Question

Given that the voltage across the capacitor as a function of time is V(t)=q0Ce−t/(ROC)V(t)=q0Ce−t/(RC), what is the current I(t) flowing through the resistor as a function of time (for t>0t>0)? It might be helpful to look again at Part A of this problem. Express your answer in terms of ttt and any quantities given in the problem introduction.

Answer 1

Answer:

$i(t) = -\frac{q_0C}{R}e^{-t/RC}$

Explanation:

If the voltage across a capacitor is given by

$V(t) = q_0Ce^{-t/RC}$

then by definition of the capacitance (C = Q/V), the charge on the capacitor is

$q(t) = CV(t) = q_0C^2e^{-t/RC}$

Now we can relate the charge on the capacitor to the current through the capacitor by

$i(t) = \frac{dq(t)}{dt} = -\frac{q_0C^2}{RC}e^{-t/RC}$

Since this is a closed current with capacitor and resistor are connected in series, the current through the capacitor is equal to that of resistor.