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: 488
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top