# Current of RC Circuit After Switch is Opened

1. Mar 9, 2017

### Math-U-Up

1. The problem statement, all variables and given/known data
See attached image for circuit diagram and givens.
The switch has been closed for a very long time.
a) What is the charge on the capacitor?
b) The switch is opened at t=0 s. What current initially flows?
2. Relevant equations
V = IR
Q=CV
I=I0e-t/(RC)
3. The attempt at a solution
Since the switch has been open for a long time, no current flows through the 10 Ohm resistor. Therefore,
I = 100V/(60 Ohm + 40 Ohm) = 1 Amp, V = 1Amp*40 Ohm, and Q = 2uF * 40V = 80uC.

So my question is: Is the current that initially flows after the switch opens at t=0 s just due to the battery? So it would just be 1Amp? I'm having trouble understanding how to find the current at part b.

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2. Mar 9, 2017

### BvU

Hello MUU,

First step is to establish where the current flows. Through an open switch ? Or elsewhere ?

3. Mar 9, 2017

### collinsmark

Hello @Math-U-Up. Welcome to PF!

I think you might have misread the problem statement. According to the wording in the problem statement, initially, the switch has been closed for a very long time. That means that current has been flowing through the battery for a very long time.

At time $t = 0$, the switch is opened, meaning that after $t = 0$ no current flows through the battery, and any current flowing in the circuit is the result of whatever residual charge is left in the capacitor.

At least that's the way that I interpret the problem.

(A "closed" switch means that the switch can conduct current. An "open" switch does not conduct current.)

4. Mar 9, 2017

### BvU

The idea in PF is that the thread poster discovers that after a small nudge from a helper ...