Problems on capacitors

  • Thread starter Clari
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  • #1
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Hi there!

I have great problem in tackling some problems on this topic, I hope you can help me. :confused:

1. For a V-t graph, it has the shape of a sine curve but well beyond the x-axis. It has 3 complete cycle in 6ms as shown in the graph.
Notes: V is the potential difference applied between the terminals in a circuit varies with time t.

a. Sketch a graph of how the charge on the capactior changes over 6ms.
I sketch it the same as the V-t graph, since Charge (Q) = CV.....Q is proportional to V. Am I right?

b. Sketch a graph of the current in this circuit, marking clearly on this graph the I=0 value.
I have thought of it for quite a long time, but can't figure it out yet.. :cry:

I have attached the document which have the graph.
 

Attachments

  • Problems (Capacitors).doc
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Answers and Replies

  • #2
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a. yes

b.Some clues:

Take the first 90 degrees of the voltage cycle. As the voltage increases from zero towards its maximum, what will the current do?

Over next 180 degrees, the voltage at the supply is reducing in comparison to the capacitor to a lower threshold, so the capacitor will have a greater PD then the supply that needs to correct itself. The voltage reduces slowly then drops at speed then slows down when it reaches the bottom threshold again. So, imagine what the currrent will be doing, when the voltage is changing slowly will the current be large or small, when the voltage is dropping fast, what will the current be doing?
 
  • #3
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Delta,Thank you very much for helping out. :)
But I don't really get what you mean by
Delta said:
...The voltage reduces slowly then drops at speed then slows down when it reaches the bottom threshold again. "...

Here are my answers to your guiding questions:(lol, I am not sure if it is correct)

Over the first 90 degrees of the voltage cycle, as V increases from 0 to max., the current drops from max. to 0. Over next 180 degrees, the voltage at the supply is reducing in comparison to the capacitor. So the current flows in the opposite direction until the voltage rises again...

umm...But I am frustrated about the I=0 value, if I draw the I-t graph silimilar to the V-t graph(not in phase), then I will never be 0. :frown:
 
Last edited:
  • #4
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Clari said:
I don't really get what you mean by
"The voltage reduces slowly then drops at speed then slows down when it reaches the bottom threshold again."
From V=maximum (at 90deg) the voltage falls slowly therefore will the current be high or low?

When V is half way to its lowest value, there is larger rate of change widening the potential between the supply and capacitor, therefore will the current be high or low?

at V=minimum (at 270deg) the voltage changes more slowly as it reaches the lower part of the wave, so would the current now be high or low?

Note when I say "low" I mean close to zero and "high" means a larger current in either direction. Up to you to work out if its is positive or negative current (charging/discharging....????).
Clari said:
Over the first 90 degrees of the voltage cycle, as V increases from 0 to max., the current drops from max. to 0.
Good.
Clari said:
Over next 180 degrees, the voltage at the supply is reducing in comparison to the capacitor. So the current flows in the opposite direction until the voltage rises again...
therefore the current will be positive or negative?
Clari said:
if I draw the I-t graph silimilar to the V-t graph(not in phase)
... bit of a key concept here
Clari said:
then I will never be 0.
Once the voltage has settled to a maximum or minimum value the capacitor is the same as the supply, therefore there is no PD between the two. Would there be in current?
 

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