Question #311100

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 ).


1
Expert's answer
2022-03-14T18:35:10-0400

Solution


The voltage across a discharging capacitor is given by

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

Using


V0=12 V{V_0}=12 \ V and τ=2 s\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 V0=12 V{V_0}=12 \ V and τ=2 s\tau = 2\ s , between 𝑡=0 𝑠𝑡 = 0 \ 𝑠 and 𝑡=10 𝑠𝑡 = 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=2m=-2.


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



Solution (c)


v = 12\,e - 2 t\


Its derivative is


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


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



Solution (d)


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



Solution (e)


v = 12\,e - 2 t\


Its first derivative is


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


And the second derivative is


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





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