- #1

thedarkone80

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## Homework Statement

The capacitors in the figure are charged and the switch closes at t=0s. At what time has the current in the 8 ohm resistor decayed to half the value it had immediately after the switch was closed?

(The circuit looks something like this; 2 capacitors are in series at 60microfarads each, another is in parallel with the other two at 20microfarads. Above and to the right is a switch, and to the right of the switch is an 8 ohm resisitor. Below and to the right of the 8 ohm resistor is two resistors in parallel, one at 30 ohm and another at 20 ohms.)

## Homework Equations

Q=Qoe^(-t/RC)

I=Ioe^(-t/RC)

From knight edition 2, a strategic approach

## The Attempt at a Solution

Value of capacitors=50*10^-6 F

Value of resistors= 23 ohms

Using I=Ioe^(-t/RC)

(We dont really need to find the values of I and Io because we know that the current has to decay to 1/2 its original value

ln(0.5)= (-t/RC)

I think my point of error lies here, but im not sure...

ln(0.5)= (-t/(8ohms*(50*10^-6)))

-t=2.77*10^-4 s

This answer is incorrect. Considering that the textbook calls this a challanging problem, im assuming that there is more work needed to answer this question than what I have done. Can someone please help me?