3 The equation for the instantaneous voltage across a discharging capacitor is given by \( v=V_{o} e^{-\frac{t}{\tau}} \), where \( V_{o} \) is the initial voltage and \( \tau \) is the time constant of the circuit.
The tasks are to:
a) Draw a graph of voltage against time for \( V_{o}=12 \mathrm{~V} \) and \( \tau=2 s \), between \( t=0 s \) and \( t=10 s \).
b) Calculate the gradient at \( t=2 \mathrm{~s} \) and \( t=4 \mathrm{~s} \).
c) Differentiate \( v=12 e^{-\frac{t}{2}} \) and calculate the value of \( \frac{d v}{d t} \) at \( t=2 \mathrm{~s} \) and \( t=4 \mathrm{~s} \).
d) Compare your answers for part b and part c.
e) Calculate the second derivative of the instantaneous voltage \( \left(\frac{d^{2} v}{d t^{2}}\right) \)
