Answer to Question #257772 in Calculus for Wah

Question

A capacitor circuit has been charged up to 12V and the instantaneous voltage is 𝒗 = 𝟏𝟐 (𝟏 − 𝒆 − 𝒕 𝜏)

The tasks are to: 

a) Draw a graph of voltage against time for 𝑣 = 12𝑉 and 𝜏 = 2𝑠, between 𝑡 = 0𝑠 and 𝑡 = 10𝑠.

b) Calculate the gradient at 𝑡 = 2𝑠 and 𝑡 = 4𝑠

c) Differentiate 𝒗 = 𝟏𝟐 (𝟏 − 𝒆 − 𝒕 𝜏) 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^2𝑣/𝑑𝑡^2 ). 

Answer 1

Given: The equation for instantaneous voltage is

$V=12(1-e^{-t/\tau})$

a)

$$V=12(1-e^{-t/2}), t\in [0, 10]$$

b)

$$grad|_{t=2s}\approx\dfrac{12(1-e^{-(2+\Delta t/2})-12(1-e^{-2/2})}{\Delta t}$$

$$\approx\dfrac{12e^{-1}(1-e^{-\Delta t/2})}{\Delta t}\approx6/e\approx2.2073\ V/s$$

$$grad|_{t=4s}\approx\dfrac{12(1-e^{-(4+\Delta t/2})-12(1-e^{-4/2})}{\Delta t}$$

$$\approx\dfrac{12e^{-2}(1-e^{-\Delta t/2})}{\Delta t}\approx6/e^2\approx0.8120\ V/s$$

c) 

$$V'(t)=12(-1/2)(-e^{-t/2})=6e^{-t/2}$$

 

$$V'(2)=6/e\approx2.2073\ V/s$$

$$V'(4)=6/e^2\approx0.8120\ V/s$$

d) The result is the same.

e)

$$V''(t)=-3e^{-t/2}$$