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## 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.