Calculate the exact value of the average power dissipated in the 2.5 k? resistor

Question

Calculate the exact value of the average power dissipated in the 2.5 k? resistor.

What is the percentage of error in the estimated value of the average power?

Part A Problem 16.33 Use the first three nonzero terms in the Fourier series representation of 2(t) to estimate the average power dissipated in the 2.5 k resistor The periodic current shown in (Figure 1) is applied to a 2.5 resistor Express your answer to three significant figures and include the appropriate units P= Value Units Submit My Answers Give Up Part B Calculate the exact value of the average power dissipated in the 2.5 k2 resistor Express your answer to three significant figures and include the appropriate units Figure 1 P= Value Units i (A) Submit My Answers Give Up Part C What is the percentage of error in the estimated value of the average power? T/2 0 T/2 Express your answer using three significant figures vec Lt Error

Explanation / Answer

(t)= 10t/T, 0<t<T/2,

= 5, T/2<t<T,

Fourier series representation of i(t) is

i(t) = C + summation[ Acos(2nt/T) + Bsin(2nt/T)] ,

Where. C= (1/T) integration[ i(t) dt] , -T/2<t<T/2

A =(2/T) integration[ i(t)cos(2nt/T)dt] , -T/2<t<T/2

B=(1/T) integration[ i(t)sin(2nt/T)dt] , -T/2<t<T/2

We find. C=15/4

B=-5/(n)

A= 5[cos(n)-1]/(n)2

for olny 3 terms n =1,2

A= -5/^2. , A=0 for n=2

B= -5/ , for n=1, B=-5/(2), fOR n = 2

p= power using fourier series= 2.5[ c^2 +(B^2 + A^2)/2].,for n=1 + 2.5[ 2.5 [c^2 +(B^2 + A^2)/2].,for n=2

Pa= actual power = 2.5c^2

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