Power in AC Circuits.pptx

Power in AC Circuits

• _{Power in Resistive Components} • _{Power in Capacitors}

• _{Power in Inductors}

• _{Circuits with Resistance and Reactance} • _{Active and Reactive Power}

• _{Power Factor Correction} • Power Transfer

• _{Three-Phase Systems} • _{Power Measurement}

Introduction

• _{The instantaneous power dissipated in a component is a}

product of the instantaneous voltage and the instantaneous current

p = vi

• _{In a resistive circuit the voltage and current are in phase –}

calculation of p is straightforward

• _{In reactive circuits, there will normally be some phase shift}

between v and i, and calculating the power becomes more complicated

Power in Resistive Components

• Suppose a voltage v = V_p sin t is applied across a resistance R. The resultant current i will be

• _{The result power p will be}

• _{The average value of (1 - cos 2t) is 1, so}

where V and I are the r.m.s. voltage and current

16.2 t I R t V R v

i   P sin  _P sin

) 2 2 cos 1 ( ) (sin sin

sin t I t V I 2 t V I t

V vi

p   _P   _P   _{P P}   _{P P}  

VI I

V I

V

P  _{P P}  P  P 

Power in Capacitors

• From our discussion of capacitors we know that the current leads the voltage by 90. Therefore, if a voltage v = V_p sin t is applied across a capacitance C, the current will be given by i = I_p

cos t

• _Then

• _{The average power is zero}

Power in Inductors

• _{From our discussion of inductors we know that the current lags}

the voltage by 90. Therefore, if a voltage v = V_p sin t is applied across an inductance L, the current will be given by i = -I_p cos t

• _Therefore

• _{Again the average power is zero}

Circuit with Resistance and

Reactance

• When a sinusoidal voltage v = V_p sin t is applied across a circuit with resistance and reactance, the current will be of the general form i = I_p sin (t - )

• _{Therefore, the instantaneous power, p is given by}

• _{The expression for p has two components}

• _{The second part oscillates at 2} _{and has an average value of zero over}

a complete cycle

• _{this is the power that is stored in the reactive elements and then returned to}

the circuit within each cycle

• _{The first part represents the power dissipated in resistive}

components. Average power dissipation is

) 2 cos( 2 1 cos 2

1 _{    }

 V I V I t

p _{P P P P}

 

 (cos ) cos

2 2 ) (cos 2 1 VI I V I V

• _{The average power dissipation given by}

is termed the active power in the circuit and is measured in watts (W)

• _{The product of the r.m.s. voltage and current VI is termed the}

apparent power, S. To avoid confusion this is given the units of volt amperes (VA)

 ) cos (cos

2 1

VI I

V

• _{From the above discussion it is clear that}

• _{In other words, the active power is the apparent power times the}

cosine of the phase angle.

• _{This cosine is referred to as the power factor}   cos cos S VI P   factor Power amperes) volt (in power Apparent watts) (in power Active _  cos factor

Power  

Active and Reactive Power

• _{When a circuit has resistive and reactive parts, the}

resultant power has 2 parts:

• _{The first is dissipated in the resistive element. This is the}

active power, P

• _{The second is stored and returned by the reactive element.}

This is the reactive power, Q , which has units of volt amperes reactive or var

• _{While reactive power is not dissipated it does have an}

effect on the system

• _{for example, it increases the current that must be supplied}

and increases losses with cables

• _{Consider an}

RL circuit

• _{the relationship}

• _Therefore

Active Power P = VI cos  watts

Reactive Power Q = VI sin  var

Apparent Power S = VI VA

Power Factor Correction

• _{Power factor is particularly important in high-power} applications

• _{Inductive loads have a lagging power factor} • _{Capacitive loads have a leading power factor} • _{Many high-power devices are inductive}

• _{a typical AC motor has a power factor of 0.9 lagging} • _{the total load on the national grid is 0.8-0.9 lagging} • _{this leads to major efficiencies}

• _{power companies therefore penalise industrial users who}

introduce a poor power factor

• _{The problem of poor power factor is tackled by adding additional}

components to bring the power factor back closer to unity

• _{a capacitor of an appropriate size in parallel with a lagging load can ‘cancel}

out’ the inductive element

• _{this is power factor correction}

• _{a capacitor can also be used in series but this is less common (since this alters}

Power Transfer

• _{When looking at amplifiers, we noted that maximum}

power transfer occurs in resistive systems when the load resistance is equal to the output resistance

• _{this is an example of matching}

• _{When the output of a circuit has a reactive element}

maximum power transfer is achieved when the load impedance is equal to the complex conjugate of the output impedance

• _{this is the maximum power transfer theorem}

• _{Thus if the output impedance Zo = R + jX, maximum power transfer}

Three-Phase Systems

• _{So far, our discussion of AC systems has been}

restricted to single-phase arrangement

• _{as in conventional domestic supplies}

• _{In high-power industrial applications we often use} three-phase arrangements

• _{these have three supplies, differing in phase by 120}  • _{phases are labeled red, yellow and blue (R, Y & B)}

Power Measurement

• _{When using AC, power is determined not only by the}

r.m.s. values of the voltage and current, but also by the

phase angle (which determines the power factor)

• _{consequently, you cannot determine the power from}

independent measurements of current and voltage

• _{In single-phase systems power is normally measured}

using an electrodynamic wattmeter

• _{measures power directly using a single meter which}

effectively multiplies instantaneous current and voltage

• _{In three-phase systems we need to sum the power taken from the}

various phases

• _{in three-wire arrangements we can deduce the total power from}

measurements using 2 wattmeter

• _{in a four-wire system it may be necessary to use 3 wattmeter}

• _{in balanced systems (systems that take equal power from each phase) a single}

Key Points

• _{In resistive circuits the average power is equal to VI, where V and I}

are r.m.s. values

• _{In a capacitor the current leads the voltage by 90} and the average power is zero

• _{In an inductor the current lags the voltage by 90} and the average power is zero

• _{In circuits with both resistive and reactive elements, the average}

power is VI cos 

• The term cos  _{is called the power factor}

Calculation of AC power using

rectangular coordinates

• _{Suppose the alternating voltage across and the current in a circuit are}

represented respectively by

• _{V = V∠α = V(cos α + j sin α) = a + jb [1]} • _{and I = I∠β = I(cos β + j sin β ) = c + jd [2]}

• _{Since the phase difference between the voltage and current is (α − β ),} • _{active power is}

VI cos(α − β )

• _{= VI(cos α · cos β + sin α · sin β )}

• _{i.e. the power is given by the sum of the products of the real}

components and of the imaginary components. Reactive power is

VI sin(α − β )

= VI(sin α · cos β − cos α · sin β )

= bc − ad [4]

• _{If we had proceeded by multiplying equation [1] by equation [2], the}

result would have been

• _{The terms within the brackets represent neither the active power nor}

the reactive power. The correct expressions for these quantities are derived by multiplying the voltage by the conjugate of the current, the conjugate of a complex number being a quantity that differs only in the sign of the imaginary component; thus the conjugate of c + jd is

c − jd. Hence

(a + jb)(c − jd) = (ac + bd) + j(bc − ad)

= (active power) + j(reactive power)