CHAPTER 22 ALTERNATING CURRENT - Physics …people.physics.tamu.edu/adair/phys202/CHAPTER 22...

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1 CHAPTER 22 ALTERNATING CURRENT Alternating Currents Phasors Reactance Impedance Power in AC Circuits R-L-C Circuits Resonance

Transcript of CHAPTER 22 ALTERNATING CURRENT - Physics …people.physics.tamu.edu/adair/phys202/CHAPTER 22...

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CHAPTER 22

ALTERNATING CURRENT

Alternating Currents

Phasors

Reactance

Impedance

Power in AC Circuits

R-L-C Circuits

Resonance

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For Resistor

Consider voltage that is sinusoidal.

� = ������

Note that V is the maximum value of voltage and v varies from 0

to a maximum of V.

Also, a current that is sinusoidal.

= �����

Note that is the maximum value of voltage and varies from 0 to

a maximum of .

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Fig 22.1

Can represent using phasors. Phasors are just tools to help

visualize.

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The

phasor, the vector from the origin to the curved line, has the

length equal to the magnitude of the current or voltage or

whatever the phasor represents.

The projection on the horizontal axis represents the

instantaneous value of the quantity.

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AVERAGE VALUES

Values for the ac quantities used to make calculations of averages

are called root-mean-square values.

They can be obtained from the max values, the amplitude, by:

�� = √2

��� = �√2

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Resistor in ac circuit with current

= �����

Instantaneous voltage across resistor is

�� = � = ������

�� = �������

��������ℎ cos ��������

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Inductor in ac Circuit with current = �����

�� = � ΔΔ�

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�� = � ΔΔ�

Current changes rapidly - Voltage large

Current doesn’t change - Voltage Zero

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It can be shown that

(Calculus by differentiating, ! " )

�� = −������ = �����(�� + &)

Where & is the phase angle. For inductor that is 900.

Inductive Reactance,

Defined as (� = ��

By analogy voltage for inductor �� = (�

Now Capacitor in ac Circuit with current = �����

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)* = +*

Therefore

Δ)*Δt = 1. ΔqΔt = 1. = 1. �����

Thus a sinusoidal curve as seen in Fig. 22.8.

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Thus the voltage curve lags the current curve by 900.

And we have (can be derived by calculus)

�0 = 1�. ����

Or

�0 = 1�. cos(�� − 903)

Or �0 = �0cos(�� − 903)

Capacitive reactance

(0 = 1�.

And �0 = (0

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IMPEDANCE

Impedance of a circuit is the sum of the resistance and reactance

in the circuit. The symbol for impedance is Z.

4 = 5�6 + (6

4 = 5�6 + ((� − (0)6

4 = 7�6 + 8�� − 1�.96

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Consider

(Across Resistor)

What is reactance of the capacitor?

(0 = 1�. = 1:2500 <�� �= > (5.0@10ABC) = 80Ω

What is the current in the circuit?

= ��� = 1.2��� :2500 <�� �= > �300Ω

= 4@10AH��� :2500 <�� �= > �

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What is the impedance?

4 = 7�6 +8�� − 1�.96

4 = 7�6 −8 1�.96

4 = 5(300Ω)6 −(80Ω)6 = 290Ω

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Power in ac circuits

Remember this about power for pure resistance

IJKL = 12 � = 1√2 1√2 � = 1√2 1√2� = �� ���

But with capacitors or inductors

MN)O = PQRSTUVW = R√Q S√QTUVW = SXYVRXYVTUVW

cosW is called The Power Factor

Also I�Z[<C����< = �\

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R-L-C Circuits

Let � = 300Ω � = 60^_ . = 0.50`C

� = 50� � = 10,000 <�� �=

What is the reactance of each component, the reactance of the

circuit and the impedance of the circuit?

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(� = �� = :10,000 <�� �= >(60@10AH) = 600Ω

(0 = 1�. = 1:10,000 <�� �= > (0.50@10ABC) = 200Ω

The reactance of the circuit is

( =(� −(0 = 600Ω − 200Ω = 400Ω

The impedance of the circuit is

4 = √�6 +(6 =5(300Ω)6 +(400Ω)6

4 = 5(9@10b) +(1.6@10c) = √2.5@10c

4 = 500Ω

The current is

= d\ = c3dc33e = 0.10f

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The phase angle &(the angle between the voltage and the current

phasor) is

& = arctan(� −(0�

(� −(0 = 600Ω − 200Ω = 400Ω

� = 300Ω

& = arctan400300 = 533

Power factor is

I = cos 533 = 0.60

Or

I = �4 = 300Ω500Ω = 0.60

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Average Power

IjKL =12 � cos& = �� ��� cos&

IjKL =12(0.10f)(50�) cos 533

IjKL = 1.5k

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Resonance

Series

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When Z is minimum

4 = 7�6 +8�� − 1�.96

Z minimum when (� = (0

�3� = lmn0

�3 = 1√�.

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Resonance

.�<<[�� = �4

At resonance 4�^�^�^ thus �^�@^�^

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Parallel

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When (� = (0

Thus again

�3 = 1√�.

Resonance for parallel circuit

4 = 7�6 +8�� − 1�.96

Thus 4 = � its maximum value.

= �4 = �� ��^�^�^��o�[