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Homework Help: RC Circuit - Currents at Timed Intervals

  1. Jul 2, 2008 #1
    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. jcsd
  3. Jul 2, 2008 #2


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    Homework Helper

    Can you show how far you got? What part were you stuck on?
  4. Jul 2, 2008 #3
    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).
  5. Jul 2, 2008 #4


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    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?
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