# Another DC Circuit (But more complicated with a SWITCH and Capacitor!)

1. Apr 10, 2010

### sweetdion

1. The problem statement, all variables and given/known data
For the circuit shown below, the capacitors are initially uncharged. At t=0, the switch S is closed.

a) Determine the current in each resistor immediately after the switch is closed.
b) Determine the current in each resistor a very long time after the switch is closed.
c) Determine the voltage across the capacitor a very long time after the switch is closed.

After the switch has been closed for a very long time, it is reopened.

d) Determine the initial current I0 through the 100 ohm resistor, as a function of time after the switch is reopened.

e) Determine I(t), the current through the 100 ohm resistor, as a function of time after the switch is reopened.

f) How long after the switch is reopened does the charge on the capacitor fall to 10% of its fully charged state?

2. Relevant equations
V=IR
I through a capacitor is I = Cdv/dt = Q/C
Kirchoff's Loops Rules

3. The attempt at a solution

a) Immediately after the switch is closed, the capacitor behaves ls a short circuit and hence no current flows through the 100 ohm resistor.

I in the 200 ohm resistor = V/R=15V/200ohms=.075A

b) long after the switch is closed, the capacitor is fully charged and behanves as an open circuit. All of the current passes through both resistors.

Left Loop Clockwise:
15V-200ΩI1-100ΩI1=0
Right Loop Counter clockwise: Q/C-100ΩI1-0

I1 is the only current passing through both the resistors

I= 20A

as for the rest of the parts, I want to make sure I get A and B right before I move on.

2. Apr 10, 2010

### ehild

I do not understand. What is that I=20 A? A few hundred ohm and 15 V emf, how can be the current 20 A?

ehild

3. Apr 11, 2010

### Phrak

I = C dv/dt is correct.

But I = Q/C is not. Don't use it. V = Q/C is the correct equation.

4. Apr 11, 2010

### sweetdion

I solved the top equation and that's what I got. Did I label my currents right?

5. Apr 11, 2010

### ehild

Do you mean 15V-200ΩI1-100ΩI1=0? That is 15 V = 300Ω * I1. Do you mean I1= 20 A?

ehild

6. Apr 11, 2010

### sweetdion

Yeah, that's what I meant.

7. Apr 11, 2010

### ehild

Amazing. Tell me please, how much is 300*20???

ehild

8. Apr 11, 2010

### sweetdion

A lot, 6000. Which, is wrong.

I'm asking for your help because I don't understand what I'm doing wrong, ehild.

9. Apr 11, 2010

### ehild

You have an equation. 300 I = 15. This equation is correct.
Solve it for I. Is not it 15/300 A?

ehild

10. Apr 11, 2010

### sweetdion

Oh. Now I see what I was doing wrong. Simple Math error. I = 0.05A

11. Apr 11, 2010

### sweetdion

Moving on to part C now.

When the switch is closed for a very long time, the capacitor starts discharging through the resistors. After a long time, the voltage across the capacitor becomes equal to the voltage across the 100 Ω resistor.

V= IR=(0.05A)(100Ω)=5V

12. Apr 11, 2010

### Melawrghk

That's right. After a very long time, the capacitor becomes essentially an open circuit.

13. Apr 11, 2010

### sweetdion

Thanks Melawrghk. :)