A sinusoidal current is given by i=Icos(wt) Part A et ti and t2 be the two small

Question

A sinusoidal current is given by i=Icos(wt)

Part A et ti and t2 be the two smallest positive times at which the rectified current is zero. Express ti and tz in terms of w. Express your answers in terms of the variable w and appropriate constants. Enter your answers separated by a comma. ti,t2 Submit My Answers Give Up Part B Find the area under the recifled i versus t curve between t, and t by computing the integral J idt. Since dy idt, his area equals the charge that fows during the ti to t2 time interval. Express your answer in terms of the variables I and w. Submit My Answers Give UR Part C Set the result in part (b) equal to Ia (t2 -ti) and calculate Iav in terms of the current amplitude / Express your answer in terms of the variable I and appropriate constants. Submit My Answers Give Up Continue

Explanation / Answer

Given that
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A sinusoidal current is given by i=I cos(wt)

Since Current at t1 and t2 is zero

this must happen that cos wt must be 0 as I cannot be 0

Thus Cos wt = 0

or Wt = (2n+1) Pi/2, where n= 0,1,2,3,4, .....

for smalles value of n,

Wt1 = ( 0 +1) * pi/2

Wt1 = pi/2

thus t1 = pi/2W

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for next , n=1

Wt2 = (2*1+1) pi/2

Wt2 = 3 pi/2

t2 = 3pi/2W

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part B:

Area A = integration from t1 to t2   i dt

Area A = I cos wt dt from ti to t2

A = I * (sinwt)/W from pi/2W   to 3pi/2W

A = (I/W) *( sin W (3pi/2W) - SIn w(pi/2W))

A = I/W(-1-1)

A = -2I/W   as A cannot be -ve, it must be 2I/W (ANSWER)
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part C:

A = Irav *(t2-t1)

2I/W = Irav *(t2-t1)

2I/W = Irav*(3pi-pi)/2W

Irav = 2I/Pi (ANSWER)
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