LC and AC Circuits

Physics 122B Electricity and Magnetism

Physics 122B Electricity and Magnetism

Martin Savage

Lecture 24 (Knight: 33.9, 34.1-5)

LC and AC Circuits

Lecture 24 (Knight: 33.9, 34.1-5)

LC and AC Circuits

March 7, 2007 Physics 122C - Lecture 23 2

Lecture 24 Announcements


Lecture HW is due tonight at 10 PM.

Midterm Exam 3 is this coming Friday. Covers explicitly everything not covered in the previous exam…and assumes understanding of all previous material.

Lecture question and lab question are multiple-choice, tutorial is long answer.


LC CircuitsLC Circuits A charged capacitor bears a certain resemblance to a stretched spring (remember the rubber diaphragm), storing energy even when the charge is not moving. An inductor similarly resembles a moving mass (remember the flywheel), storing energy only when charge is in motion. We know that a mass and spring can make an oscillator. What about a capacitor and inductor. Consider the circuit shown in the diagram. What happens when the switch is closed? The capacitor discharges by creating a current in the inductor. But where does the energy go that had been stored in the inductor? There are no dissipative elements in the system. Therefore, when the charge of the capacitor goes to zero, all of its previous energy must reside in the inductor. The current in the inductor falls while charging the capacitor in the opposite direction. And so on …


The Oscillation CycleThe Oscillation Cycle


LC circuits (2)LC circuits (2)

QLdIdt + C

1= 0

d2Q

dt2 Q+ LC1

= 0

Q = Sin( t + )

1

L C2 =


Example:An AM Radio Oscillator

Example:An AM Radio Oscillator

You have a 10mH inductor. What capacitor should you use with it to make an oscillator with a frequency of 920 kHz? (This frequency is near the center of the AM radio band.

5 -1 6 -12 2 (9.20 10 s ) 5.78 10 sf

112 6 -1 2 2

1 13.0 10 F 30 pF

(5.78 10 s ) (1.0 10 H)C

L


Plumber’s LC AnalogyPlumber’s LC Analogy

P1

Valve = Switch

Rubber Diaphragm = Capacitor Flywheel = InductorPressure = PotentialWater Flow = Current

Valve

RubberDiaphragm

The “plumber’s analogy” of an LC circuit is a rubber diaphragm that has been stretched by pressure on the top (P1) side. When the valve starts the flow, the diaphragm forces water past the flywheel, which begins to spin. After the diaphragm has become flat, the momentum of the flywheel forces the diaphragm to be stretched in the other direction, and the cycle repeats.

P2

P3

Flywheel

V1

V3

V2


Chapter 33 - Summary (1)









AC Sources and PhasorsAC Sources and Phasors You can think of an AC generator as a battery-like object with an emf that varies sinusoidally as E (t) = E0cos t, where E0 is the maximum emf and is the angular frequency, with =2f, where f is the frequency in Hz. Alternatively, the emf and other oscillatory quantities can be represented by a phasor diagram. The phasor is a vector of length E0 that rotates counterclockwise around the origin with angular frequency , so that the angle it makes with the horizontal axis at any time is t. The projection of the phasor on the horizontal axis at any time gives the emf.


Resistor AC CircuitsResistor AC Circuits Consider an AC current iR through a resistor. Ohm’s Law gives the potential drop across the resistor, which we will call the resistor voltage vR.

R Rv i R

If the resistor is connected in an AC circuit as shown, then Kirschoff’s loop law tells us that:

0soruce R RV V v =E

0( ) cos Rt t v E E0 cos cosR

R R

vi t I t

R R

E

In the phasor diagram, the phasors for vR and iR are parallel.

source


Example:Finding Resistor Voltages

Example:Finding Resistor Voltages

In the circuit shown, find (a) the peak voltage across each resistor, and(b) the instantaneous resistor voltages at t=20 ms.

eq 1 2 (5 ) (15 ) 20 R R R

0

eq eq

cos (100 V)cos 2 (60 Hz)cos (5.0 A)cos 2 (60 Hz)

(20 )R

R R

tv ti I t t

R R

E

10

2

25 V for R =5

75 V for R =15 RV I R

2( 20 ms) (5.0 A)cos 2 (60 Hz)(2.0 10 s) 1.545 ARi t

1

2

7.7 V for R =5

23.2 V for R =15 R Rv i R


Capacitor AC Circuits (1)Capacitor AC Circuits (1)

Consider an AC current iC through a capacitor as shown. The capacitor voltage vC = E = E0cos t = VCcos wt. The charge on the capacitor will be q = CvC = CVCcos t.

cos sinC C C

dq di CV t CV t

dt dt cos( / 2)C Ci CV t

The AC current through a capacitor leads the capacitor voltage by /2 rad or 900.


Capacitor AC Circuits (2)Capacitor AC Circuits (2)

The AC current through a capacitor leads the capacitor voltage by /2 rad or 900. The phasors for vC and iC are perpendicular, with the iC phasor ahead of the vC phasor.

This is analogous to the behavior of the position and velocity of a mass-and-spring harmonic oscillator.


We can then use a form of Ohm’s Law to relate the peak voltage VC, the peak current IC, and the capacitive reactance XC in an AC circuit:

Capacitive ReactanceCapacitive Reactance For AC circuits we can define a resistance-like quantity, measured in ohms, for capacitance. It is called the capacitive reactance XC:

1 1

2CX C f C

and CC C C C

C

VI V I X

X


QuestionQuestion

The instantaneous value of the emf E represented by this phasor is:

(a)Increasing;(b) Decreasing;(c) Constant;(d) It is not possible to tell without knowing t.


Example: Capacitive Reactance

Example: Capacitive Reactance

What is the capacitive reactance of a 0.10 F capacitor at a 100 Hz audio frequency and at a 100 MHz FM radio frequency?

-1 -7

1(100 Hz) 15,900

2 (100 s )(1.0 10 F)CX

8 -1 -7

1(100 MHz) 0.0159

2 (1.0 10 s )(1.0 10 F)CX


Example: Capacitive Current

Example: Capacitive Current

A 10 F capacitor is connected to a 1000 Hz oscillator with a peak emf of 5.0 V. What is the peak current in the capacitor?

-1 -5

1(1000 Hz) 15.9

2 (100 s )(1.0 10 F)CX

(5.0 V)0.314 A

(15.9 )C

CC

VI

X


Voltage DividersVoltage Dividers

rVout

The circuit indicates a potentiometer, a resistor with a sliding contact. The overall resistance of the unit is R, while the resistance from the sliding tap to the bottom is r. What is the voltage Vout delivered between the output terminals?

/I RE out

rV I r

R E

Thus, the potentiometer divides the input voltage and delivers some fraction of it proportional to r/R. This is a voltage divider.


Analyzing an RC CircuitAnalyzing an RC Circuit

Draw the current vector I at some arbitrary angle. All elements of the circuit will have this current.

Draw the resistor voltage VR in phase with the current. Draw the capacitor voltage VC 900 behind the current. Make sure all phasor lengths scale properly.

Draw the emf E0 as the vector sum of VR and VC. The angle of this phasor is t, where the time-dependent emf is E0 cost.

The phasors VR and VC form the sides of a right triangle, with E0 as the hypotenuse. Therefore,E0

2 = VR

2+VC2.


RC Filter CircuitsRC Filter Circuits

Now consider a circuit that includes both a resistor and a capacitor. Because the capacitor voltage VC and the resistor voltage VR are 900 apart in the phasor diagram, they must be added like the sides of a right triangle:

2 2 2 2 20

22 2 2 2 2

( ) ( )

( )

C R C

C

V V IR IX

R X I R C I

E

0

22I

R C

E 0

22R

RV IR

R C

E

0

22

/c C

CV IX

R C

E


Frequency DependenceFrequency Dependence

0

22R

RV IR

R C

E

0

22

/c C

CV IX

R C

E

Define the crossover frequency where VR=VC as C:1

C RC

0 0At / 2 0.707C R CV V E E


Filters and TransmissionFilters and Transmission An RC filter is a circuit that passes a signal with attenuation of some frequencies. Define the transmission of an RC filter asT = vout/vin with C = 1/(RC):

LoPass 2 2 2

/1/( )

( ) 1 ( / )C

C

CT

R C

HiPass 2 2 2

1

( ) 1 ( / )C

RT

R C

High PassLow Pass

Cross-over Point

Note log scale


Example: Designing a Filter

Example: Designing a Filter

For a science project you have built a radio to listen to AM radio broadcasts at frequencies near 1 MHz. The basic circuit is an antenna, which produces a very small oscillating voltage when it absorbs energy from an electromagnetic wave, and an amplifier. Unfortunately, your neighbor’s short wave broadcast at 10 MHz interferes with your reception. You decide to place a filter between the antenna and the amplifier. You have a 500 pF capacitor. What frequency should you select for the filter’s cross over frequency? What value of resistance should be used in the filter?

1 2 (1.0 MHz)(10.0 MHz) 3.16 MHzCf f f

6 -1 10

1 1 1100

2 2 (3.16 10 s )(5.0 10 F)C C

RC f C

Tlow ( 1) = T high ( 2)


QuestionQuestion

Which of these RC filter circuits has the largest cross-over frequency C?


AC Inductor CircuitsAC Inductor Circuits Consider an AC current iR through an inductor. The changing current produces an instantaneous inductor voltage vL. L

L

div L

dt

If the inductor is connected in an AC circuit as shown, then Kirschoff’s loop law tells us that:

0soruce L LV V v =E 0( ) cos Lt t v E E

In the phasor diagram, the inductor current iL lags the voltage vL by 900, so that iL peaks T/4 later than vL.

cosL LL

v Vdi dt tdt

L L

cos sin cos cos2 2

L L LL L

V V Vi tdt t t I t

L L L


We can then use a form of Ohm’s Law to relate the peak voltage VL, the peak current IL, and the inductive reactance XL in an AC circuit:

Inductive ReactanceInductive Reactance For AC circuits we can define a resistance-like quantity, measured in ohms, for inductance. It is called the inductive reactance XL:

2LX L f L

and LL L L L

L

VI V I X

X


Example: Current and Voltage

of an Inductor

Example: Current and Voltage

of an Inductor A 25 H inductor is used in a circuit thatis driven at 100 kHZ. The current throughthe inductor reaches a peak value of 20 mAat t=5.0 s. What is the peak inductor voltage, and when, closest to t=5.0 s, does it occur?

5 -1 52 (1.0 10 s )(2.5 10 H) 16 LX L

2(2.0 10 A)(16 ) 0.320 VL L LV I X

The voltage peaks ¼ cycle before the current, which peaks at 5 s. For f = 100 kHz, T = 10 s, so T/4 = 2.5 s. Therefore, the voltage peaks at t =(5.0-2.5) s = 2.5 s.


Analyzing an LRC CircuitAnalyzing an LRC Circuit

Draw the current vector I at some arbitrary angle. All elements of the circuit will have this current.

Draw the resistor voltage VR in phase with the current. Draw the inductor and capacitor voltages VL and VC 900 before and behind the current, respectively.

Draw the emf E0 as the vector sum of VR and VL-VC. The angle of this phasor is t, where the time-dependent emf is E0 cost.

The phasors VR and VL-VC form the sides of a right triangle, with E0 as the hypotenuse. Therefore, E0

2

= VR2+(VL-VC)2.


The Series RLC CircuitThe Series RLC Circuit The figure shows a resistor, inductor, and capacitor connected in series. The same current i passes through all of the elements in the loop. From Kirchhoff’s loop law, E = vR + vL + vC.

Because of the capacitive and inductive elements in the circuit, the current i will not in general be in phase with E, so we will have i = I cos(t-) where is the phase angle between current and voltage. If VL>VC then the current i will lag E and >0.2 2 2 2 2 2

0 ( ) ( )R L C L CV V V R X X I E

0 0

2 2 2 2( ) ( 1/ )L C

IR X X R L C

E E


Impedance and Phase Angle

Impedance and Phase Angle

We can define the impedance Z of the circuit as:

2 2

2 2

( )

( 1/ )

L CZ R X X

R L C

Then /I ZE

From the phasor diagram ,we see that the phase angle f of the current is given by:

tan L CL C

R

I X XV V

V IR

1 1 1/tan tanL CX X L C

R R

0 cosRV E


ResonanceResonance0

2 2( 1/ )I

R L C

E

The current I will be a maximum when L=1/C. This defines the resonant frequency of the system 0:

0

1

LC

0

2222 01

I

R L

E


Example:Designing a Radio Receiver

Example:Designing a Radio Receiver

An AM radio antenna picks up a 1000 kHz signal with a peak voltage of 5.0 mV. The tuning circuit consists of a 60 H inductor in series with a variable capacitor. The inductor coil has a resistance of 0.25 , and the resistance of the rest of the circuit is negligible.

(a)To what capacitance should the capacitor be tuned to listen to this radio station.

(b) What is the peak current through the circuit at resonance?(c) A stronger station at 1050 kHz produces a 10 mV antenna

signal. What is the current in the radio at this frequency when the station is tuned to 1000 kHz.

2 -6 6 20

-10

1 1

(60 10 H)(6.28 10 rad/s)

4.23 10 F 423 pF

CL

so L CX X Z R 3

1 0 / (5.0 10 V) /(0.25 ) 0.020 A 20 mAI R E

' ' 396 ' 1/ ' 358 L CX L X C

02 2 2

'0.26 mA

( ' ')L C

IR X X

E

0 1/ 1000 kHz = 1 MHzLC




Lecture HW is due tonight at 10 PM.

Midterm Exam 3 is this coming Friday. Covers explicitly everything not covered in the previous exam…and assumes understanding of all previous material.

Lecture question and lab question are multiple-choice, tutorial is long answer.

LC and AC Circuits

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