Lecture7
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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.
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7 - laplace.pptx
This document provides an overview of the Laplace transform and its applications in analysing electrical circuits. It begins with learning outcomes, then defines the Laplace transform and its properties. Examples are provided to illustrate how to take the Laplace transform of common functions and how circuit elements are represented using the Laplace transform. The document demonstrates using the Laplace transform to analyse transient responses of circuits.
Electrical Circuit - CSE132
This document discusses key concepts related to electrical circuits including:
1) It defines average/active power, reactive power, and apparent power as well as the relationship between power, voltage, and current using mathematical equations.
2) It explains the significance of power factor as the ratio of real power to apparent power and how power delivered is affected by the power factor.
3) It describes how to use phasor and impedance diagrams to represent voltages, currents, and phase relationships in AC circuits and how to use them to calculate unknown values.
Chapter 5 operational amplifier
The document discusses operational amplifiers and their applications. It describes the basic op-amp configuration, ideal op-amp model, and applications such as inverting amplifier, non-inverting amplifier, summing amplifier, differential amplifier, integrator, differentiator, and voltage follower. It also discusses offset adjustments and multiple op-amp circuits.
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Concept of ac circuit for electrical engineering and this help to study about...
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BEF 23803 - Lecture 8 - Conservation of Complex Power.ppt
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Lecture7
- 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?