Why Do Capacitors in Parallel Share the Same Voltage?

  • Thread starter Thread starter oneplusone
  • Start date Start date
  • Tags Tags
    Voltage
AI Thread Summary
Capacitors in parallel share the same voltage due to their direct connection via a conductor, which allows current to flow until equilibrium is reached. When capacitor C1, charged to 20.0 V, is connected to uncharged capacitor C2, the voltage across both capacitors becomes equal. The relationship Q1/C1 = Q2/C2 holds true because the voltage across both capacitors must be the same when they are connected. Current flows until the voltage across the resistive elements in the circuit stabilizes, confirming that the final voltages are equal. Understanding this principle is essential for solving related capacitor problems effectively.
oneplusone
Messages
127
Reaction score
2

Homework Statement



C1 = 6.00 uF
C2 = 3.00 uF
##\Delta V = 20.0## V

Capacitor c1 is first charged by the closing of s_1. Switch S_1 is then opened and the carved capacitor is connected to the uncharged capacitor b the closing of S_2. Calculate the initial charge acquired by C_1 and the final charge on each capacitor.


Homework Equations



C=Q/V

The Attempt at a Solution



See attached.

===============


I get everything in this solution besides when they set Q_1/C_1 = Q_2/C_2.
Is it ALWAYS true that the voltage across two points are the same?
 

Attachments

  • Screen Shot 2014-03-12 at 4.10.03 PM.png
    Screen Shot 2014-03-12 at 4.10.03 PM.png
    14.8 KB · Views: 498
Physics news on Phys.org
If you connected 2 capacitors, remember that their plates are connected by a conductor.

Now what happens when you connect two points with a conductor?
 
Current is formed, so there is a voltage?
And since there is no loss of electrons (charge) and the intensity (current) is the same, the voltage is the same?
Is that correct?
 
Well, as long as there is current flowing, you can't really make a problem out of it, at least not here, because it's not in a steady state.

But yes, when current does stop flowing in the circuit, the voltages are the same on both capacitors.

Now, there are formulae for the final charge/voltage in such an arrangement. I'd suggest you go through your textbook or look through the forums. I've forgotten the formulae myself, but I do remember that they exist.
 
oneplusone said:
Current is formed, so there is a voltage?
And since there is no loss of electrons (charge) and the intensity (current) is the same, the voltage is the same?
Is that correct?
You could include the resistance of the wire as a resistor in your crcuit, making it 3 elements. Current will flow through the resistor until the voltage across the resistor falls to zero. At the start, there are different voltages on each side.
 
oneplusone said:

Homework Statement



C1 = 6.00 uF
C2 = 3.00 uF
##\Delta V = 20.0## V

Capacitor c1 is first charged by the closing of s_1. Switch S_1 is then opened and the carved capacitor is connected to the uncharged capacitor b the closing of S_2. Calculate the initial charge acquired by C_1 and the final charge on each capacitor.


Homework Equations



C=Q/V

The Attempt at a Solution



See attached.

===============


I get everything in this solution besides when they set Q_1/C_1 = Q_2/C_2.
Is it ALWAYS true that the voltage across two points are the same?

When S2 is closed the two capacitors are in parallel so by definition their voltage drops must be the same.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top