- #1

goonking

- 434

- 3

## Homework Statement

## Homework Equations

## The Attempt at a Solution

so when t=0, the capacitor is a closed circuit.

The Req of the circuit is (20k)(30k)/(20k+30k)= 12000 ohms.

U = I Req = (7.5x10

^{-3}A) (12000 ohms) = 90 Volts

so there is a voltage drop of 90 volts across the 20k resistor (R1), when t=0.

RC= Req C = (12000 ohms) (0.1 x10

^{-6}F) = 0.0012 sec

∴ U

_{R1}= U

_{o}(e

^{-t/RC}) = U

_{R1}= 90V (e

^{-t/0.0012})Ok, so then the current flowing through the capacitor would be the same current flowing through the 30k resistor when t=0.

Using current division, 7.5x10

^{-3}A [20 / (20+30)] = 0.003 A

i

_{C1}= i

_{o}(e

^{-t/RC}) = (0.003A)(e

^{-t/0.0012})

This is my final answer, I've erased my initial attempt as it was wrong. Could someone verify if this is correct?

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