TOPIC 2- sinusoidal steady state

Power in R-L, R-C or R-L-C
series

For an R–L, R–C or R–L–C series a.c. circuit,

the average power P is given by:

POWER FACTOR

Exercise

Problem 1.

A series circuit of resistance 60Ω and inductance
75mH is connected to a 110V, 60 Hz supply.
Calculate the power dissipated. (P=165W)

POWER TRIANGLE

POWER FACTOR

POWER FACTOR

EXERCISE

Problem 1.
A pure inductance is connected to A 150V, 50 Hz supply,
and the apparent power of the circuit is 300VA. Find the
value of the inductance. (L=0.24H)

Problem 2.
A transformer has a rated output of 200 kVA at a power
factor of 0.8. Determine the rated power output and the
corresponding reactive power. (P=160kW, Q=120kVAR)

EXERCISE

Problem 3.

A load takes 90kW at a power factor of 0.5
lagging. Calculate the apparent power and the
reactive power.
(S=180kVA, Q=156kVAR)

Problem 4. EXERCISE

The power taken by an inductive circuit when

connected to a 120V, 50 Hz supply is 400W and the

current is 8A. Calculate

(a) the resistance, (R=6.25Ω)

(b) the impedance, (Z=15Ω)
(c) the reactance, (XL=13.64Ω)

(d) the power factor, (P.F=0.4167)

(e) the phase angle between voltage and current.
(θ=65.37° LAGGING)

*** Do EXERCISE 87 on CHAPTER 15.
PRACTICE MAKES PERFECT!!!

POWER FACTOR IMPROVEMENT

⮚ The design of any power transmission system is very
sensitive to the magnitude of the current in the lines as
determined by the applied loads.

⮚ Increased currents result in increased power losses (by
a squared factor since P = I2R) in the transmission lines
due to the resistance of the lines.

⮚ Heavier currents also require larger conductors,
increasing the amount of copper needed for the system,
and they require increased generating capacities by the
utility company.

⮚ Since the line voltage of a transmission system is fixed,
the apparent power is directly related to the current level.

⮚ In turn, the smaller the net apparent power, the

smaller the current drawn from the supply. Minimum
current is therefore drawn from a supply when S = P
and QT = 0.

⮚ The process of introducing reactive elements
to bring the power factor closer to unity is
called power-factor correction. Since most

loads are inductive, the process normally involves
introducing elements with capacitive terminal
characteristics having the sole purpose of improving
the power factor.

Increasing the power
factor without altering
the voltage or current
to the load is called
Power Factor
Correction

Effect of capacitor on total current Power triangle of power factor correction

Concept of Work, Power and Energy

Reference : CHAPTER 15

Electrical and Electronic Principles and Technology
Third edition

John Bird BSc(Hons), CEng, CSci, CMath, FIET, MIEE,
FIIE, FIMA, FCollT

by
TS ZULKIFLI & TS SHALIZAN/JKE/PKK/2018

The Concept

WORK : Unit JOULE (J) W=Fs

Work done or energy transferred when a
force of 1 newton is exerted through a
distance of 1m in the force direction.

ENERGY : Unit JOULE (J) W=Pt

Ability to do work or capacity for doing work

POWER : Unit WATT (W) P=W/t

The rate at which energy is used or rate of doing
work

***ONE WATT is the amount of power when 1Joule of energy is used in a second

EXERCISE

P=20W
W=300J

W=1.4kWh W=5kWh W=500kWh
W=2kWh

Power Supply Efficiency

EFFICIENCY

Efficiency is the ratio of the output power delivered
to a load to the input power to a circuit.

POWER LOSS
Internal power dissipation – operate power supply circuitry

***The HIGHER EFFICIENCY = The LOWER POWER LOSS

EXERCISE

PROBLEM 1

Efficiency = 80%, PLoss = 5W
PROBLEM 2

POUT = 46W