The charge, Q, on a capacitor which starts discharging at time t = 0 is given by

Question

The charge, Q, on a capacitor which starts discharging at time t = 0 is given by the function below. Q = {Q_0 for t lessthanorequalto 0 Q_0e^-t/RC for t > 0, R and C are positive constants depending on the circuit and Q_0 is the charge at t = 0 where Q_0 notequalto 0. The current, I, flowing in the circuit is given by I = dQ/dt. Find the current I for t 0. Use upper case R and C if necessary. Use Q_0 or the subscript feature in the CalcPad if necessary. Is it possible to define I at t = 0? yes no Is the function Q differentiable at t = 0? yes no

Explanation / Answer

(a) I = dQ/dt

Q is a constant, and the derivative of a constant is 0. So for t<0,
I = 0.

(b) I = dQ/dt
You have to use the chain rule, and the fact that d(e^x) = e^x.

I = (-Q/RC)e^(-t/RC)

(c) No , We cant define I at t=0

(d) For Q(t) to be differentiable at t = 0,
-->it must be continuous and
-->the left handed and right handed derivative limits must be equal. In other words, are the derivatives of Q and Q equal as t approaches 0.

Is it continuous?

At t = 0, Q = Q

To find Q at 0, imagine that the t>0 constraint didn't apply. Plug 0 in for t to get Q = Q(e^0) = Q

Since the piecewise functions are equal at t=0, Q is continuous.

Does d(Q(0)) = d(Q(0))?

d(Q(0)) = 0

d(Q(t)) = (-Q/RC)e^(-t/RC)
d(Q(0)) = (-Q/RC)e^(-0/RC)
= -Q/RC

d(Q(0)) d(Q(0)), therefore Q(t) is not differentiable at t = 0.

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