# RC circuit current

## Homework Statement

Three resistors, three capacitors, a battery, and two switches are connected in the circuit shown below. The values of all circuit elements are given in the figure. Originally, the switches S1 andS2 are open (as shown) and all of the capacitors are uncharged. At time t = 0, both switches are closed.
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam2/fa06/fig17.gif [Broken]

What is the current through resistor R1 immediately after the switches are closed?

## The Attempt at a Solution

Immediately (t=0) after the switches are closed, the capacitors should be uncharged. So they acts like wires. They have no V, that means there is no V across R3 as well since they are in parallel. So I redrew the circuit, having only the battery, R1 and R2 in series. And solve for the current, and that should be the current through R1.
9 V / (R1 + R2) = I1
But this isn't the answer apparently.

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Pythagorean
Gold Member
Just because there's no voltage drop across the capacitors is not a good enough reason to neglect R3. R3 is still connected to the battery and the other resistors.

I would bitterly complain that there is no reason to assume C1 is uncharged before T0. Afterall, this is an academic idealism, n stuff. Not that it changes the solutions...

R3 can be neglected. At t0 it is in parallel with 0 ohms.

Try drawing your circuit again with the switches and capacitors as shorts.

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I am working on the past physics exams, the answer to this problem is 4.5 Amp (according to my textbook). Does that mean I only redraw the circuit with the battery and the R1?

I would bitterly complain that there is no reason to assume C1 is uncharged before T0. Afterall, this is an academic idealism, n stuff. Not that it changes the solutions...

R3 can be neglected. At t0 it is in parallel with 0 ohms.

Try drawing your circuit again with the switches and capacitors as shorts.

I am sorry, maybe I didn't state the whole problem. Originally, the switches are open and all the capacitors are uncharged. At t=0 both switches are closed.

Pythagorean
Gold Member
R3 can be neglected. At t0 it is in parallel with 0 ohms.

what's the logic behind this?

what's the logic behind this?

I operate logic free. Actually it's just habbit.

$$i(t) = C \frac{dv}{dt}$$

For a step change in voltage, the change in current is infinite.

The dynamic resistance is $\Delta v / \Delta i \$ . With delta i, infinite the resistance at t = +0 is zero.

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I am working on the past physics exams, the answer to this problem is 4.5 Amp (according to my textbook). Does that mean I only redraw the circuit with the battery and the R1?

What happened to R2?

What happened to R2?

This is what I thought too, I redrew the circuit with the battery and R1 R2 in series. So to solve for the current through R1 I used the equation V/ R, R = R1+R2, so V/ R12.

But that's not the right answer apparently.

If there's a short circuit in the circuit, all the current will go through it. No current through R2 or R3, no voltage drop.

If there's a short circuit in the circuit, all the current will go through it. No current through R2 or R3, no voltage drop.

My mistake!

Listen to Kruum, and I'll take off :)

Pythagorean
Gold Member
I operate logic free. Actually it's just habbit.

$$i(t) = C \frac{dv}{dt}$$

For a step change in voltage, the change in current is infinite.

The dynamic resistance is $\Delta v / \Delta i \$ . With delta i, infinite the resistance at t = +0 is zero.

And of course, $$R_eq = \frac{1}{\frac{1}{R_3}+\frac{1}{0}} = 1/inf = 0$$. I didn't even really know that mathematically, but it's obviously a short if you just think of it practically.

Thanks for the proof!

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