+ Post New Thread
Results 1 to 6 of 6

5th November 2018, 12:34 #1

Advertisement

5th November 2018, 12:56 #2
 Join Date
 Jan 2008
 Location
 Bochum, Germany
 Posts
 44,658
 Helped
 13587 / 13587
 Points
 256,096
 Level
 100
Re: Capacitance question
Circuit is antiparallel.

Advertisement

5th November 2018, 13:14 #3
Awards:
 Join Date
 Apr 2014
 Posts
 15,150
 Helped
 3448 / 3448
 Points
 74,693
 Level
 66
Re: Capacitance question
Hi,
two capacitors connected together.
Are they now in parallel or in series? Its a chickenandegg problem.
But one knows:
* both voltages before connection
* both charges before connection
* both capacitances befroe connection
* and how they are connected.
> No need to know whether they are in series or in parallel.
Klaus
   Updated   
added:
Circuit is antiparallel.
If we consider both capacitors as "power supply" or as "battery" and one connects "+" of the one with "" of the other then we call it (straight) series connection.
But what if we connect the still open "" with the still open "+" will it now become "anti" and "parallel"?
Couldnīt we call it "short circuited series connection"?
But as said above: Maybe one should not care about the naming. The definition of the task is clear.
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

Advertisement

5th November 2018, 14:06 #4
 Join Date
 Jan 2008
 Location
 Bochum, Germany
 Posts
 44,658
 Helped
 13587 / 13587
 Points
 256,096
 Level
 100
Re: Capacitance question
You can of course interpret the circuit as series connection, if you like.
I think that parallel (in the other thread started by the author) respectively antiparallel (this thread) is closely related to the achieved charge balancing, which is sum in parallel and difference in antiparallel case.

5th November 2018, 14:35 #5
 Join Date
 Nov 2012
 Posts
 2,993
 Helped
 727 / 727
 Points
 16,347
 Level
 30
Re: Capacitance question
Please draw the diagram yourself and you will see whether it is series or parallel.
Consider the negative voltage as zero (it does not matter because what matters is the potential difference).
Step 1: 0 V end of Cap A is connected to the 40V end of Cap B. Now the 40V end of Cap A will appear to be 80V with respect to the 0V end of Cap B. Just like connecting two cells in series.
Step 2: The free end of Cap A (now at 80V) is next connected to the free end of Cap B (now at 0V). Because they are at different potentials, some current (or charge in fact) will flow. This is like shorting a series connection of two cells.
Step 3: You now need to determine the new voltage across the two connections (the capacitors now look to be in parallel) the two ends.
The text book analysis is clear and correct.

Advertisement

5th November 2018, 17:58 #6
 Join Date
 Mar 2012
 Posts
 379
 Helped
 74 / 74
 Points
 3,015
 Level
 12
Re: Capacitance question
It is true that an ambiguity can occur if only two capacitors are involved. Your two capacitor circuit can be considered as connected either in series or parallel. If three or more capacitors are connected in series, then no amount of topological manipulation can make them into a parallel circuit. The following method works for any number of capacitors connected in series, with the higher voltage connected to the next capacitor's lower voltage. This is what your problem does. What is happening here? Well, you know that the sum of the voltages of the capacitor string before the head and tail are connected together is the sum of the voltages of each capacitor. After the final connection, the sum of the capacitor voltages in the string has to be zero. That means that some the the capacitor voltages in the string have the opposite polarity of their neighbors. You can conclude that the charge imbalance of each capacitor has changed. Some capacitors have lost a little their former charge imbalance, and some have lost so much imbalance that they passed the neutral point and are imbalanced the opposite way, so as to show a opposite voltage with respect to their former voltages. The following is the equation to calculate the charge imbalance change.
(Q1x)/C1 +(Q2x)/C1...(Qnx)/Cn = 0, where x is the charge imbalance change. x = 40.6 uC for your problem.
In your problem the charge imbalance of the 0.4 uF capacitor changes from 24 uC to 16.6 uC. That corresponds to a 41.54 voltage. The 2.2 uF capacitor changes from 132 uC to 91.4 uC. That corresponds to a +41.54 voltage. The voltage totals are zero. Any questions?
RatchHopelessly Pedantic
+ Post New Thread
Please login