## 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 thr

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.

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1 year 2021-09-05T01:23:50+00:00 1 Answers 289 views 0

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

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.