Question

The equation for the instantaneous voltage across a discharging capacitor is given by, where is the initial voltage and is the time constant of the circuit

The tasks are to:

A) draw a graph of voltage against time for and, between and.
B) calculate the gradient at and
C) differentiate and calculate the value of at and
D) compare your answers with part b and part c
E) calculate the second derivative of the instantaneous voltage

Answer 1

a)

b) $v=V_0\times e^{\frac{-t}{\tau}}$ ;

1) $\frac{\partial v}{\partial V_0}= e^{\frac{-t}{\tau}}$ ;

2)$\frac{\partial v}{\partial t}= -\frac{V_0}{\tau}\times e^{\frac{-t}{\tau}}$ ;

3) $\frac{\partial v}{\partial t}= -V_0\times t \times e^{\frac{-t}{\tau}}$ ;

c) $v^{\prime}=-6\times e^{-\frac{t}{2}}$

$v^{\prime}(2)=-2.21$ ;

$v^{\prime}(4)=-0.81$ ;

d) Derivative equal to the gradient from t

e) $v=\frac{V_0}{\tau^2}\times e^{-\frac{t}{\tau}}$