Answer to Question #257772 in Calculus for Wah

Question #257772

A capacitor circuit has been charged up to 12V and the instantaneous voltage is ๐’— = ๐Ÿ๐Ÿ (๐Ÿ โˆ’ ๐’† โˆ’ ๐’• ๐œ)

The tasks are to: 

a) Draw a graph of voltage against time for ๐‘ฃ = 12๐‘‰ and ๐œ = 2๐‘ , between ๐‘ก = 0๐‘  and ๐‘ก = 10๐‘ .

b) Calculate the gradient at ๐‘ก = 2๐‘  and ๐‘ก = 4๐‘ 

c) Differentiate ๐’— = ๐Ÿ๐Ÿ (๐Ÿ โˆ’ ๐’† โˆ’ ๐’• ๐œ) and calculate the value of ๐‘‘๐‘ฃ ๐‘‘๐‘ก at ๐‘ก = 2๐‘  and ๐‘ก = 4๐‘ .

d) Compare your answers for part b and part c. 

e) Calculate the second derivative of the instantaneous voltage ( ๐‘‘^2๐‘ฃ/๐‘‘๐‘ก^2 ). 


1
Expert's answer
2022-02-22T13:52:21-0500

Given: The equation for instantaneous voltage is

V=12(1โˆ’eโˆ’t/ฯ„)V=12(1-e^{-t/\tau})

a)


V=12(1โˆ’eโˆ’t/2),tโˆˆ[0,10]V=12(1-e^{-t/2}), t\in [0, 10]


b)

gradโˆฃt=2sโ‰ˆ12(1โˆ’eโˆ’(2+ฮ”t/2)โˆ’12(1โˆ’eโˆ’2/2)ฮ”tgrad|_{t=2s}\approx\dfrac{12(1-e^{-(2+\Delta t/2})-12(1-e^{-2/2})}{\Delta t}

โ‰ˆ12eโˆ’1(1โˆ’eโˆ’ฮ”t/2)ฮ”tโ‰ˆ6/eโ‰ˆ2.2073 V/s\approx\dfrac{12e^{-1}(1-e^{-\Delta t/2})}{\Delta t}\approx6/e\approx2.2073\ V/s

gradโˆฃt=4sโ‰ˆ12(1โˆ’eโˆ’(4+ฮ”t/2)โˆ’12(1โˆ’eโˆ’4/2)ฮ”tgrad|_{t=4s}\approx\dfrac{12(1-e^{-(4+\Delta t/2})-12(1-e^{-4/2})}{\Delta t}

โ‰ˆ12eโˆ’2(1โˆ’eโˆ’ฮ”t/2)ฮ”tโ‰ˆ6/e2โ‰ˆ0.8120 V/s\approx\dfrac{12e^{-2}(1-e^{-\Delta t/2})}{\Delta t}\approx6/e^2\approx0.8120\ V/s

c) 

Vโ€ฒ(t)=12(โˆ’1/2)(โˆ’eโˆ’t/2)=6eโˆ’t/2V'(t)=12(-1/2)(-e^{-t/2})=6e^{-t/2}


 

Vโ€ฒ(2)=6/eโ‰ˆ2.2073 V/sV'(2)=6/e\approx2.2073\ V/s

Vโ€ฒ(4)=6/e2โ‰ˆ0.8120 V/sV'(4)=6/e^2\approx0.8120\ V/s

d) The result is the same.


e)


Vโ€ฒโ€ฒ(t)=โˆ’3eโˆ’t/2V''(t)=-3e^{-t/2}


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