RC Circuit capacitor charge Help

In summary, when the switch is open, the current through the resistors is zero and the voltage drop across the resistors is Q/V = Q/R1 = Q/R2 = Q/C1 = Q/C2.
  • #1
cp51
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0
[SOLVED] RC Circuit Help

Homework Statement


What is the potential of point a with respect to point b in the circuit shown when switch S is open? Which point, a or b is at a higher potential? b.) What is the potential (after a long time) of point a with respect to ground when switch S is closed? How much does the charge in each capacitor change when S is closed?
R1 = 6 Ohms, C1 = 6microFarads, C2 = 3 microFarads, R2 = 3 ohms, C3 = 6 microFarads
V = 18V

Im sorry for the poor drawing

Homework Equations


V = IR
C = Q/V


The Attempt at a Solution


Ok, I know the solution to this question, I just don't know why...
A.) Apparently this circuit simplifies to a circuit of just the capacitors, the resistors drop out and you end up with C1 and C2 in parallel, connected to C3. Why do the resistors drop out?
From this you can combine all 3 resistors into 1 resistor Ctotal = 3.6 microFarad, and so
Q = CV = (3.6microF)(18V) = 6.48x10^-5
then combining C1 and C2 to get C12 = 9microF, i can get the voltage drop across the capacitors = Q/9microF = 7.2V

B.) When the switch is closed, aparently all the voltage drop is across the bottom capacitor, C3... why?

Thank you very much for any help
 

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  • #2
What is the current in each resistor when the switch is open? What is the voltage drop across each resistor?
 
  • #3
well, as I understand it:

Once C1 and C2 are fully charged no current flows through the two branches because the capacitors act as an open circuit. So that would mean that the current through the reisitors is 0 and therefore no voltage drop.

However, because the two capacitors are now charged, isn't there a volage difference between the two capacitors? So does that cause current to flow through o C3?? If so, doesn't that also require current to flow through R2? That is how I see it, which is apparently wrong cause it isn't the answer... heh...
 
  • #4
cp51 said:
well, as I understand it:

Once C1 and C2 are fully charged no current flows through the two branches because the capacitors act as an open circuit. So that would mean that the current through the reisitors is 0 and therefore no voltage drop.

However, because the two capacitors are now charged, isn't there a volage difference between the two capacitors? So does that cause current to flow through o C3?? If so, doesn't that also require current to flow through R2? That is how I see it, which is apparently wrong cause it isn't the answer... heh...
After the switch is closed, there is a direct path from the source V to C3 through the resistors R1 and R2, so C3 will be charged until its voltage reaches V. During the transient, capacitors C1 and C2 discharge trough R2 into C3.
 
Last edited:
  • #5
ok, I am still not getting why when the switch is open, the circuit simplifies to a circuit without resistors...
 
  • #6
cp51 said:
ok, I am still not getting why when the switch is open, the circuit simplifies to a circuit without resistors...

If there is no current, there is no voltage drop on the resistor.
 
  • #7
omg, I am so dumb... i totally didnt think of it that way... Thanks.

I started to look at it as charge and stuff, and I kinda thought i figured it out, but man...

Thanks a lot.
 

1. How does a capacitor charge in an RC circuit?

In an RC circuit, a capacitor charges when a voltage source is connected to the circuit. The capacitor initially has no charge, so it acts as a short circuit and allows a large current to flow through the circuit. As the capacitor charges, the voltage across it increases and the current decreases until it reaches its maximum capacity.

2. What is the equation for calculating the charge on a capacitor in an RC circuit?

The equation for calculating the charge on a capacitor in an RC circuit is Q = CV, where Q is the charge, C is the capacitance of the capacitor, and V is the voltage across the capacitor.

3. How does the value of the capacitor affect the charging time in an RC circuit?

The value of the capacitor affects the charging time in an RC circuit. A larger capacitor will take longer to charge because it can store more charge, while a smaller capacitor will charge faster because it can store less charge.

4. What happens to the current as a capacitor charges in an RC circuit?

As a capacitor charges in an RC circuit, the current decreases. This is because as the capacitor charges, it acts as an insulator and restricts the flow of current through the circuit.

5. How does the addition of a resistor affect the charging of a capacitor in an RC circuit?

The addition of a resistor in an RC circuit affects the charging of a capacitor by limiting the current flow and slowing down the charging process. This is because the resistor creates a voltage drop, reducing the voltage across the capacitor and decreasing the rate of charging.

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