Alternating Current

Chapter 31 Alternating Current

PowerPoint® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman Lectures by James Pazun Modified by P. Lam 8_8_2008 Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Topics for Chapter 31 & 32 • Alternating current source • Voltage, current, power in AC circuit • Complex impedance and its application to the L- R-C series circuit • AC Transformer

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Alternating current (AC) source An AC generator is a conducting coil rotating in a magnetic field. The magnetic flux through the coil varies as a cosine function, hence the induced emf varies as sine function.

Given : i(t) = Icost Given : i(t) = Icost Given : i(t) = Icost di Q 1 VC = = Isint VL = L = LIsint C C VR = iR = IRcost dt 1 I & VR are "in phase" = LI cos()t + /2 = Icos()t /2 C 0 VL leads I by 90 0 VC lags I by 90 Compare the voltage equation for these three circuit elements. What are the units for L and 1/(C)? Answer : They have unit of resistance.

L X L = inductive reactance (an effective resistance of an inductor; it increases with the frequency)

1/(C) XC = capacitive reactance (an effective resistance of a capacitor; it decreases as the frequency increases)

• The capacitor blocks the low frequencies and lets the high frequencies go to the tweeter. XC=1/(C) • The inductor blocks high frequencies and lets low frequencies go to the woofer. XL=L

Given : i(t) = Icost Given : i(t) = Icost Given : i(t) = Icost Q 1 di VC = = Isint VR = iR = IRcost V = L = LIsint C C L dt I & V are in phase 1 R LI cos t /2 = Icos()t /2 = () + C P(t) = i(t)VR (t) 0 0 VL leads I by 90 VC lags I by 90 = I 2R(cost)2 P(t) = i(t)V (t) = always positive P(t) = i(t)V (t) C L 1 2 2 2 2 Paverage = I R < (cost) > = LI cost sint = I cost sint C 2 1 = positive and negative = I R = positive and negative 2 2 Paverage = LI < cost sint >= 0 1 2 2 Paverage = I < cost sint >= 0 Irms < [i(t)] > An inductor takes energy C = from the source and then A capacitor takes energy I gives it back, net power from the source and then = 2 dissipation is zero. gives it back, net power 2 dissipation is zero. Paverage = IrmsR Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley The L-R-C circuit Given : The source (t) = cost, find i(t) This is a hard problem. We will do an easier problem first. Given : i(t) = Icost

Find VR (t),VL (t),VC (t),and the source voltage (t)

Answer : di Q 1 V (t) = IRcost, V (t) = L = LI sint, V (t) = = Isint R L dt C C C (t) = ? Kirchoff 's rule :

 VR VL VC = 0 (t) = VR + VL + VC

(t) = IR{}cost []X L XC sint Combine cost and sint into a single function.

R 2 []X X Let = cos; Z = R2 + []X X = impedance, and L C = sin Z L C Z then,(t) = IZ{}cos cost sin sint = IZ cos()t + To answer our original question: Given :(t) = cost, then i(t) = cos()t Z

X = L L Z XL- XC

XC=1/(C) R

2 2 Z = R + []X L XC = impedance R []X X cos = ,sin = L C Z Z

Note : If []X L XC > 0 = positive voltage leads current

If []XL XC < 0 = negative voltage lags current Given :i(t)=Icost then (t)=IZ cos()t + Given :(t)=cost then i(t)= cos()t Z Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example: source frequency = natural frequency (resonance)

Given :(t) =10cost,R = 2,L = 2mH,C = 80μF

(1) What is the natural frequency (o ) of this circuit?

(2) Suppose the frequency () of the AC source = o , what the current i(t) through the circuit? (3) What are the voltages across the R, L, and C?

Answer : 1 1 1 rad rad (1) = = = x104 = 2500 o LC (2x103 )(80x106 ) 4 s s

(2)At resoance = o . L X = L = = 5 L o C 1 L X = = = 5 C C C o XL=L 2 2 Z = R + []X L XC = R = 2 and = 0 10 Z i(t) = cos()t = cos2500t Z 2

(3)VR = iR =10cos2500t = (t) di X =1/( C) VL = L = oLIsinot = 25sin2500t R dt C Q 1 VC = = sinot =+25sin2500t C oC

Note(1) : VL cancels VC

• Minimum impedance occurs at resonance => maximum current (and =0)

• >o => >0

• <0

X = L L Z XL- XC

XC=1/(C) R

Given :(t) =10cost,R = 2,L = 2mH,C = 80μF

(1) What is the natural frequency (o ) of this circuit?

(2) Suppose the frequency () of the AC source = 2o , what the current i(t) through the circuit? (3) What are the voltages across the R, L, and C?

Given :(t) =10cos(120t)

Find i(t),VR (t),VC (t)

d = N 1 1 1 dt d = N 2 2 2 dt = magnetic flux per turn. The iron core ensure that the magnetic flux per turn is the same for coil 1 and coil 2 d d N 1 = 2 2 = 2 dt dt 1 N1 N "Step up" transformer : 2 >1 N1 N "Step down" transformer : 2 <1 N1 Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Transformers - energy conservation

Suppose we have a step up transformer which steps up the 10 volts in the primary coil into 1,000 volts in the secondary coil, is there a violation of conservation of energy?

Answer is No. Suppose the secondary coil is connect to

a 1,000 resistor => I2 =1A, then the primary coil must carry at least 100A to provide the necessary power to generate the 1000 V in the secondary coil. In a step up transformer, the primary coil has fewer turns but the conductor is much thicker to accommodate higher current.

Power in = Power out (if there are no losses)

V1I1 = V2I2 In real situation, there are losses

V1I1 > V2I2

• Induced surface current on the the iron coil (eddy current) heats up the iron core => energy is loss. • Use laminated core to reduce eddy current.