Answer to Question #311100 in Calculus for DrackfromBTS

Question

The equation for the instantaneous voltage across a discharging capacitor is given by 𝑣 = 𝑉𝑂𝑒 − 𝑡 𝜏 , where 𝑉𝑂 is the initial voltage and 𝜏 is the time constant of the circuit. The tasks are to: a) Draw a graph of voltage against time for 𝑉𝑂 = 12𝑉 and 𝜏 = 2𝑠, between 𝑡 = 0𝑠 and 𝑡 = 10𝑠. b) Calculate the gradient at 𝑡 = 2𝑠 and 𝑡 = 4𝑠. c) Differentiate 𝑣 = 12𝑒 − 𝑡 2 and calculate the value of 𝑑𝑣 𝑑𝑡 at 𝑡 = 2𝑠 and 𝑡 = 4𝑠. d) Compare your answers for part b and part c. e) Calculate the second derivative of the instantaneous voltage ( 𝑑 2 𝑣 𝑑𝑡2 ).

Answer 1

Solution

The voltage across a discharging capacitor is given by

$v = {V_0}\,e - \tau t\$

Using

${V_0}=12 \ V$ and $\tau = 2\ s$

We have

$v = 12\,e - 2 t\$

This is a linear equation and its plot is shown below (graph of $v = {V_0}\,e - \tau t\$ voltage against time for ${V_0}=12 \ V$ and $\tau = 2\ s$ , between $𝑡 = 0 \ 𝑠$ and $𝑡 = 10\ 𝑠$ .

Solution (b)

The above plot is a straight line. It means it has a constant gradient. From the equation $v = 12\,e - 2 t\$ , the gradient is $m=-2$.

Therefore, the gradient at 𝑡 = 2𝑠 is $m=-2$ and 𝑡 = 4𝑠 is $m=-2$.

Solution (c)

$v = 12\,e - 2 t\$

Its derivative is

$\frac{dv}{dt} = -2$

Therefore, $\frac{dv}{dt}$ at 𝑡 = 2𝑠 and 𝑡 = 4𝑠, is $\frac{dv}{dt} = -2$

Solution (d)

From part (b) and part (c), we can say that $\frac{dv}{dt}$ gives the gradient of the curve/ line $v = 12\,e - 2 t\$ , which is constant $m=-2$ for all values of $t$ .

Solution (e)

$v = 12\,e - 2 t\$

Its first derivative is

$\frac{dv}{dt} = -2$

And the second derivative is

$\frac{{{d^2}v}}{{d{t^2}}} = 0\$