# Chapter 33. Alternating Current Circuits

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1 Chapter 33 Alternating Current Circuits 1

2 Capacitor Resistor + Q = C V = I R R I + + Inductance d I Vab = L dt

3 AC power source The AC power source provides an alternative voltage, Notation - Lower case symbols will indicate instantaneous values - Capital letters will indicate fixed values The output of an AC power source is sinusoidal v = sin ωt v is the instantaneous voltage v(t) is the imum output voltage of the source ω is the angular frequency of the AC voltage 3

4 AC voltage The angular frequency is ω = πƒ = v = V ( ωt + φ) π T cos ƒ is the frequency of the source T is the period of the source The voltage is positive during one half of the cycle and negative during the other half The current in any circuit driven by an AC source is an alternating current that varies sinusoidally with time Commercial electric power plants in the US use a frequency of 60 Hz 4

5 Resistor in AC circuit Consider a circuit consisting of an AC source and a resistor The AC source is symbolized by v = v R = sin ωt v R is the instantaneous voltage across the resistor The instantaneous current in the resistor is vr ir = = sin ωt = I sin ωt R R 5

6 Resistor in AC circuit v = v = sin( ωt) R vr ir = = sin ωt = I sin ωt R R The current and the voltage are in phase Resistors behave essentially the same way in both DC and AC circuits 6

7 Resistor in AC circuit: Phasor diagram v = v = sin( ωt) R vr ir = = sin ωt = I sin ωt R R A phasor is a vector whose length is proportional to the imum value of the variable it represents The vector rotates at an angular speed equal to the angular frequency associated with the variable The projection of the phasor onto the vertical axis represents the instantaneous value of the quantity it represents 7

8 rms current and voltage ir = I sin ωt The average current in one cycle is zero rms stands for root mean square T 1/ T 1/ ir dt ( ωt) dt T Irms = = I sin T π 1/ I I sin () τ dτ I 0 1 = = = π Alternating voltages can also be discussed in terms of rms values Vrms = =

9 rms current and voltage: power The rate at which electrical energy is dissipated in the circuit is given by P = i R where i is the instantaneous current The average power delivered to a resistor that carries an alternating current is P av = I rms R 9

10 Inductors in AC circuit v + v L = 0, or di v L = 0 dt di v = L = sinωt dt L ωl il = sin ωt dt= cos ωt i L π = sin ωt I = ωl ωl This shows that the instantaneous current i L in the inductor and the instantaneous voltage v L across the inductor are out of phase by ( π / ) rad = 90 o. 10

11 Inductors in AC circuit v = ωt sin sin π i = L I ωt I = V ωl 11

12 Inductors in AC circuit v = ωt sin i = sin L I ωt π I = V ωl The phasors are at 90 o with respect to each other This represents the phase difference between the current and voltage Specifically, the current lags behind the voltage by 90 o 1

13 Inductors in AC circuit v = ωt sin i = sin L I ωt π I = V ωl The factor ωl has the same units as resistance and is related to current and voltage in the same way as resistance The factor is the inductive reactance and is given by: X L = ωl As the frequency increases, the inductive reactance increases I = X L 13

14 Capacitors in AC circuit v + v c = 0 and so v = v C = sin ωt v c is the instantaneous voltage across the capacitor The charge is q = C v C =C sin ωt The instantaneous current is given by dq ic = = ωccos ωt dt π ic = ωc Vsin ωt + The current is (π/) rad = 90 o out of phase with the voltage 14

15 Capacitors in AC circuit v = sin C V ωt = sin π ic ωc V ωt + 15

16 Capacitors in AC circuit v = sinωt C = sin π ic ωc V ωt + The phasor diagram shows that for a sinusoidally applied voltage, the current always leads the voltage across a capacitor by 90 o This is equivalent to saying the voltage lags the current 16

17 Capacitors in AC circuit v = sinωt C = sin π ic ωc V ωt + The imum current I = ωc V = (1/ ωc) The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance and is given by 1 XC and I = ωc X C 17

18 v =sinωt i = I sinωt L I = V R v =sinωt π il = Isin ωt ωl I = = X L I v = sinωt C = sin π ic I ωt + = = (1/ ωc) XC 18

19 RLC series circuit The instantaneous voltage would be given by v = sin ωt The instantaneous current would be given by i = I sin (ωt - φ) φ is the phase angle between the current and the applied voltage Since the elements are in series, the current at all points in the circuit has the same amplitude and phase 19

20 RLC series circuit The instantaneous voltage across the resistor is in phase with the current The instantaneous voltage across the inductor leads the current by 90 The instantaneous voltage across the capacitor lags the current by 90 0

21 RLC series circuit The instantaneous voltage across each of the three circuit elements can be expressed as v = I R sin ωt = sin ωt R π vl = I XL sin ωt + = L cos ωt π vc = I XC sin ωt = C cos ωt R 1

22 RLC series circuit v = I R sin ωt = sin ωt R π vl = I XL sin ωt + = L cos ωt π vc = I XC sin ωt = C cos ωt R In series, voltages add and the instantaneous voltage across all three elements would be v = v R + v L + v C Easier to use the phasor diagrams

23 RLC series circuit i = I sin ωt v = I R sin ωt = sin ωt R π vl = I XL sin ωt + = L cos ωt π vc = I XC sin ωt = C cos ωt v = v + v + v = R L C = V sin ωt + cos ωt V cos ωt = R L C = V sin ( ωt + φ) R Easier to use the phasor diagrams 3

24 RLC series circuit The phasors for the individual elements: The individual phasor diagrams can be combined Here a single phasor I is used to represent the current in each element In series, the current is the same in each element 4

25 RLC series circuit Vector addition is used to combine the voltage phasors L and C are in opposite directions, so they can be combined Their resultant is perpendicular to R 5

26 RLC series circuit From the vector diagram, can be calculated ( ) ( I ) ( I I ) V = V + V = R + X X R L C L C ( ) V = I R + X X L C 6

27 RLC series circuit ( ) V = I R + X X L C The current in an RLC circuit is I = = R + ( XL XC) Z Z is called the impedance of the circuit and it plays the role of resistance in the circuit, where ( ) Z R + XL XC 7

28 RLC series circuit I = Z ( ) Z R + XL XC impedance triangle 8

29 RLC series circuit: impedance triangle ( ) Z R + XL XC The impedance triangle can also be used to find the phase angle, φ 1 XL XC φ = tan R The phase angle can be positive or negative and determines the nature of the circuit R Also, cos φ = Z i = I sin ωt v = V sin ( ωt + φ ) 9

30 RLC series circuit ( ) Z R + XL XC φ X X R 1 L C = tan 30

31 Power in AC circuit I = Z I rms I = The average power delivered by the generator is converted to internal energy in the resistor P av = ½ I cos φ = I rms rms cos φ cos φ is called the power factor of the circuit We can also find the average power in terms of R 1 1 V R Pav = Irms R = I R = R = Z R + X X ( ) L C 31

32 Resonances in AC circuit P av R R Z R + X X = = ( ) L C Resonance in P ( occurs at the frequency ω av ω ) o where the current has its imum value To achieve imum current, the impedance must have a minimum value This occurs when X L = X C or 1 XL = ω0l= XC = ω0c Solving for the frequency gives ω o = 1 LC The resonance frequency also corresponds to the natural frequency of oscillation of an LC circuit 3

33 Resonances in AC circuit P av R R R = = = 1 R + ωl ωc ω o Z R + ( XL XC) = 1 LC P av ( ω ) = 0 R Resonance occurs at the same frequency regardless of the value of R As R decreases, the curve becomes narrower and taller Theoretically, if R = 0 the current would be infinite at resonance Real circuits always have some resistance 33

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