# Homework Help: D.C voltage and capacitors

1. Apr 16, 2012

### jsmith613

1. The problem statement, all variables and given/known data

see attachment

2. Relevant equations

3. The attempt at a solution

here are the MS points :

Capacitor stores charge/charges up (1)
(If voltage is constant) capacitor doesn’t discharge (1)

I don't understand how the answer makes sense. The question asks how placing a capacitor causes the power source to produce a constant voltage. but how does putting the capacitor there suddenly make the power source decide its voltage?

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• ###### CapacitorQ.png
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2. Apr 16, 2012

### Staff: Mentor

Although not stated in the problem, the power supply will incorporate some device that prevents current from flowing backwards into the supply. This is the 'rectifier' portion of the supply which takes an AC source voltage and turns it into a pulsating DC voltage as depicted on your graph. Rectifiers are usually comprised of diodes in some arrangement.

What will happen to the voltage across the capacitor if current is not allowed to flow back into the supply?

3. Apr 16, 2012

### jsmith613

I don't know, sorry :(

4. Apr 16, 2012

### Staff: Mentor

Suppose that the capacitor starts out uncharged and the supply voltage curve begins at a point where it happens to be at 0V. What will happen to the voltage on the capacitor as time progresses and the supply voltage increases towards its next peak?

Last edited: Apr 16, 2012
5. Apr 16, 2012

### jsmith613

well I guess the capacitor will charge up to a maximum and then discharge as the power supply reaches a minimum again?

6. Apr 16, 2012

### Staff: Mentor

I didn't mention the next part of the curve, so no

But the capacitor voltage will follow that of the supply up to the peak of the supply voltage curve as you stated. What direction is the current flowing between the supply and the capacitor during this period?

7. Apr 16, 2012

### jsmith613

current is flowing towards the capacitor (from the supply)

8. Apr 16, 2012

### Staff: Mentor

Correct. Now time moves on and the supply voltage begins its descent from the peak. What direction will the current want to flow?

9. Apr 16, 2012

### jsmith613

from the capacitor toward the supply.

10. Apr 16, 2012

### Staff: Mentor

Correct, the capacitor would like to discharge in order to follow the source voltage. But as I stated, it's prevented from doing so because the supply contains components that prevent current from flowing back into it. So what happens to the capacitor charge and voltage?

11. Apr 16, 2012

### jsmith613

well both stay constant

12. Apr 16, 2012

### Staff: Mentor

Right. And a constant voltage is just DC by another name

To summarize:

Note that this assumes that there is no load connected across the capacitor (some device you want to power). If there is a load, the capacitor will supply current to it during the times that the source voltage is below the peak. This in turn will make the capacitor voltage "sag" a bit while it discharges into the load, but it will get topped up again when the supply voltage catches up. This is known as "ripple" on the average DC supplied by the circuit. So, not perfect DC but much better than the "raw" supply voltage. Bigger capacitors are better at smoothing the output.

#### Attached Files:

• ###### Fig1.gif
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13. Apr 16, 2012

### jsmith613

thanks so much for this :)
I get it now

just as a last question if the capacitor was larger the "sag" would be smaller. why?

14. Apr 16, 2012

### Staff: Mentor

Assume that the load happens to be a resistor. When the supply voltage is less than that of the capacitor voltage, the capacitor and load resistor are essentially "alone". What happens to the capacitor voltage?

15. Apr 16, 2012

### jsmith613

the capacitor will discharge through the resistor

16. Apr 16, 2012

### Staff: Mentor

Yes. What governs the time it takes to discharge, or if you will, the rate at which it discharges? When will it stop discharging and start charging up again? It might help if you were to sketch the power supply voltage and a discharge curve on the same graph.

17. Apr 16, 2012

### jsmith613

well
Q = Q0e-t/RC

where RC = "discharge" constant....oh right...larger capacitance = larger time to discharge

yes

thanks again

18. Apr 18, 2012

### jsmith613

sorry to bother you again but I noticed that the ms said:

Capacitor stores charge/charges up (1)
(If voltage is constant) capacitor doesn’t discharge (1)

the second statment "If voltage is constant" capacitor doesn't discharge
HOW is this relevant to our situation...the voltage is NOT constant?

19. Apr 18, 2012

### Staff: Mentor

The statements, while true, are at best incomplete descriptions of capacitor behavior. Of course capacitors change voltage as they charge or discharge: V = Q/C.

Perhaps the second statement would have been better phrased as "if the capacitor isn't charging or discharging it's voltage is constant".

I think the idea was to get across the point that if the capacitor is charged up by the supply and is not subsequently discharged by the supply when the supply voltage drops, then the capacitor will maintain the last peak voltage that the supply provided.

The capacitor serves as a "reservoir" of charge for periods when the power supply itself is not able to provide charge (current) to a load.