lecture 21 - university of illinois · 2012. 7. 19. · physics 212 lecture 21, slide 9 i max x l i...

Physics 212 Lecture 21, Slide 1

Physics 212 Lecture 21

Resonance and power in AC circuits


Imax XL

Imax XC

Imax R

L

R

C

emax

Imax R

emax = Imax Z

Imax(XL-XC)

Imax XL

Imax XC

Imax R

w

XL = wL

XC = 1/wC

R

f XL-XC


Peak AC Problems

• “Ohms” Law for each element – PEAK values

– Vgen = Imax Z

– VResistor = Imax R

– Vinductor = Imax XL

– VCapacitor = Imax XC

• Problem A generator with peak voltage 15 volts and angular

frequency 25 rad/sec is connected in series with an 8 Henry inductor, a 0.4 mF capacitor and a 50 ohm resistor. What is the peak current through the circuit?

07

L

R

C

200 LX Lw

1100 CX Cw

22 112 L CZ R X X

AZ

VI

gen13.0max

XL

XC

R


Peak AC Problems

• “Ohms” Law for each element

– Vgen = I Z

– VResistor = I R

– Vinductor = I XL

– VCapacitor = I XC

• Typical Problem A generator with peak voltage 15 volts and angular frequency 25 rad/sec is

connected in series with an 8 Henry inductor, a 0.004 Farad capacitor and a 50 ohm resistor. What is the peak current through the circuit?

Which element has the largest peak voltage across it?

A) Generator E) All the same.

B) Inductor

C) Resistor

D) Capacitor

max maxV I X

12

L

R

C

LX Lw

22

L CZ R X X

1CX Cw

200 LX Lw

1100 CX Cw

22 112 L CZ R X X

AZ

VI

gen13.0max

XL

XC

R


Peak AC Problems

• “Ohms” Law for each element

– Vgen = I Z

– VResistor = I R

– Vinductor = I XL

– VCapacitor = I XC

• Typical Problem A generator with peak voltage 15 volts and angular frequency 25 rad/sec is

connected in series with an 8 Henry inductor, a 0.4 mF capacitor and a 50 ohm resistor. What is the peak current through the circuit?

What happens to the impedance if we decrease the angular frequency to 20 rad/sec?

A) Z increases

B) Z remains the same

C) Z decreases (XL-XC): (200-100) (160-125)

14

L

R

C

LX Lw

22

L CZ R X X

1CX Cw

XL

XC

R

XL

XC

R

Z25 Z20


Resonance

Light-bulb Demo


Resonance

R is independent of w

XL increases with w

LLX w

is minimum at resonance

22 )( CL XXRZ

XC increases with 1/w 1

CCXw

Resonance: XL = XC 0

1

LCw

10

Frequency at which voltage across inductor and capacitor cancel

Resonance in AC Circuits

frequency

Imp

ed

an

ce

R XC

Z

XL w0

Z = R at resonance


Imax XL

Imax XC

Imax R

Case 1

Imax XL

Imax XC

Imax R

Case 2

Resonance: XL = XC

Z = R

Same since R doesn't change

Checkpoint 1a

Consider two RLC circuits with identical generators and resistors. Both circuits

are driven at the resonant frequency. Circuit II has twice the inductance and 1/2

the capacitance of circuit I as shown above.

Compare the peak voltage across the resistor in the two circuits

A. VI > VII B. VI = VII C. VI < VII


Voltage in second circuit will be twice that of the first because of the 2L compared to L

Checkpoint 1b




Compare the peak voltage across the inductor in the two circuits

A. VI > VII B. VI = VII C. VI < VII Imax XL

Imax XC

Imax R

Case 1

Imax XL

Imax XC

Imax R

Case 2


The peak voltage will be greater in circuit 2 because the value of XC doubles.

Checkpoint 1c




Compare the peak voltage across the capacitor in the two circuits

A. VI > VII B. VI = VII C. VI < VII Imax XL

Imax XC

Imax R

Case 1

Imax XL

Imax XC

Imax R

Case 2


The voltage across the inductor and the capacitor are equal when at resonant frequency, so there is no lag or lead.

Checkpoint 1d




At the resonant frequency, which of the following is true?

A. Current leads voltage across the generator

B. Current lags voltage across the generator

C. Current is in phase with voltage across the generator

Imax XL

Imax XC

Imax R

Case 1

Imax XL

Imax XC

Imax R

Case 2


Z

Quality factor Q

In general

U

UQ

max2

Umax = max energy stored U = energy dissipated in one cycle at resonance


Larger Q, sharper resonance


Series LCR circuit

1. The current leads generator voltage by 45o.

2. Assume V = Vmaxsinwt What does the phasor diagram look like at t = 0?

V = Vmax sinwt V is horizontal at t = 0 (V = 0)

C

R

L V ~

(A) (B) (C) (D) X X X

L C RV V V V VL < VC if current leads generator voltage


The current leads generator voltage by 45o

C

R

L V ~

(A) decrease w (B) increase w (C) Not enough info

Original w

f

Increase w

At resonance (w0) At resonance

XL = XC

XL increases

XC decreases

How should we change w to bring circuit to resonance?

Series LCR circuit


1. The current leads generator voltage by 45o.

2. Suppose XC = 2XL at frequency w.

C

R

L V ~

f

By what factor should we increase w to bring circuit to resonance w0 ?

2C LX X2 2

0

1 1 12

2 2L

C LCw w w

w

02

w

w

02 2

w

w

02

w

w

04

w

w(A) (B) (C) (D)

Series LCR circuit


Consider the harmonically driven series LCR circuit shown. At some frequency w, XC = 2XL = R Vmax = 100 V R = 50 k

C

R

L V ~

At resonance XL = XC Z R

maxmax 0 3

100( ) 2 mA

50 10

VI

R xw

If we change w, what is the maximum current that can flow?

maxI 2 mA maxI 2 mA maxI 2 2 mA (A) (B) (C)

Series LCR circuit


L

R

C

0

1T

P t P t dtT

Average, over one cycle, of any periodic function :

P(t) varies in time. We want the average power.

Instantaneous power

P(t) = I(t)V(t)

For harmonically-varying functions like sin wt or cos wt , T = 2π/ω .

0

1T

P t V t I t dtT

Circuit element:

,

0

1sin cos 0

T

L CP t t t dtT

w w For L or C, voltage and

current are /2 out of phase:

I sin wt, V cost wt

22 2

0

1 1sin

2

T

R peak peakP t I R I t Rdt I RT

w

Average power for R is not zero since V and I are in phase.

Define root mean square (RMS)

2 2 21

2peak RMSI R I R I R

1.707

2RMS peak peakI I I

1 1cos

2 2

peak

peak peak peakI R IZ

ee f

R

f XL-XC Z

Average power dissipated in R for LCR circuit

Maximum power delivered when f = 0 … at resonance !

1cos

2peak peakPower I e f

R

f XL-XC Z

Imax w0L

Imax / w0C

Imax R = emax

w0L = 1/w0C

f = 0

At resonance,

0 max 0

max

C L

R R

LI LV VQ

V V RI R

w w


Power Transmission

• If you want to deliver 1500 Watts at 100 Volts over transmission lines w/ resistance of 5 Ohms. How much power is lost in the lines?

– Current Delivered: I = P/V = 15 Amps

– Loss from power line = I2 R = 15*15 * 5 = 1,125 Watts!

• If you deliver 1500 Watts at 10,000 Volts over the same transmission lines. How much power is lost?

– Current Delivered: I = P/V = 0.15 Amps

– Loss from power line = I2R = 0.15*0.15*5 = 0.113 Watts

DEMO


Transformer

Iron core maintains same B throughout.

Primary coil captures flux P = NP B

Secondary coil captures flux S = NS B

Faraday’s Law:

VS = d S /dt = NS B

VP = d P /dt = NP B p p

s s

V N

V N

How do we transform 10,000 V into 120 V for your toaster ?

Time-varying voltage in primary causes

time-varying B in iron-core.

Flux per turn = B (Area) = B

7200 V / 240 V

lecture 21 - university of illinois · 2012. 7. 19. · physics 212 lecture 21, slide 9 i max x l i...

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