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Capacitor

Part (a) of the following question :

IMG_0248.jpg

I think this is the answer (red lines):

Attachment not found


EDIT : For part (c) (i) and (ii)

The average energy dissipated is : 5.2*10^-3 J

The dissipation takes place over 3*10^-3 s

Hence, the rate = 1.73 W ?

Using P=V^2/ R

The average voltage across the resistor is 10 V

100/1.73 = 57.7 ohms
(edited 11 years ago)
Reply 1
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bmqib
13
how'd you get 5.2, the energy dissipitated over 3ms is 7.2-3.2 mJ? the rest should be correct
Original post by bmqib
how'd you get 5.2, the energy dissipitated over 3ms is 7.2-3.2 mJ? the rest should be correct


They've asked for the AVERAGE rate of energy dissipation.
Reply 3
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bmqib
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Original post by Ari Ben Canaan
They've asked for the AVERAGE rate of energy dissipation.


hmm, but while the capacitor is discharging. can you tell me how you got 5.2?
Original post by bmqib
hmm, but while the capacitor is discharging. can you tell me how you got 5.2?


An average would imply summing the two extremes (7.2 and 3.2) and dividing by two to give 5.2.
Reply 5
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bmqib
13
Original post by Ari Ben Canaan
An average would imply summing the two extremes (7.2 and 3.2) and dividing by two to give 5.2.


but wouldn't that give the average energy stored in the capacitor during the cycle. the energy across the resistor is the energy the capacitor supplies to resistor over the given time?
Reply 6
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bmqib
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the capacitor doesn't discharge fully
The capacitor discharges 4mJ of energy into the resistor every 3 seconds.
The average rate of discharge is just this amount divided by the time.

The current in the diode is constant (not rising) during the charging of the capacitor if the slope of the increase in pd across the capacitor is uniform. (I can't tell from the diagram on my screen but it looks fairly straight.)
(edited 11 years ago)
Original post by Stonebridge
The capacitor discharges 4mJ of energy into the resistor every 3 seconds.
The average rate of discharge is just this amount divided by the time.

The current in the diode is constant (not rising) during the charging of the capacitor if the slope of the increase in pd across the capacitor is uniform. (I can't tell from the diagram on my screen but it looks fairly straight.)


Actually, the part of the graph representing discharge of the capacitor is not a straight line. I placed a ruler next to it and it definitely deviates from a straight line.

Also the slope of the increase is not constant. It's decreasing slowly..... Its actually the first quarter of a sine curve.

Would that mean my answers are correct ?

EDIT : Why would the current in the diode be constant ? We are dealing with an AC source....
(edited 11 years ago)
The current in the diode is the rate of flow of charge.
The charge on the capacitor = CV so the charge on it increases with the pd.
The current is equal to the rate of increase of this charge. (dQ/dt)
If the pd (and charge) increases uniformly, the current is constant.
I guess that is what the question intended. Your answer is definitely not correct.
If the pd line is slightly curved (such that its gradient gets less) then the current will decrease slightly.
I have no way of knowing, without the mark scheme, if this is what they intended.
Original post by Stonebridge
The current in the diode is the rate of flow of charge.
The charge on the capacitor = CV so the charge on it increases with the pd.
The current is equal to the rate of increase of this charge. (dQ/dt)
If the pd (and charge) increases uniformly, the current is constant.
I guess that is what the question intended. Your answer is definitely not correct.
If the pd line is slightly curved (such that its gradient gets less) then the current will decrease slightly.
I have no way of knowing, without the mark scheme, if this is what they intended.


Hmmm.... That does make sense.... Unfortunately, I don't have the mark scheme.

This is from a very old paper...
Original post by Ari Ben Canaan
Hmmm.... That does make sense.... Unfortunately, I don't have the mark scheme.

This is from a very old paper...


I suspect they intended the current to be constant.
You can actually work out what it is because the final charge on the capacitor = C x 12V and the initial charge is C x 8V
The current during charging will be the difference in charge divided by the time taken, (1ms) though this is not asked for.
Original post by Stonebridge
I suspect they intended the current to be constant.
You can actually work out what it is because the final charge on the capacitor = C x 12V and the initial charge is C x 8V
The current during charging will be the difference in charge divided by the time taken, (1ms) though this is not asked for.


Doesn't your previous comment apply to the shape of the current vs. time graph only for the capacitor ? Is it possible that the current vs. time graph for the diode is different ?
(edited 11 years ago)
diode17.gif13.9KB
Original post by Ari Ben Canaan
Doesn't your previous comment apply to the shape of the current vs. time graph only for the capacitor ? Is it possible that the current vs. time graph for the diode is different ?


Your diagram above is the voltage vs time curve for the output.
The question paper has given a similar diagram and also represents voltage vs time, for the capacitor. (Should be the same as they are in parallel)
The current vs time graph for the diode will be as I said before, the gradient of the voltage curve for the charging part of the cycle. The current through the diode is what charges the capacitor. It's complicated by the fact that a small amount of this current will flow through R.
That's why I said that I can't be 100% certain what they want. I suspect they want a straight line (constant value), because otherwise the question gets very complicated.
Given that they are asking for mean values in the other parts of the question, I think this is probably correct.
You just have to take the data as given in the question. Charging and discharging are really exponentials, as you know, but if the time interval is very short they approximate to a straight line. If the charging part in this question is meant to be near to a straight line, then the current will be constant.
I agree it still leaves the question about the AC sinusoidal source, but we just have to accept the graph they draw and make deductions from that.

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