RC Circuit - Charge on Capacitor

In summary, to find the charge on the 4 μF capacitor in this circuit with a 20 V battery and an internal resistance of 1, we can use the equation Q = V*C, where V is the voltage across the capacitor. To find V, we can use Kirchhoff's laws and the voltage divider rule to determine the equivalent resistance of the circuit and the voltage drop across the capacitor. Using these values, we can then calculate the charge on the capacitor, which is approximately 45.71 μC.
  • #1
calvert11
32
0

Homework Statement


In the figure below the battery has an emf of 20 V and an internal resistance of 1.
Assume there is a steady current flowing in the circuit.
http://img48.imageshack.us/img48/9211/44724253.gif [Broken]
Find the charge on the 4 μF capacitor.
Answer in units of μC.

Homework Equations



Q = CE

The Attempt at a Solution


No time is given, so I can't use the equation for charge as a function of t.

From the figure, Resistors 1 and 5 seem to be in series. The circuit can't be simplified further.

But the charge on the capacitor can't be Q=CE since a current is still flowing. Can I have some advice on how to get started?

I tried using Kirchhoff, but I didn't get anywhere. I think I have too many unknowns, not enough equations (am I wrong?)

I'd be very grateful for any help.
 
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  • #2
It's true there is current flowing in the system, but there is still a voltage across the capacitor.

Q = V*C

Determine then the voltage across the capacitor and the charge on the capacitor will be known.
 
  • #3
LowlyPion said:
It's true there is current flowing in the system, but there is still a voltage across the capacitor.

Q = V*C

Determine then the voltage across the capacitor and the charge on the capacitor will be known.
Okay, using I= V/R the current for equivalent resistor 6 is 3.333 and for resistor 8 it's 2.5.

Voltage at resistor 5 is 3.333*5=16.6665
Voltage at resistor 8 is 2.5*8 = 20
Voltage through the capacitor is 20 - 16.6665 = 3.3335, which is wrong.

What about my approach is incorrect?
 
  • #4
calvert11 said:
Okay, using I= V/R the current for equivalent resistor 6 is 3.333 and for resistor 8 it's 2.5.

Voltage at resistor 5 is 3.333*5=16.6665
Voltage at resistor 8 is 2.5*8 = 20
Voltage through the capacitor is 20 - 16.6665 = 3.3335, which is wrong.

What about my approach is incorrect?

For 1 thing V = I*∑R = 20/14

That makes Vc = 8*20/14

Q = V*C = 4μf*160/14 = 640/14 μf
 
  • #5
LowlyPion said:
For 1 thing V = I*∑R = 20/14

That makes Vc = 8*20/14

Q = V*C = 4μf*160/14 = 640/14 μf
That's incredibly helpful. But to clarify, resistors 1, 5, and 8 are actually in series because no current flows through the wire with the capacitor?
 
  • #6
calvert11 said:
That's incredibly helpful. But to clarify, resistors 1, 5, and 8 are actually in series because no current flows through the wire with the capacitor?

There was current when it charged of course. But basically with the capacitor all you are concerned with is its voltage. And that is determined by the Voltage Divider of the resistors in series.
 

1. What is an RC circuit?

An RC circuit is a type of electrical circuit that contains a resistor (R) and a capacitor (C) connected in series or in parallel. The circuit is used to control the flow of electric current and store electrical energy.

2. How does an RC circuit work?

In an RC circuit, the capacitor stores electrical charge when connected to a power source. As the capacitor charges, the voltage across it increases until it reaches the same voltage as the power source. The resistor limits the flow of current, causing the capacitor to charge at a slower rate.

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

The equation for calculating charge on a capacitor in an RC circuit is Q = Q0(1 - e-t/RC), where Q is the charge at any time t, Q0 is the initial charge on the capacitor, R is the resistance in the circuit, C is the capacitance of the capacitor, and e is the base of the natural logarithm.

4. How does the resistance and capacitance affect the charge on a capacitor in an RC circuit?

The resistance and capacitance affect the charge on a capacitor in an RC circuit by determining the rate at which the capacitor charges. A higher resistance or capacitance will result in a slower charging rate, while a lower resistance or capacitance will result in a faster charging rate.

5. What happens to the charge on a capacitor in an RC circuit after the power source is disconnected?

After the power source is disconnected, the charge on the capacitor will remain constant. This is because the capacitor will act as a temporary battery, maintaining the voltage across it until it is discharged by connecting it to a circuit with a lower or equal voltage.

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