# Capacitor circuit exam question

1. Nov 26, 2012

### ofeyrpf

1. The problem statement, all variables and given/known data
Attached is the problem. This is the June 2010 AS Physics exam unit 4 question 16 (Edexcel)

2. Relevant equations
See the attachments.

3. The attempt at a solution
Q 16. (a) The voltage is always positive and so although the current must vary it is always positive and so the current is always traveling in the same direction. Therefore the graph shows a dc current. No problem here.
(b) (i) "The capacitor stores charge" is the answer in the marking scheme, how does this answer the question? Does this mean that the voltage is then always constant and doesn't vary?
(ii) E = 1/2 CV^2 this is easy!
(c) (i) The capacitor charges and discharges. The graph appears to show the voltage increase exponentially as would be expected for a capacitor, but it also seems to discharge linearly. But should a capacitor not discharge exponentially? Also is it true that a capacitor in parallel with a power supply will away charge and then discharge continuously? This seems to be a different type of RC circuit where the capacitor and resistor are in parallel not in series.
(ii) The marking scheme suggests using the V=V_0 exp(-t/RC) where t is substituted with 6x10^-3 seconds. How is this read from the graph? The graph seems to repeat itself every 10 ms, taking 4 ms to charge and 6 ms to discharge. I presume this is where the 6 ms comes from but why? The voltage seems to change from 3v to 5V so that R=6x10^-3/(10x10^-6)ln(3/5). C is known (C = 10x10^-6F) and so R can be found.
(iii) Increase the capacitor, this is fine.

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2. Nov 27, 2012

### CWatters

b) Their suggested answer is rather minimal but correct. Suppose the capacitor has voltage V on it when it is connected. If/when the output from the power supply is higher than that voltage charge will flow into the capacitor raising it's voltage. In fact the voltage will follow the waveform upto the peak. The question assumes you recognise the wave form as the output from a bridge rectifier. In which case when the supply voltage falls again the diodes in the rectifier prevent charge in the capacitor flowing back into the power supply. In fact because no load is shown the charge on the capacitor has nowhere to go so. So the net result is the capacitor retains (stores) a constant charge and maintains a constant DC voltage equal to the peak voltage.

Correct.

That's not quite correct. It may look exponential but during that phase it's actually following the SIN curve shape of the power supply voltage.

That depends on the load. If the load were a resistor then yes it would discharge exponentially. If the load was a constant current sink then it would discharge linearly. In addition the ripple voltage is sometimes approximated to a linear discharge to make the sums simpler. This is a reasonable thing to do when the ripple voltage is small relative to the DC voltage. For example if there is no load the ripple is zero and the output is a constant DC - which is linear, so it's reasonable to assume it approaches linear at small loads/low ripple.

If you connected a capacitor to an ideal AC voltage source or an idealvoltage source with a wave form as per the question then the capacitor voltage would charge and discharge and the voltage on it would follow the supply wave form exactly. In other words the capacitor wouldn't do what is required to make the output DC! You can see this on the rising part of the wave form where it follows the SIN curve.

HOWEVER ...

This isn't an ideal voltage source, there are diodes in it which prevent the voltage source sinking current. See above description I gave for b).

3. Nov 27, 2012

### CWatters

Can I suggest you superimpose the original wave form Fig 1 on the one in figure 3. That should make things clearer.

I think that's correct.

Just for info... If you were to approximate it to a linear discharge you could use...

Q=VC
dQ/dt = C dv/dt

dQ/dt = I so

I=Cdv/dt

dv/dt is the slope of the discharge "curve" so from the graph...

dv = 2V
dt = 6 * 10^-3

and

I = 10 * 10^-6 * 2 / 6 * 10^-3
= 3.33mA

R = V/I = 4/(3.33 *10^-3) = 1201 Ohms which is between 1000 and 2000 Ohms.

I used 4V as that is midway between 5V and 3V but it's not very critical.

4. Nov 27, 2012

### ofeyrpf

Hi CWatter,

Thanks for the reply and for looking at this in details. Your help is really appreciated.

(b) I understand now that the capacitor charges to the peak voltage and then the charge has nowhere to go and so the voltage across the capacitor is constant.

(c) I did not know that a capacitor only discharged exponentially if the load was a resistor. When you say a "current sink" does that mean something like the anode of a battery, if it were possible to just hook up to one side of a battery?

Whe you say "ripple voltage" does that mean the max to min voltage range?

Charge can only be discharged into the load, providing the load has a current sink and is not just a resistor. If the load is just a resistor, there would just be a constant voltage across the resistor once the capacitor has charged. So the second diagram shows the voltage across the load and that it follows the original waveform from the power supply, which is altered somewhat by the capacitor.

If I=dv/dt C is equivalent to y = mx, where I is the y axis and C the x axis, but capacitance, C is constant so there would be no change along the x axis.

I understand the rest of the last post but would never have known