- #1

majormaaz

- 62

- 1

## Homework Statement

Consider the circuit in the figure below, in which V = 85 V, R = 40 , and the switch has been closed for a very long time.

(a) What's the charge on the capacitor?

(b) The switch is opened at t = 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value?

[Here's the link for the picture if it doesn't come up]

http://www.webassign.net/knight/p31-77alt.gif

## Homework Equations

C = Q/V

V = IR

Q = C*EMF*e^(-t/RC)

## The Attempt at a Solution

For part (a), I thought that since the circuit's been connected like that for a long time, the voltage across the capacitor would be the same as the battery's voltage - i.e. 85 Volts. From there, a simple Q = CV would yield 1.7 e -4 C = 170 μC, but that's not right.

For part (b), I was thinking that since it's a parallel circuit, I'd have to use Kirchoff's Law and the Loop Rule to get those 2 loop equations and the Junction equation, but I'm not sure as to if there's another method to get to the answer.

I would appreciate any help!

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