# RC Circuit - Currents at Timed Intervals

## Homework Statement

An uncharged 1.15 uF capacitor is in series, through a switch, with a 4.50-M Ohms resistor and a 24.0- V battery (with negligible internal resistance.) The switch is closed at t = 0 and a current of Ii immediately appears. (Which I calculated to be 5.33 uA.) How long will it take for the current in the circuit to drop to 0.37 Ii?

## Homework Equations

V = Voltage of Battery - I*r (This will not apply here because r is negligible)
V = IR
I(t) = Voltage of Battery/Resistivity*exp^(-t/time constant)
Time constant = RC
4.50 MOhms = 4.50E6 Ohms
1.15 uF = 1.15E-6 F

## The Attempt at a Solution

I calculated the time constant and got 5.175 s. I plugged in the information above into the current based on time equation, but couldn't solve for time.

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alphysicist
Homework Helper
Can you show how far you got? What part were you stuck on?

I can try.

(.37 A)(t) = 24.0V/ (4.50E6 Ohms)*exp^(-t/5.175 s)
.37 A(t) = 5.33E-6 V/Ohms*exp^(-t/5.175 s)

Here's where I have trouble. I know by using ln I could cancel out the exp. However, by having to apply it to both sides, I would get ln(.37 A)(t).

alphysicist
Homework Helper
I(t) does not mean I times t; I(t) means "the value of I when time = t". (It also indicates what I depends on.)

So I(2) is I when t=2.

So the t should not be on the left hand side.

However, they don't say that the current is 0.37; they say the current is 0.37 Ii (that is, 37 percent of the initial current). Can you see that you did not have to calculate the initial current?