Question

The same capacitor circuit is now charged up to 12V and the instantaneous voltage is v=12(1- e^-t/2)

The same capacitor circuit is now charged up to 12V and the instantaneous voltage is v=12(1- e^-t/2)

a) Differentiate v with respect to t to give an equation for dv/dt?

b) Calculate the value of dv/dt at t=2s and t=4s?

c) Find the second derivative (d^2 v/dt^2)?

Answer 1

Given: The equation for instantaneous voltage is;

$V=12(1-e^{-t/2})$

Solution:

a) $dV/dt=12(-1/2)(-e^{-t/2})$

$=6e^{-t/2}$ (Answer)

b) $dV/dt=6e^{-t/2}$

Thus, at $t=2s$ , $dV/dt=6e^{-2/2}$

$=6e^{-1}$

$= 2.207 Vs^{-1}$ (Answer)

And at $t=4s$ , $dV/dt=6e^{-4/2}$

$=6e^{-2}$

$=0.812 Vs^{-1}$ (Answer)

c) $d^2V/dt^2=d/dt(dV/dt)$

$= d/dt(6e^{-t/2})$

$=6(-1/2)(e^{-t/2})$

$=-3e^{-t/2}$ (Answer)