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 ).
Solution
The voltage across a discharging capacitor is given by
v = {V_0}\,e - \tau t\
Using
and
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 and , between and .
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 .
Therefore, the gradient at 𝑡 = 2𝑠 is and 𝑡 = 4𝑠 is .
Solution (c)
v = 12\,e - 2 t\
Its derivative is
Therefore, at 𝑡 = 2𝑠 and 𝑡 = 4𝑠, is
Solution (d)
From part (b) and part (c), we can say that gives the gradient of the curve/ line v = 12\,e - 2 t\ , which is constant for all values of .
Solution (e)
v = 12\,e - 2 t\
Its first derivative is
And the second derivative is
\frac{{{d^2}v}}{{d{t^2}}} = 0\
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