdgarside]

Carl

I have attached the circuit diagram.

Regards
Paul G

 

[crutschow]

Do you understand how a rectifier and filter work to rectify an AC waveform? The rectifier allows current to pass only in one direction, and the capacitor filters the rectified waveform to give a smoother DC level. Between the rectified peaks the current to the load is provided by the capacitor to the load thus the capacitor somewhat discharges until recharged by the next input peak of the AC. The rate of discharge is determined by the value of the capacitance and the load.

Does that help you understand what is happening so you can answer the questions?

 

[pdgarside]

Thanks

I do understand how the rectifier works, is it fair to say that the smaller the capacitor the smoother the waveform? Due to the capacitor charging quicker.

Is it also correct to say that the voltage drops when fitting a smaller capacitor due to the volt meter reading an average, therefore the peaks and the dips are reduced due to the capacitor charging up when higher peak and discharging at the dip.

With appreciation for assisting in my understanding.

Regards, Paul

 

[crutschow]

It's true a smaller capacitor will charge somewhat more rapidly but that is a small effect due to the low impedance of the transformer and diodes. It however discharges more rapidly between the peaks due to the load, making the ripple larger. Thus a larger capacitor reduces the ripple.

If you do the multsim simulation you will see how the ripple varies with load and capacitance. Look at it with the oscilloscope, not a voltmeter.

Edit: Yes, the voltmeter will read a lower voltage with a smaller cap since it is measuring the average (or perhaps the RMS) value