Physics 2113 Jonathan Dowling Lecture 35: MON 16 NOV Electrical ...

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Physics 2113 Jonathan Dowling Lecture 35: MON 16 NOV Electrical Oscillations, LC Circuits, Alternating Current II

Transcript of Physics 2113 Jonathan Dowling Lecture 35: MON 16 NOV Electrical ...

Page 1: Physics 2113 Jonathan Dowling Lecture 35: MON 16 NOV Electrical ...

Physics 2113 Jonathan Dowling

Lecture 35: MON 16 NOV Electrical Oscillations, LC Circuits, Alternating Current II

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Damped LCR Oscillator Ideal LC circuit without resistance: oscillations go

on forever; ω = (LC)–1/2

Real circuit has resistance, dissipates energy:

oscillations die out, or are “damped”

Math is complicated! Important points:

–  Frequency of oscillator shifts away from

ω = (LC)-1/2

–  Peak CHARGE decays with time constant =

–  τQLCR=2L/R

–  For small damping, peak ENERGY decays with

time constant

–  τULCR= L/R

C

R L

0 4 8 1 2 1 6 2 0 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0

E

t i m e ( s )

Umax =Q2

2Ce−RtL

U

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31.2.1. Which one of the following choices will damp oscillations in an LC circuit?

a) increase the inductance b) increase the emf c) increase the circuit resistance d) increase the capacitance

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31.2.1. Which one of the following choices will damp oscillations in an LC circuit?

a) increase the inductance b) increase the emf c) increase the circuit resistance d) increase the capacitance

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2

2

If we add a resistor in an circuit (see figure) we mustmodify the energy equation, because now energy is

being dissipated on the resistor: .

E B

RL

dU i Rdt

qU U U

= −

= + =

Damped Oscillations in an CircuitRCL

22

2 2Li dU q dq diLi i R

C dt C dt dt+ → = + = −

( )

2

2

2

/2

2

2 2

10. This is the same equation as that

of the damped harmonics osc 0, which has theillator:

The angular f

solution

re( ) c que

:

os

.bt mm

d x dxm b kxdt dt

x

dq di d q d q dqi L R qdt dt dt dt dt

t x e t

C

ω φ−

= → = → + + =

+ + =

′= +

( )2

/22

2

2ncy

For the damped circuit the solution is:

The angular freque1

( ) cos . .4

ncy

.

4

Rt L Rq

k b

RC

t Qe t

m

L

L

C L

m

ω φ ω

ω

− ′ ′ −

+

=

= =

(31-6)

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/ 2Rt LQe−

/ 2Rt LQe−

( )q tQ

Q−

( )q t ( )/ 2( ) cosRt Lq t Qe tω φ− ′= +

2

2

14R

LC Lω′ = −

The equations above describe a harmonic oscillator with an exponentially decaying

amplitude Qe−Rt /2 L. The angular frequency of the damped oscillator

′ω = 1LC

− R2

4L2 is always smaller than the angular frequency ω = 1LC

of the

undamped oscillator. If the term R2

4L2 ≪1

LC we can use the approximation ′ω ≈ω .

τ RC = RC                τ RL = L / R            τ RCL = 2L / R

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R2

4L2 ≪1

LC?

103 ≪108! ′ω = 1

LC− R2

4L2 ≅ 1LC

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t(s)

Q(t)

Q(0) / 2 = 50%

1 2 3 • • •13

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Summary •  Capacitor and inductor combination produces

an electrical oscillator, natural frequency of oscillator is ω=1/√LC

•  Total energy in circuit is conserved: switches between capacitor (electric field) and inductor (magnetic field).

•  If a resistor is included in the circuit, the total energy decays (is dissipated by R).

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Alternating Current:

To keep oscillations going we need to drive the circuit with an external emf that produces a current that goes back and forth.

Notice that there are two frequencies involved: ω at which the circuit would oscillate “naturally”. The other is ωd the driving frequency at which we drive the oscillation.

However, the “natural” oscillation usually dies off quickly (exponentially) with time. Therefore in the long run, circuits actually oscillate with the frequency at which they are driven. (All this is true for the gentleman trying to make the lady swing back and forth in the picture too).

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We have studied that a loop of wire, spinning in a constant magnetic field will have an induced emf that oscillates with time,

E = Em sin(ωd t)

That is, it is an AC generator.

Alternating Current:

AC’s are very easy to generate, they are also easy to amplify and decrease in voltage. This in turn makes them easy to send in distribution grids like the ones that power our homes.

Because the interplay of AC and oscillating circuits can be quite complex, we will start by steps, studying how currents and voltages respond in various simple circuits to AC’s.

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31.6: Forced Oscillations:

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31.9: The Series RLC Circuit, Resonance:

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Example 1 : Tuning a Radio Receiver Driven RLC With EMF Antennal

The inductor and capacitor in a car radio have one program at L = 1 mH & C = 3.18 pF. Which is the FM station?

FM radio stations: frequency is in MHz.

ω =1LC

=1

1×10−6 × 3.18 ×10−12rad/s

= 5.61×108 rad/s

f = ω2π

= 8.93×107Hz= 89.3 MHz

What is the wavelength Of the radio wave from The tower?

WRKF 89.3

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Example 1 : Tuning a Radio Receiver Driven RLC at Resonance With EMF Antenna

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31.6: Forced Oscillations:

In the power grid we make sure that all circuits are far away From any resonances!

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AC Driven Circuits:

1) A Resistor:

0=− Rvemf

vR = emf = Em sin(ωd t)

iR = vRR

= Em

Rsin(ωd t)

Resistors behave in AC very much as in DC, current and voltage are proportional (as functions of time in the case of AC), that is, they are “in phase”.

For time dependent periodic situations it is useful to represent magnitudes using Steinmetz “phasors”. These are vectors that rotate at a frequency wd , their magnitude is equal to the amplitude of the quantity in question and their projection on the vertical axis represents the instantaneous value of the quantity.

Charles Steinmetz

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AC Driven Circuits:

2) Capacitors:

vC = emf = Em sin(ωd t)

qC = Cemf = CEm sin(ωd t)

iC = dqCdt

= ωdCEm cos(ωd t)

iC = ωdCEm sin(ωd t + 900)

iC = Em

Xsin(ωd t + 900)

reactance"" 1 whereC

Xdω

=

im = Em

X

looks like i = VR

Capacitors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude 1/ωd C (this idea is true only for maximum amplitudes, the instantaneous story is more complex).

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AC Driven Circuits:

3) Inductors:

vL = emf = Em sin(ωd t)

Ldtvi

dtidLv L

LL

L∫=⇒=

iL = − Em

Lωd

cos(ωd t)

= Em

Lωd

sin(ωd t − 900)

iL = Em

Xsin(ωd t − 90

0)

im = Em

X dLX ω= where

Inductors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude wd L (this idea is true only for maximum amplitudes, the instantaneous story is more complex).

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Power Station

Transmission lines Erms =735 kV , I rms = 500 A Home

110 V

T1 T2

Step-up transformer

Step-down transformer

R = 220Ω 1000 km=ℓ

Energy Transmission Requirements

The resistance of the power line R = ρℓA

. R is fixed (220 Ω in our example).

Heating of power lines Pheat = Irms2 R. This parameter is also fixed

(55 MW in our example).Power transmitted Ptrans = ErmsIrms (368 MW in our example).

In our example Pheat is almost 15 % of Ptrans and is acceptable.

To keep Pheat small we must keep Irms as low as possible. The only way to accomplish this

is by increasing Erms. In our example Erms = 735 kV. To do that we need a device

that can change the amplitude of any ac voltage (either increase or decrease).

(31-24)

Ptrans=iV = “big” Pheat=i2R = “small” Solution: Big V!

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Thomas Edison pushed for the development of a DC power network.

George Westinghouse backed Tesla’s development of an AC power network.

The DC vs. AC Current Wars

Nikola Tesla was instrumental in developing AC networks.

Edison was a brute-force experimenter, but was no mathematician. AC cannot be properly understood or exploited without a substantial understanding of mathematics and mathematical physics, which Tesla possessed.

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The most common example is the Tesla three-phase power system used for industrial applications and for power transmission. The most obvious advantage of three phase power transmission using three wires, as compared to single phase power transmission over two wires, is that the power transmitted in the three phase system is the voltage multiplied by the current in each wire times the square root of three (approximately 1.73). The power transmitted by the single phase system is simply the voltage multiplied by the current. Thus the three phase system transmits 73% more power but uses only 50% more wire.

The Tesla Three-Phase AC Transmission System

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Against General Electric and Edison's proposal, Westinghouse, using Tesla's AC system, won the international Niagara Falls Commission contract. Tesla’s three-phase AC transmission became the World’s power-grid standard.

Niagara Falls and Steinmetz’s Turning of the Screw

Transforming DC power from one voltage to another was difficult and expensive due to the need for a large spinning rotary converter or motor-generator set, whereas with AC the voltage changes can be done with simple and efficient transformer coils that have no moving parts and require no maintenance. This was the key to the success of the AC system. Modern transmission grids regularly use AC voltages up to 765,000 volts.

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The transformer is a device that can change the voltage amplitude of any ac signal. It consists of two coils with a different number of turns wound around a common iron core.

The Transformer

The coil on which we apply the voltage to be changed is called the " " andit has turns. The transformer output appears on the second coil, which is knownas the "secondary" and has turns

P

S

NN

primary

. The role of the iron core is to ensure that the magnetic field lines from one coil also pass through the second. We assume that if voltage equal to is applied across the primary then a voltagPV e appears on the secondary coil. We also assume that the magnetic field through both coilsis equal to and that the iron core has cross-sectional area . The magnetic flux

through the primary

S

P

V

B A

Φ ( ).

The flux through the secondary ( ).

PP P P

SS S S S

d dBN BA V N Adt dt

d dBN BA V N Adt dt

Φ= → = − = −

ΦΦ = → = − = −

eq. 1

eq. 2(31-25)

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( )

( )

If we divide equation 2 by equati

.

on 1 we get:

S P

S P

PP P P P

SS S S S

SS S

P PP

d dBN BA V N Adt dt

d dBN BA V N Adt dt

dBN AV NdtdBV NN Adt

V VN N

ΦΦ = → = − = −

ΦΦ = → = − = −

−= =−

=→

eq. 1

eq. 2

The voltage on the secondary .

If 1 , we have what is known as a " " transformer.

If 1 , we have what is known as a " " transformer.

Both type

SS P

P

SS P S P

P

SS P S P

P

NV VN

NN N V VNNN N V VN

=

> → > → >

< → < → <

step up

step down

s of transformers are used in the transport of electric power over large distances.

S P

S P

V VN N

=

(31-26)

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PISI

If we close switch S in the figure we have in addition to the primary current a current in the secondary coil. We assume that the transformer is " " i.e., it suffers no losses due to heatin

P

S

II ideal,

g. Then we have: (eq. 2).

If we divide eq. 2 with eq. 1 we get:

In a step-up transformer ( ) we have that .In a step-down transformer

.

(

P

P P S S

S SP P

P S S P

PS

P S S

PS

S P S P

I N

V I V IV IV I

V N V NN

I N

I IN

N N I IN

=

= →

<

=

=

>) we have that .S P S PN I I< >

We have that:

(eq. 1).

S P

S P

S P P S

V VN N

V N V N

=

→ =

S P

S P

V VN N

=S S P PI N I N=

(31-27)

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Example, Transformer:

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31.11.2. The ac adapter for a laptop computer contains a transformer. The input of the adapter is the 120 volts from the ac wall outlet. The output from the transformer is 20 volts. What is the turns ratio of the transformer?

a) 0.17 b) 6 c) 100 d) This cannot be determined without knowing how many turns one

of the coils in the transformer has.

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31.11.2. The ac adapter for a laptop computer contains a transformer. The input of the adapter is the 120 volts from the ac wall outlet. The output from the transformer is 20 volts. What is the turns ratio of the transformer?

a) 0.17 b) 6 c) 100 d) This cannot be determined without knowing how many turns one

of the coils in the transformer has.