Question

In the circuit shown in Fig.26.14, the capacitors are all initially uncharged, the connecting leads have no resistance, the battery has no appreciable internal resistance, and the switch S is originally open.

a) Just after closing the swtich S, what is the current in the 15.0 ohms resistor ?

here is the link to the circuit
http://session.masteringphysics.com/problemAsset/1072341/2/26.14.jpg

Answer 1

Hi! Can you imagine what will happen "just after closing the switch"?

Answer 2

when you clos the switch there is no current flow to 15 ohms resistor so it's 0 ???

Answer 3

@theEric

Answer 4

Thanks for @ ing me. I didn't see the notification because this problem was up in a different tab.

Actually, the capacitors are initially \(\rm{uncharged}\), so they don't oppose the flow of current at all. They just charge up. They only start to oppose the current when the charge builds up. A simplified capacitor example might help.

Answer 5

So, initially, this capacitor is uncharged.

Since the battery is connected, current will flow.

Answer 6

Electrons flow from the negative side of the battery. They are loose, and are drawn by more positive electrical potential. But they can't go across the capacitor. They are just attracted to the other side, and stay put there.

Answer 7

Eventually, so much charge accumulates that the electrons just aren't motivated to move anymore. The electrical potential is more negative now at the capacitor - just the same as the potential of the negative side of the battery.

Answer 8

So that's when you have problems. Initial, there is no charge, so no opposition to current.

So for your problem, treat the capacitors as creating no impedance.

Answer 9

ok thank you:)

Answer 10

You're welcome! :)