# RC circuit and a battery

1. Jul 23, 2009

### songoku

1. The problem statement, all variables and given/known data
Consider the electrical circuit consisting of an E = 6 V battery, two switches S1 and S2, two resistors R1 = 4 ohm and R2 = 2 ohm, and a capacitor C = 2 $$\mu$$F. The internal resistance of the battery may be ignored. Initially the switches are both open and the capacitor has no charge. Close the switch S1 at a certain time. At a sufficient long time after the switch S1 is closed, the capacitor is fully charged and the circuit becomes steady.

a. Just after the switch S1 is closed, what is the current flowing through the resistor R1
b. How much charge is stored in the capacitor C
c. During the period in which the capacitor is charged, how much work is done by the battery
d. During the period in which the capacitor is charged, how much thermal heat is emitted from the resistor R1

Keeping the switch S1 closed, the switch S2 is also closed. At a sufficiently long time after the switch S2 is closed, the circuit becomes steady again.

e. How much charge is stored in the capacitor C long after the switch S2 is closed

2. Relevant equations
Q = CV
V = IR
W = 1/2 QV

3. The attempt at a solution
a. I = V/R1 = 6/4 A

b. Q = VC = 6 x 2 = 12 $$\mu$$C

c. W battery = 2 x W capacitor = QV = 12 x 6 = 72 $$\mu$$J

d. W dissipated = 1/2 W battery = 36 $$\mu$$J

e. I = V / R total = 6 / 6 = 1 A
V across Capacitor = I x R2 = 2 V
Q = CV = 2 x 2 = 4 $$\mu$$C

Do I get it right?

thx

2. Jul 23, 2009

### jackqpublic

b. is wrong
you have to consider the V in the capacitor not in the whole system.

3. Jul 23, 2009

### songoku

I think because the capacitor has been fully charged, the voltage across it will be the same as battery.

Am I wrong?

And i'm not sure about my answer on c and d. I've read it somewhere but i don't know the reason why the amount of work done on capacitor is equal to the heat dissipated on the resistor

thx

4. Jul 24, 2009

### alphysicist

You can calculate it directly. The power dissipated in a resistor is:

P = I2R

and here power is the time derivative of the energy dissipated (call that W), and you can also plug in the equation for the current in a charging RC circuit:

$$\frac{dW}{dt}=\left(\frac{V}{R}\ e^{-t/(RC)}\right)^2 R$$

You can then move the dt to the right side, and integrate both sides. The left side just becomes total energy dissipated by the resistor W, and the right side (integrated over the entire charging time from t=0 to t=infinity) becomes equal to the energy stored in the capacitor.

5. Jul 24, 2009

### songoku

When integrating the right side, is V constant?

thx

6. Jul 24, 2009

### alphysicist

Yes, that V is the battery voltage. Remember that the original equation for the current is:

$$I(t) = I_0 e^{-t/(RC)}$$

where I0 is the initial current at t=0, which you found in part a to be the battery voltage divided by the resistance of the RC circuit.

7. Jul 24, 2009

### songoku

Integrating both side I get W = 1/2 CV^2

Thx a lot alphysicist ^^

8. Jul 24, 2009

### alphysicist

You're welcome!