ee252solved

EE 252 Solved Problems 7.1. Given a circuit with an applied voltage: v = 150 sin ( t + 10 ) and a resulting current i = 5 sin ( t - 50 ), determine the power triangle. Answer: V = (150/ 2) 10 = 106 10 and I = (5/ 2) -50 = 3.54 -50 . Then S = VI* = (106 10 )(3.54 50 ) = 375 60 = 187.5 + j325 from which P = Re VI* = 187.5 w Q = Im VI* = 325 vars lagging S = |VI*| = 375 va P = 187.5 60 Q = 325 lagging S = 375

pf = cos 60 = 0.5 lagging.

7.2. Given the series circuit of Fig. 7-11, determine the power triangle.

I

50< - 90

P = 300

3

53.1

j6

AC

Q = 400 lagging S = 500

-j2 Fig. 7-11

Fig. 7-12

Answer: From Fig. 7-11, Z = 3 + j6 – j2 = 5 53.1 and I = V/Z = (50 -90 )/(5 53.1 ) = 10 -143.1 .

Then S = VI* = (50 -90 )(10 143.1 ) = 500 53.1 = 300 + j400 The components of the power triangle shown in Fig. 7-12 are P = 300 w, Q = 400 vars lagging, S = 500 va, pf = cos 53.1 = 0.6 lagging Alternate method. Substituting I = 10amperes in the power equation of each element, P = I2R = 102(3) = 300 w, Qj6 = 102(6) = 600 vars lagging, Q-j2 = 102(2) = 200 vars leading and Q = Qj6 + Q-j2 = 600 – 200 = 400 vars lagging. 7.4. In the circuit shown in Fig. 7-13, the total effective current is 30 amperes. Determine the power relations. Letting IT = 30 0 , I2 = 30 0 ( (5 – j3)/(9 – j3) ) = 18.45 -12.55 and I1 = 30 0 ( 4/(9 – j3) ) = 12.7 18.45 . Then P = I22R4 + I21R5 = (18.45)2(4) + (12.7)2(5) = 2165w Q = I21X = (12.7)2(3) = 483 vars leading S = P – jQ = 2165 – j483 = 2210 -12.6 , S = 2210 va pf = P/S = 2165/2210 = 0.98 leading The above results can also be found by calculating the equivalent impedance Zeq = ((5 – j3)4)/(9 – j3) = 2.4 – j0.533. Then P = I2TR = 302(2.4) = 2160 w and Q = 302(0.533) = 479.7 vars leading. 5 -j3

I1 IT

I2

4 Fig. 7-13

7.5.Determine the power triangles for each branch of the parallel circuit of Fig. 7-15 and add them to obtain the power triangle for the entire circuit. Answer: P 1 = 86.6 43.4

I2

30 S1

AC

I1

Z 1= 4 < 3 0

Q 1 = 50 lagging

Z 2= 5 < 6 0

V = 20