The document discusses various types of power in AC circuits including active power, reactive power, apparent power and power factor. It defines these terms and discusses how to calculate them. Reactive power is caused by inductive or capacitive loads. Apparent power is the total power and accounts for both active and reactive power. Power factor is the ratio of active to apparent power and indicates how efficiently current is being used in a circuit. Power factor correction can be used to improve efficiency by adding capacitors to counteract the effects of inductive loads.

1. KF1063 Introduction to Electrical Engineering, ®mbi, bb KF 1063 LECTURE 7

4. Z = R (purely resistive) KF1063 Introduction to Electrical Engineering, ®mbi, bb ACTIVE POWER P = VI = I 2 R = V 2 / R (Watt)

8. KF1063 Introduction to Electrical Engineering, ®mbi, bb Determine the total P T and Q T for the circuit. Sketch the series equivalent circuit. R = P T / I 2 = 1200/20 2 = 3  X eq = X L = Q T / I 2 = 1600/20 2 = 4  ACTIVE/REACTIVE POWER - Example

9. KF1063 Introduction to Electrical Engineering, ®mbi, bb For load consisting of series resistance and reactance, Z = R  j X = Z /  θ , the power produced is called Apparent Power or Complex Power ), S or P S with unit Volt-Amp (VA) APPARENT POWER θ positive, inductive load θ negative, capacitive load S = VI (VA) P = VI kos θ = I 2 R = V R 2 /R (W) = S kos θ (W) Q = VI sin θ = I 2 X = V x 2 / X (VAR) = S sin θ S =  ( P 2 + Q 2 ) = VI * V /0  Power Triangle S = P + j Q L S = P – j Q C S = VI S = VI

10. KF1063 Introduction to Electrical Engineering, ®mbi, bb POWER TRIANGLE - Example Sketch the power triangle.

12. Leading p.f. (final) = cos θ J ; Q J = P tan θ J Q C = Q – Q J Q C = V 2 / X C ; X C = 1/ j  C = V 2 / Q C KF1063 Introduction to Electrical Engineering, ®mbi, bb POWER FACTOR - Correction

13. Figure 7.18, 7.19, 7.20 KF1063 Introduction to Electrical Engineering, ®mbi, bb Given: V s = 117  0 V, R L = 50  , j X L = 86.7   = 377 rad/s Z L = 50 + j86.7 = 100  1.047  I L = V L / Z L = (117  0 )/(100  1.047) = 1.17  – 1.047 A S = V L I L * = 137  1.047) = 68.4 + j118.5 VA Q C = – 118.5 VAR X C = V L 2 /118.5 = – j115  C = 1/  X c = 23.1  F Find the complex power for the circuit. Correct the circuit power factor to p.f. = 1 using parallel reactance. POWER FACTOR - Example

14. KF1063 Introduction to Electrical Engineering, ®mbi, bb POWER MEASUREMENT Wattmeter only reads active power on the load side

15. KF1063 Introduction to Electrical Engineering, ®mbi, bb POWER MEASUREMENT - Example What is the reading of wattmeter, W? P T = 10 + 40 + 700 = 750 W

16. KF1063 Introduction to Electrical Engineering, ®mbi, bb P.F. MEASUREMENT - Example S = VI = (9.615)(240) = 2.308 kVA p.f. = 1.5 kW/2.308 kVA = 0.65 Q = 1.754 kVAR If p.f. = 1 (correction) X c = V 2 / Q then C = 80.761  F What happen if C = 80  F?