RC Circuit - Currents at Timed Intervals

1. Jul 2, 2008

purduegirl

1. The problem statement, all variables and given/known data

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?

2. Relevant 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

3. 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.

2. Jul 2, 2008

alphysicist

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

3. Jul 2, 2008

purduegirl

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).

4. Jul 2, 2008

alphysicist

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?