Please show how they got to these answers. Given the phasor voltage V = 115 2/45

Question

Please show how they got to these answers.

Given the phasor voltage V = 115 2/45 degree V across an impedance Z = 16.26/19.3 degree Ohm, obtain an expression for the instantaneous power, and compute the average power if omega = 50 rad/s. Ans: 767.5 + 813.2 cos(100t + 70.7 degree ) W; 767.5 W.

Explanation / Answer

given Z =16.26 angle(19.3) convert into rectangular form Z = 15.1+ i 5.288 and V = 115srqt(2) angle (45) given value is in rms value vm = 115*sqrt(2)*sqrt(2) =115*2 =230 and express in cosine form V = 230 cos(wt+45) current I = V/Z = 115sqrt(2) angle(45)/16.26 angle(19.3) = 10 .002 angle(45-19.3) =10.002 angle(25.7) instantaneous power = V*I = 115sqrt(2) cos(wt+45)* 10.002 cos(wt+25.7) =115sqrt(2)*10.002 {cos(wt+45)*cos(wt+25.7)} =1626.67 {cos(wt+45)*cos(wt+25.7)} we know cosAcosB = 1/2{COS(A-B)+COS(A+B)} = 1626.27/2 {COS(45-25.7)+ COS(2wt+45+25.7)} put w= 50 = 813.335 cos(19.3)+ 813.3 cos(100t+45+25.7) instanteous power =765.5 + 813.3cos(100t+70.7)--------------->(1) above equation (1) is called instanteous power- 2) we know average for dc = itself and average for cos term = 0 in istanteous power equation both dc and ac present average power = 765.5+ 0 =756.5 -------------------------->2 equation 2 represents average power hence proved please rate i wrote answers for both questions in detail

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