l c r i m r i m l i m c m lecture 20

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L C R I m R I m L I m C m Lecture 20

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Page 1: L C   R     I m R I m  L I m  C  m Lecture 20

LC

R

ImR

Im L

Im C

m

Lecture 20

Page 2: L C   R     I m R I m  L I m  C  m Lecture 20

Today . . .

• Phasors– phase shifts between VL, VR, VC

• Driven Series LRC Circuit:• General solution• Resonance condition

» Resonant frequency

» “Sharpness of resonance” = Q• Power considerations

» Power factor depends on impedance

Text Reference: Chapter 31.4-6 Examples: 31.5 through 31.8

Page 3: L C   R     I m R I m  L I m  C  m Lecture 20

Last Time. . .

• The loop equation for a series LRC circuit driven by an AC generator is:

• Here all unknowns, (Im,) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: m sint.

• To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram.

LC

R

tdt

dQR

C

Q

dt

QdL m sin

2

2

• Assume a solution of the form: )sin( tII m

Page 4: L C   R     I m R I m  L I m  C  m Lecture 20

Phasors: LRCThe general solution for the driven LRC circuit:

LX L

CXC

1

22CL XXRZ

R

XL

XC

Z

XL - XC

R

XX CL tan

ZI mm

ZIm

the loop

eqn

)sin( tII m

LC

R

ImR

m

ImXC

ImXL

Page 5: L C   R     I m R I m  L I m  C  m Lecture 20

Lagging & LeadingThe phase between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances.

R

XX CL tanZ

I mm

LX L

CXC

1

R

XL

XC

Z

XL > XC

> 0 current LAGS

applied voltage

R

XL

XC

Z

XL < XC

< 0 current LEADS

applied voltage

XL = XC

= 0 current

IN PHASE WITH applied voltage

R

XL

XC

Z

1

Page 6: L C   R     I m R I m  L I m  C  m Lecture 20

Lecture 20, ACT 1• The series LRC circuit shown is driven

by a generator with voltage = m sint. The time dependence of the current I which flows in the circuit is shown in the plot.

– How should be changed to bring the current and driving voltage into phase?

(a) increase (b) decrease (c) impossible

1A

LC

R

0

2 4 6

1

Io

Im

-Im0 t

• Which of the following phasors represents the current I at t=0?1B(a) (b) (c) II

I

Page 7: L C   R     I m R I m  L I m  C  m Lecture 20

Lecture 20, ACT 1

(a) increase (b) decrease (c) impossible

1A

• From the plot, it is clear that the current is LEADING the applied voltage.

I

XC

XL

Therefore, the phasor diagram must look like this:

Therefore, LC XX • To bring the current into phase with the applied voltage, we need to increase XL and decrease XC.• Increasing will do both!!

• The series LRC circuit shown is driven by a generator with voltage = m sint. The time dependence of the current I which flows in the circuit is shown in the plot.

– How should be changed to bring the current and driving voltage into phase?

LC

R

0

2 4 6

1

Io

Im

-Im0 t

Page 8: L C   R     I m R I m  L I m  C  m Lecture 20

Lecture 20, ACT 1

(a) increase (b) decrease (c) impossible

– Which of the following phasors represents the current I at t=0?

(a) (b) (c)

1A

1B

II

I

• The projection of I along the vertical axis is negative here!

• The sign of I is correct at t=0.• However, it soon will become negative!

• This one looks just right!!• = - 30 ° = measures angle of rel to I

• The series LRC circuit shown is driven by a generator with voltage = m sint. The time dependence of the current I which flows in the circuit is shown in the plot.

– How should be changed to bring the current and driving voltage into phase?

LC

R

0

2 4 6

1

Io

Im

-Im0 t

Page 9: L C   R     I m R I m  L I m  C  m Lecture 20

Resonance• For fixed R, C, L the current Im will be a maximum at the

resonant frequency 0 which makes the impedance Z purely resistive.

This condition is obtained when:

C

Lo

o 1

LC

o

1

• Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself!

• At this frequency, the current and the driving voltage are in phase!

0tan

R

XX CL

22CL

mmm

XXRZI

i.e.:

reaches a maximum when: X XCL

Page 10: L C   R     I m R I m  L I m  C  m Lecture 20

ResonanceThe current in an LRC circuit depends on the values

of the elements and on the driving frequency through the relation

Im

00

o

0Rm

R=Ro

R=2Ro

R

XL

XC

Z

XL - XC

Plot the current versus , the frequency of the voltage source: →

R

XX CL tan

cos φR

I mm

ZI mm

cos

RZ

Page 11: L C   R     I m R I m  L I m  C  m Lecture 20

Power in LRC Circuit• The power supplied by the emf in a series LRC circuit

depends on the frequency . It will turn out that the maximum power is supplied at the resonant frequency 0.

• The instantaneous power (for some frequency, ) delivered at time t is given by:

• The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle.

• To evaluate the average on the right, we first expand the sin(t- ) term.

)sin(sin)()()( tIttIttP mm

Remember what this stands for

)sin(sin)( ttItP mm

Page 12: L C   R     I m R I m  L I m  C  m Lecture 20

Power in LRC Circuit• Expanding,

• Taking the averages,

)sincoscos(sinsin)sin(sin ttttt

0cossin tt

• Generally:

2

0

22

2

1sin

2

1sin xdxx

• Putting it all back together again,

sintcost

t0

0

+1

-1

(Product of even and odd function = 0)

sin2t

t0

0

+1

-1

tttItP mm cossinsinsincos)( 2

cos2

1)( mmItP

01/2

Page 13: L C   R     I m R I m  L I m  C  m Lecture 20

Power in LRC Circuit

• This result is often rewritten in terms of rms values:

mrms 2

1 mrms II

2

1 cos)( rmsrmsItP

• Power delivered depends on the phase, the“power factor”

• Phase depends on the values of L, C, R, and

• Therefore...

Page 14: L C   R     I m R I m  L I m  C  m Lecture 20

Power in RLC

• Power, as well as current, peaks at = 0. The sharpness of the resonance depends on the values of the components.

• Recall:

cos)( rmsrmsItP

cos

RI mm

• Therefore, 22 2

rms( ) cosrms

P t I RR

We can write this in the following manner (which we won’t try to prove):

2222

22

)1()(

xQx

x

RtP

rms

…introducing the curious factors Q and x...

Page 15: L C   R     I m R I m  L I m  C  m Lecture 20

The Q factor

where Umax is max energy stored in the system and U is the energy dissipated in one cycle

A parameter “Q” is often defined to describe the sharpness of resonance peaks in both mechanical and electrical oscillating systems. “Q” is defined as

U

UQ

max2

For RLC circuit, Umax is 2maxmax 2

1LIU

Losses only come from R:

res

RIRTIU2

2

1

2

1 2max

2max

period

This givesR

LQ res

And for completeness, note res

x

Page 16: L C   R     I m R I m  L I m  C  m Lecture 20

Power in RLC

<P>

00

o

0

2

Rrms

R=Ro

R=2Ro

Q=3

FWHM

For Q > few, fwhm

Q res

Page 17: L C   R     I m R I m  L I m  C  m Lecture 20

FWHM and Q

• FWHM– Full Width Half Maximum

• Q– Quality of the peak

– Higher Q = sharper peak = better quality

<P>

00

o

0

2

Rrms

R=Ro

R=2Ro

Q=3

FWHM

Page 18: L C   R     I m R I m  L I m  C  m Lecture 20

Q Amplification

On Resonance:

2

LC

R

RV IR IR

LL L

XV IX Q

R

CC C

XV IX Q

R

On resonance, the voltage across the reactive elements is amplified by Q!Necessary to pick up weak radio signals, cell phone transmissions, etc.

Page 19: L C   R     I m R I m  L I m  C  m Lecture 20

Lecture 20, ACT 2• Consider the two circuits shown where CII

= 2 CI. – What is the relation between the quality

factors, QI and QII , of the two circuits?

(a) QII < QI (b) QII = QI (c) QII > QI

(a) PII < PI (b) PII = PI (c) PII > PI

– What is the relation between PI and PII , the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency?

2B

2A

LC

R

LC

R

I II

Page 20: L C   R     I m R I m  L I m  C  m Lecture 20

Lecture 20, ACT 2

(a) QII < QI (b) QII = QI (c) QII > QI

2A

• We know the definition of Q:R

LQ 0

• At first glance, it looks like Q is independent of C.• At second glance, we see this cannot be true, since the resonant frequency o depends on C!

LC

10 Doubling C decreases o by sqrt(2)!

Doubling C decreases Q by sqrt(2)!

Doubling C increases the width of the resonance!

• Consider the two circuits shown where CII = 2 CI.

– What is the relation between the quality factors, QI and QII , of the two circuits?

LC

R

LC

R

I II

Page 21: L C   R     I m R I m  L I m  C  m Lecture 20

Lecture 20, ACT 2

– What is the relation between PI and PII , the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency?

(a) PII < PI (b) PII = PI (c) PII > PI

2A

2B

• At the resonant frequency, the impedance of the circuit is purely resistive.• Since the resistances in each circuit are the same, the impedances at the resonant frequency for each circuit are equal.• Therefore, the power delivered by the generator to each circuit is identical.

• Consider the two circuits shown where CII = 2 CI.

– What is the relation between the quality factors, QI and QII , of the two circuits?

(a) QII < QI (b) QII = QI (c) QII > QI

LC

R

LC

R

I II

Page 22: L C   R     I m R I m  L I m  C  m Lecture 20

Summary

• Phasors– phase shifts between VL, VR, VC

• Driven Series LRC Circuit:

• Resonance condition» Resonant frequency

» “Sharpness of resonance” = Q

LCo

1

R

LQ res

m

y

x

VL

VR

VC

Text Reference: Chapter 31.4-6 Examples: 31.5 through 31.8

Page 23: L C   R     I m R I m  L I m  C  m Lecture 20

Next time

• Repaired Ampere’s Law and Maxwell’s Displacement Current• E&M waves • Reading assignment: Ch. 32.1, 2, and 4, 33.4

– Examples: 32.1 and 2