EPE491_Power in AC Circuits

Electrical Power and Machines EPE491

D. Johari, FKE UiTM 1

Power in AC Circuits

Introduction

There are 2 fundamentals laws that are important in AC circuitso Ohms Law

States that the current through a conductor btwn 2 points is directly proportional to the potential difference across the 2 points, and is inversely proportional to the resistancebetween them.

Mathematical equation: V IRWhere V - potential difference in volts (V)

I - current in amperes (A)

R - resistance in ohms ()o Kirchoffs Laws

Kirchoffs Voltage Law States that the algebraic sum of all voltages around any closed loop is zero

Kirchoffs Current Law States that the algebraic sum of currents at a node is zero



AC Circuit

Means voltage signal is in alternating current form (AC) that has positive & negative portions. Normally referred as sinusoidal form. General equation: V(t) = Vm sin t

Where Vm - the maximum voltage

- angular frequency in rad/s = 2ff - supply frequency in Hz

T - period = 1/f

AC Signal

AC Signal

Two AC Signals with Different Phases

Vm sin (t + ) is a signal that leads the original signal by an angle of



Phaseo When capacitors or inductors are involved in an AC cct, the current & voltage do not peak

at the same time.

o The period diff btwn the peaks expressed in degrees is said to be the phase difference.

o This phase relation is often represented graphically in a phasor diagram.

Phasor diagram between V and I

RMS Voltageo Stands for root-mean-square.

o In AC cct, current & voltages are generally stated as rms values instead of maximum values.

o It is the effective voltage that is utilized in practical or theoretical analysis given by:

2m

rms

VV where Vm - the maximum voltage

Example 1

An AC signal is given as V (t) = 141.4 sin 314t. Determine the following:

a) Maximum voltage

b) RMS voltage

c) Frequency

d) Period to complete 1 cycle

e) Phase shift

is the angle between V & I = v - i



Components of AC Circuit

Ohms Law in AC Circuitso Modified to the form: V = I Z where

V - the effective or rms voltage values

I - the effective or rms current values

Z - the impedance

3 main componentso Resistor R in Ohms ()o Inductor L in Henry (H)

o Capacitor C in Farad (F)

Pure Resistive Loado The impedance, Z consists only of a resistor i.e. Z = R

Pure Resistive Circuit

o Phasor diagram

Current is in phase with voltage = v - i = 0 Therefore, v= i With v taken as reference point, the phasor diagram is shown below

o Power Factor

p.f. = cos Power factor is 1.0 or unity since = 0



Inductive Loado The impedance, Z consists of a resistor in series with an inductor i.e. Z = R + j XL

Inductive Circuit

o Phasor diagram

Current lags voltageo When a voltage is applied to an inductor, it resists the change in current. Current

builds up more slowly than the voltage, lagging it in time and phase.

= v - i > 0

o Power Factor

p.f. = cos Power factor is lagging

Capacitive Loado The impedance, Z consists of a resistor in series with a capacitor i.e. Z = R jXC

Capacitive Circuit

Where XL = inductive reactance ()= 2fL = L

L = inductance (H)f = supply frequency

Where XC = capacitive reactance ()

= CfC 1

2

1

C = capacitance (F)



o Phasor diagram

Current leads voltageo Since the voltage on a capacitor is directly proportional to the charge on it, the

current must lead the voltage in time and phase to conduct charge to the capacitor plates and raise the voltage.

= v - i < 0

o Power Factor

p.f. = cos Power factor is leading

Example 2

A resistance of 7.0 is connected in series with a pure inductance of 31.4mH and the circuit is connected to a 100V, 50Hz, sinusoidal supply. Calculate the circuit current & the phase angle.



Power in AC Circuits

The behavior of AC machines & systems are often easier to understand by working with power, rather than working with voltages and currents

Active, reactive & apparent power apply to steady-state AC circuits with sinusoidal waveforms only

o Cannot be used to describe transient (temporary) behaviour

o Cannot be used to describe DC circuits

Instantaneous Power in AC circuitso The product of instantaneous voltage & current, unit in Watts (W)

o Given by: P = V I

Active Power, Po The average value of the instantaneous power over one cycle of the voltage

o Also known as real power @ true power @ actual power. The effective power that does real work, unit in Watts (W)

o Given by: P = V I cos where = angle btwn V & Io Since V = IZ, P can also be written as P = I2Z cos

Reactive Power, Qo The circulating power in the circuit.

o Power which does no real work. Also known as the imaginary power, unit in volt-amperes-reactive [var]

o Given by: Q = V I sin o Q can also be written as Q = I2Z sin

Sources and Loadso Generator

Active source, delivers active powero Resistor

Active load, absorbs active powero Capacitor

Reactive source, delivers reactive powero Inductor

Reactive load, absorbs reactive power



Complex Power, So The product of the voltage across the load and the current through the load, unit in VA

(volts-amperes)

o Given by: S = V I*

o Apparent power the power that supplied to the load if phase angle diff btwn V & I are ignored

o Given by S = V I

o S can also be written as S = I2Z

Power Triangleo The relationship btwn S, P and Q can be represented graphically by a power triangle.

For inductive load, since > 0, Q = VI sin + = +ve Q For capacitive load, since < 0, Q = VI sin - = -ve Q Based on the discussion earlier, the summary is as follows:

Resistive Load Inductive Load Capacitive Load

I is in phase with V I lags V I leads V

= v - i = 0 = v - i > 0 = v - i < 0Q is zero Q is +ve Q is -ve

pf is unity pf is lagging pf is leading

S = (Vv)(I-i)* = (Vv)(I+ i ) = P + j Q = VI cos + j VI sin

Where S2 = P2 + Q2



Example 3

Figure below shows an AC voltage source supplying power to a load with impedance Z = 20-30. Calculate the current, I supplied to the load, the power factor of the load, and the real, reactive, apparent, and complex power supplied to the load.



Power Factoro The ratio of the active power P to the apparent power S

cosS

Ppf

o The value lies between 0 and 1

o A power factor of 1 or unity is the goal of any electric utility company.

o If less than 1, the utility company has to supply more current to the user for a given amount of power use.

This will cause more line losses. Larger capacity equipment is required.

o An industrial facility will be charged a penalty if its power factor is less than 1.

o In West Malaysia, minimum p.f specified by TNB is 0.85

Power Factor Correctiono It is the process of adjusting the characteristics of electric loads in order to improve power

factor so that it is closer to unity (1).

o Can be improved by connecting a capacitor bank in parallel with the load.

o Connecting a capacitor bank in parallel with the load means to inject reactive power to the system.



From phasor diagram for inductive load, the power factor is lagging. When capacitor is connected in parallel with the load, 2 < 1 Therefore, cos 2 > cos 1 i.e. p.f.2 > p.f.1

Calculating the Capacitance Valueo Consider the power triangle

o From diagram:

21 QQQC o The reactive power is given by:

CVX

VQ C

C

CC 2

2

o The capacitance value can then be calculated as below:

2CC

V

QC

With capacitor, I= IL+ICWithout capacitor, I= IL



Example 4

A single phase circuit is depicted in the following figure. The supply rms mode is 120V with 60Hz frequency. Calculate the corresponding capacitance value that is needed to improve the circuit power factor to 0.95 lagging. Show the answer through the aid of phasor diagram.

Date post:	25-Nov-2015
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