Question

The capacitor in the figure (Figure 1) is initially uncharged. The switch is closed at t=0.

http://session.masteringphysics.com/problemAsset/1261664/2/YF-26-72.jpg

Immediately after the switch is closed, what is the current through the resistor R1?

Immediately after the switch is closed, what is the current through the resistor R2?

Immediately after the switch is closed, what is the current through each resistor R3?

What is the final charge on the capacitor?

Answer 1

Help someone?

Answer 2

Capacitors act like shorts when a switch connecting them to a source is closed.
so just calculate the currents and voltages due to the resistive elements.

Answer 3

after time the capacitor starts building up charge and the current through it drops off... but they're only asking about the initial voltage/current here...

Answer 4

Are you talking about I=P/V?

Answer 5

I=V/R you mean?

Answer 6

whoops, yeah. but I did that earlier and got 5.25 for R1. Am I doing it right cause I got the answer wrong

Answer 7

5.25 V?

Answer 8

yeah, am I not supposed to just plug in the values? I honestly don't know what I'm doing though

Answer 9

no, 5.25 A

Answer 10

4.2A

Answer 11

how'd you get that?

Answer 12

3 ohm || 6 ohm = ..?

Answer 13

(1/3 +1/6)^-1

Answer 14

2 ohm.

Answer 15

ok, and then you use that for what?

Answer 16

in series with 8 ohms...

Answer 17

so 10 ohms is the resistance of the circuit

Answer 18

ohhh, ok

Answer 19

ok? you got this one?

Answer 20

I'm a little lost on R2

Answer 21

find the drop across R1

Answer 22

voltage drop...

Answer 23

the rest of the voltage drops across R2 || R3

Answer 24

from that you can find the current through each

Answer 25

so basically (1/6 + 1/3) ^-1 again?

Answer 26

no.