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
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
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.
Voltage and current relationship in AC circuit V R = ir = IRcos t I & V R are "in phase" V L = L di dt = LIsin t = LI cos t + /2 ( ) V L leads I by 90 0 V C = Q C = 1 C I sin t = 1 C V C lags I by 90 0 I cos( t /2)
Resistance and Reactance V R = ir = IRcos t I & V R are "in phase" V L = L di dt = LIsin t = LI cos t + /2 ( ) V L leads I by 90 0 V C = Q C = 1 C I sin t = 1 C V C lags I by 90 0 I cos( t /2) 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) X C = capacitive reactance (an effective resistance of a capacitor; it decreases as the frequency increases)
The loudspeaker, a useful application The capacitor blocks the low frequencies and lets the high frequencies go to the tweeter. X C =1/( C) The inductor blocks high frequencies and lets low frequencies go to the woofer. X L = L
Resistance dissipates power but reactance doesn t V R = ir = IRcos t I & V R are in phase P(t) = i(t)v R (t) = I 2 R(cos t) 2 = always positive P average = I 2 R < (cos t) 2 > = I 2 R 1 2 I rms < [i(t)] 2 > = <I 2 (cos t) 2 > = I 2 P average = I 2 rms R V L = L di dt = LIsin t = LI cos t + /2 ( ) V L leads I by 90 0 P(t) = i(t)v L (t) = LI 2 cos t sin t = positive and negative P average = LI 2 < cos t sin t >= 0 An inductor takes energy from the source and then gives it back, net power dissipation is zero. V C = Q C = 1 C I sin t = 1 C V C lags I by 90 0 I cos( t /2) P(t) = i(t)v C (t) = 1 C I 2 cos t sin t = positive and negative P average = 1 C I 2 < cos t sin t >= 0 A capacitor takes energy from the source and then gives it back, net power dissipation is zero.
The L-R-C circuit Given : The source (t) = cos t, find i(t) This is a hard problem. We will do an easier problem first. Find V R (t),v L (t),v C (t),and the source voltage (t) Answer : V R (t) = IRcos t, (t) =? Kirchoff 's rule : V L (t) = L di dt = LI sin t, V C (t) = Q C = 1 C I sin t V R V L V C = 0 (t) = V R + V L + V C (t) = IRcos t { [ X L X C ]sin t} Combine cos t and sin t into a single function. Let R Z = cos ; Z = [ [ ] 2 = impedance, and X X L C ] R2 + X L X C then, (t) = IZ{ cos cos t sin sin t}= IZ cos( t + ) To answer our original question: Given : (t) = cos t, then i(t) = Z cos ( t ) Z = sin
Geometric method to visualize Z and phase angle X L = L Z X L - X C X C =1/( C) R Z = [ ] 2 = impedance [ ] R 2 + X L X C cos = R Z,sin = X X L C Z Note : If [ X L X C ]> 0 = positive voltage leads current If [ X L X C ]< 0 = negative voltage lags current Given : i(t) =I cos t then (t) =IZ cos t + Given : (t) = cos t then i(t) = Z ( ) cos( t )
Example: source frequency = natural frequency (resonance) Answer : 1 (1) o = LC = 1 (2x10 3 )(80x10 6 ) = 1 rad 4 x104 = 2500 rad s s (2)At resoance = o. Given : (t) =10cos t,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? X L = o L = L C = 5 X C = 1 o C = Z = L C = 5 R 2 + [ X L X C ] 2 = R = 2 and = 0 i(t) = cos t Z ( )= 10 2 cos2500t (3)V R = ir =10cos2500t = (t) V L = L di dt = LIsin t = 25sin2500t o o V C = Q C = 1 o C sin t =+25sin2500t o Note(1) : V L cancels V C Note(2) : V L and V C > X L = L X C =1/( C) Z R
Circuit behavior at resonance and off-resonance Minimum impedance occurs at resonance => maximum current (and =0) > o => >0 < o => <0 X L = L Z X L - X C X C =1/( C) R
Example: source frequency > natural frequency (resonance) Given : (t) =10cos t,r = 2,L = 2mH,C = 80μF (1) What is the natural frequency ( o ) of this circuit? (2) Suppose the frequency ( ) of the AC source = 2 o, what the current i(t) through the circuit? (3) What are the voltages across the R, L, and C?
Use impedance and phase angle method. Given : (t) =10cos(120 t) Find i(t), V R (t), V C (t)
Transformers - transfer energy from one coil to another 1 = N 1 d 1 dt d 2 = N 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 1 dt = d 2 dt 2 1 = N 2 N 1 "Step up" transformer : N 2 N 1 >1 "Step down" transformer : N 2 N 1 <1
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 => I 2 =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) V 1 I 1 = V 2 I 2 In real situation, there are losses V 1 I 1 > V 2 I 2
Example of transformer energy loss 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.