Post a New Question


posted by on .

A capacitor of capacitance C= 10mF, is charged using a battery of potential V, through a resistor of resistance R = 250 ohms.
the charge of the capacitor is given by
Q(t)=CV[1− e ^ (−t/RC)]

Find an expression for the current passing through the resistor as a function of time, and make a graph clearly indicating the initial values. how long will it take for the current passing through the resistor to drop to 37% of its max value?

  • Physics - ,

    Current is the derivative of the capacitor charge.

    I(t) = dQ/dt = CV/(RC) exp(-t/RC)
    = (V/R) exp(-t/RC)

    Current is 1/e = 37% of the initial value when t/RC = 1. That is called the time constant.

    You will have to draw you own graph.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question