# Potential drop across capacitor after very long time

1. Oct 29, 2014

### Les talons

1. The problem statement, all variables and given/known data
For the circuit shown in the figure, the switch has been open for a very long time.

(a) What is the potential drop across the 15.0-mH inductor just after closing the switch?
(b) What is the potential drop across the 70.0-µF capacitor after the switch has been closed for a very long time?

2. Relevant equations
I = V/R (1 -e-Rt/L)
V = IR = Q/C

3. The attempt at a solution
(a) I replaced the capacitors with wires at t=0 and found the current in the outer loop from V = IR
200V = 100ohm *I
I = 2A
Then the resistance through the loop with the inductor is 75ohm, so
V = 2A *75ohm = 150V

(b) I replaced the inductors with wires from the limit at t=infinity of the equation for current through the inductor and the capacitors with an open circuit, so there is no current through the capacitors, and the current through the middle loop is
200V = I *75ohm
I = 2.6... A
I'm confused if there is no current through the loop with the 70-µF capacitor, what resistance is used to find the voltage? The voltage is not 0 or 200V (the battery supply). I also know the three branches are technically in parallel.

2. Oct 29, 2014

### Staff: Mentor

If there's no current in the branch with the 70 μF capacitor, that means the potential drop across its series resistor is what?

3. Oct 29, 2014

### Les talons

For the 38ohm resistor, would it have 200V? I thought the capacitor would act like an open switch after a very long time, so I am confused how there can be any potential drop if the circuit is broken there.

4. Oct 29, 2014

### Staff: Mentor

Nope. What does Ohm's law tell you about the potential drop across a resistor with zero current flowing through it?
Sure it'll look like an open switch (open circuit). That doesn't mean there can't be a potential difference between either side of the "switch".

5. Oct 29, 2014

### Les talons

Ah, so since V = IR, and I is 0, then there must be 0V across the 38ohm resistor.
Is that because the electrical potential difference is defined between two points, regardless of a circuit between those points?

6. Oct 29, 2014

### Staff: Mentor

Yes and yes.

7. Oct 29, 2014

### Les talons

Okay, but - I am still not sure how this helps find the voltage drop across the capacitor. It is not 0V, or 200V. Will the battery voltage be shared equally in each branch, because the three branches are in parallel?

8. Oct 29, 2014

### Staff: Mentor

Redraw the circuit replacing the inductors with wires and the capacitors with open circuits. Trim off the branch with the 75 Ohm resistor since no current can flow there and there's no potential of interest there:

The open terminals a -- b are where the capacitor of interest is located. According to your prior arguments, terminal a has the same potential as A, b the same as B.

9. Oct 29, 2014

### Les talons

At point A, assuming ideal wires, the potential is 200V, the battery. Then I found the current through the middle loop as 2.6... A, so the voltage before the 25ohm resistor is this current times the 50ohm resistor, and there is 0V through the 38ohm resistor. So then I just need to subtract these values for the potential drop?

10. Oct 29, 2014

### Staff: Mentor

Simpler than that even. The potential from A to B is identical to the potential across the 50 Ohm resistor.

Oh, keep a few extra decimal places in intermediate calculations. You wouldn't want rounding errors to creep into the significant figures of your results. In particular, keep a few more decimal places in your current value.

11. Oct 29, 2014

### Les talons

Ah, so since the branches are in parallel, they would have the same voltage drop? Also, I tried to indicate a repeating decimal with the ... after the number.

12. Oct 29, 2014

Correct
Ah. Okay.