# Capacitor and voltage source

Discussion in 'Homework Help' started by rsashwinkumar, Dec 29, 2013.

1. ### rsashwinkumar Thread Starter New Member

Feb 15, 2011
23
2
I have a basic doubt. Suppose a voltage source of value V is connected to one end of a capacitor and the other end of the capacitor is left open. (other end of voltage source is connected to ground).
The voltage (w.r.t ground) at one end of the cap is equal to V, but what will be voltage at the other end of the capacitor be equal to? Please help me out...

2. ### Ramussons Active Member

May 3, 2013
623
104
An uncharged capacitor will behave like a short. To start with, both ends of the capacitor will show V volts.

However, if in measuring the voltage there is a current flow thru' the capacitor, over time the capacitor will charge fully, the current will reduce and the voltage will go down to Zero.

Ramesh

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Because of this the voltage (if any) at the open end of the capacitor is indeterminate.

Any attempt to measure this will result in the situation described by Ramussons since the measurement device will introduce a path to earth.

4. ### WBahn Moderator

Mar 31, 2012
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5,758
It will be V plus whatever the capacitor voltage happens to be (the open end relative to the end connected to the voltage source). If the cap is more than a few μF you can easily see this by charging it to some voltage, say 9V, and then doing what you suggested and measuring the voltage relative to ground. You will see it start at or near 18V and decay slowly to 9V (assuming you are using something like a DMM with several megohms of resistance).

5. ### Ramussons Active Member

May 3, 2013
623
104
Sorry, I must disagree.

" . . .decay slowly to 9V . ." is transitional - it will finally be Zero when the capacitor charges to the Applied Voltage (in this case, V = 9 Volts)

" ...
start at or near 18V ..." not necessarily. It could start at Zero and Remain there, or, 18 Volts decaying to Zero - depending on the connected Polarity of the charged capacitor.

Ramesh

6. ### WBahn Moderator

Mar 31, 2012
20,247
5,758

You are correct. That is what I get for working with just a mental image of the circuit off the cuff -- especially since I was trying to decide whether to mention the case of a null-point galvanometer being used (and decided not to). The meter will eventually read zero with the capacitor charging to whatever is required to make that happen.

Again, you are correct. I was working with an implied orientation but I see that there is nothing that really implied it.

Thanks for catching my mistakes.

7. ### rsashwinkumar Thread Starter New Member

Feb 15, 2011
23
2
Thank you all. Got it cleared!

8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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I think we need to be a little careful here as the bald statemnent about the voltage can lead to some wrong thinking.

In the attachment I have drawn the situation with the additional pre-existing voltage on the cap as Vc.

So the voltage to earth is (B + Vc) but you should be careful using this to calculate energies and current flows with the normal capacitor relations between these quantities. Attaching the capacitor to the battery has not increased the energy stored in it.

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9. ### WBahn Moderator

Mar 31, 2012
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You need to indicate the polarity of the capacitor voltage so that it doesn't have to be reverse engineered from your equation.

10. ### studiot AAC Fanatic!

Nov 9, 2007
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But no reverse engineering is needed. My point moves on from the OP and warns not to use the value gained by blindly applying the examples given in this thread.

eg

A capacitor, C, has one side connected to earth and the other charged up to 9volts.

The capacitor is then placed in series with an earthed 9 volt battery so that there is 18 volts between the free end of the capacitor and earth.

What is the energy in the capacitor.

It is tempting, but wrong, to say 0.5 C (18)^2 J

The energy is still only 0.5 C (9)^2 J

and a Happy New Year.

11. ### WBahn Moderator

Mar 31, 2012
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With no polarity indicated, reverse engineering is very much needed. Vc can be defined as either the top place relative to the bottom plate or the bottom plate relative to the top plate. In order to determine which, someone has to start with the equation given and determine which polarity is meant.

I had mad a similar unstated assumption elsewhere in the thread and quite rightly got called on it.

I don't know that anyone was trying to claim that the energy stored in the capacitor somehow changed. I must of missed that.