AC Power: A Worked Example

1 ac power : A Worked ExampleAndrew McHutchonApril 22, 2013 The voltage in a circuit is V= 240 0and the impedance of the circuit at 50Hz is Z= 48 + the following we work through the analysis of this setup, tackling many of the calculations you will be requiredto make in the Cambridge 1A course. A notational note: V, I, and Z, are used to represent complex quantitieswith a magnitude and phase;V,I, andZare used to represent their polar setupWhat does this circuit look like? The truth is we don t know.

2 It could be made up of many different componentswhich combine to give this impedance. The imaginary component of the impedance (called thereactance) ispositive and this implies that the circuit is overallinductiverather than capacitive. This means that we can modelthe circuit using just a resistor and an inductor. But in which configuration should we place the resistor andinductor: in series or parallel (see figure 1)? It doesn t actually matter - there exists both a series combination anda parallel combination of a resistor and an inductor which has this impedance.

3 However, as the impedance is givenin rectangular form (aka Cartesian form) it is significantly easier to work with the series representation: we recallthat the impedance of a resistor is series with an inductor is given by, Z=R(s)+j L(s)(1)To find the value ofRandLwe equate real and imaginary parts,R(s)= 48 L= 36 L(s)= (2)remember = 2 f= 100 as we are told the frequency is find the equivalent parallel representation we use the equation for the impedance of a resistor in parallel withan inductor, Z=jR(p) L(p)R(p)+j L(p)(3)Equating real and imaginary parts with this equation requires several lines of maths to deduce that,R(p)

4 = 75 andL(p)= For all the subsequent analysis we could use either configuration but to save repetition we willFigure 1: Series and parallel configurations of a resistor and inductor. Either of these circuits could be used tomodel an impedance of 48 +j361 Figure 2: Argand diagram showing the complex voltage, impedance, and ourselves to just considering the series setup. Nearly all of the quantities that are calculated depend purelyon the overall impedance and so are invariant to the series or parallel setup.

5 The only differences are in calculatingindividual voltages and currents. A good test of your understanding would be to perform the calculations for theparallel configuration and ensure you get the same , phase, and power factorTo find the current in the series circuit we use Ohm s Law as usual, I= V Z=240 048 +j36=240 060 4 (4)The voltage, impedance, and current are shown on the Argand diagram in figure 2. Note that since,arg( V Z)= arg( V) arg( Z)(5)impedances with positive imaginary components will lead to currents whichlagthe angleis given by, = arg( V) arg( I) = (6) the angle from the current to the voltage.

6 Therefore, positive phase angles mean that the current lags thevoltage, and thus are calledlagging, and negative phase angles mean that the current leads the voltage, and factoris defined as, power factor =PS= cos( )(7)As this is always a positive number the tag leading or lagging is usually added to describe the phase circuit in our Example therefore has a power factor of cos( ) = , reactive, and apparent powerReal power , P, is the amount of power consumed by the resistive components.

7 It can be calculated from,P=V Icos( ) ={V2 Rfor parallel componentsI2 Rfor series components(8)2 Figure 3: The power trianglewhereRis the total resistance of the circuit and we are using the polar magnitudes. Conservation of real powermeans that,PT=P1+P2+.. In our Example P is power , Q, is the amount of power consumed or supplied by the reactive components, capacitors andinductors. It can be calculated from,Q=V Isin( ) ={V2 Xfor parallel componentsI2 Xfor series components(9)whereXis the total reactance of the circuit.}}

8 Be careful of the signs as X and Q can be positive or sign of the reactive power should be the same as that of the circuit reactance. Therefore, lagging power factorsare caused by positive reactive power and leading power factors by negative reactive power . Note that capacitorssupply reactive power (Q negative) and inductors consume it (Q positive). Conservation of reactive power meansthat,QT=Q1+Q2+.. In our Example Q is poweris the magnitude of the total power , volts times amps, supplied by the source.

9 It is givenby,S= P2+Q2=VrmsIrms(10) apparent power isnotadditive in the same way that real and apparent power are. That is, in general,ST6=S1+S2. In our Example S is 3 shows thepower triangle. This relates thereal,reactive, andapparentpowers. The phase angle can alsobe seen, confirming the relationship between the power factor and the ratio is reactive power bad?In an AC circuit voltage and current vary sinusoidally. This means that currentflows backwards and forwards between the load and the source.

10 If the load is purely resistive, the voltage and thecurrent are in phase with each other. This means that their product, the power delivered to the load, is alwaysgreater or equal to zero, power always flows into the load. However, capacitive or inductive components in theload store energy and cause the current to move out of phase with the voltage. This results in a part of the AC cyclewhere the product of voltage and current is negative, power is flowing from the load back into the source, seefigure 4.