3/31/2020 USF Physics 101 1

Physics 101

AC Circuits

3/31/2020 USF Physics 101 2

Agenda• Administrative matters

– EVB is still ill– Homework due today?

• AC Circuits– AC in R, L, C

• Phase Shifts• Filters

– Series LCR Circuit• Phasors

– Parallel LCR Circuit– Resonance

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AC Circuits

0( ) sin i.e. At 0, 0 and 0I t I t t I

Note that I could just as well use cos with a different ’ = + 90° or 2 rad.

sin lags cos by 90°

0 0 2 2

RMS RMS

V IV I

We will assume a sinusoidal voltage source which supplies a current

0 sin( )V V t

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R only:

Loop rule: 0V IR

0 0sin sinV I R t V t

I and V are in phase

Energy is transformed onto heat

22 RMS

RMS RMS RMS

VP IV I V I R

R

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L Only:

0

Loop rule: 0

and sin

dI dIV L V L

dt dtI I t

0 0cos cosdI

V L LI t V tdt

0

Identity: cos sin 2

so sin 90V V t

In an inductor the current lags the voltage in phase by 90°.

Alternatively, the voltage leads the current by 90°.

cos sin 0 0 0t t dt IV dt P On the average, no energy is transformed into heat

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Flow of charge impeded by back EMF as energy is stored in L

In analogy to Ohm’s Law

RMS RMS LV I X

= LX L The inductive reactance

Units:

Notes: V0 and I0 are peak values. Can also write

0 0 LV I X

V and I do not peak at the same time so V ≠ I XL at a particular time. For a resistor V = IR t.

XL = 0 for DC ( = 0)

(End of previous)

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Example: Inductive reactance of a coil: R = 1.00 L

What is current for (a) 120 VDC, (b) 120 VRMS at 60 Hz

(a)

0

120 V120 A

1.00

LX L

VI

R

(b)

2 2 (60 Hz)(0.300 H) 113

neglecting the 1 w.r.t. 113

120 V1.06 A

113

LX L fL n

VI

R

Cannot simply add R and XL. There are phase considerations. (Later)

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C only:

0

Loop rule: 0

and sin

Q QV V

C CdQ

I I tdt

000 0

0 0

( ) sin cos

1and cos cos

t t IQ t dQ I t dt t

QV I t V t

C C

0

Identity: cos sin 2

so sin 90V V t

In an capacitor the current leads the voltage in phase by 90°.

Alternatively, the voltage lags the current by 90°.

Again <P> = 0. On average, no energy → heat

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Flow of charge impeded by back EMF as energy is stored in C

In analogy to Ohm’s Law

1= CX

CThe capacitive reactance

0 0 CV I X

RMS RMS CV I XUnits:

Notes: V0 and I0 are peak values. Can also write

V and I are not in phase so V ≠ I XL at a particular time.

XC = for DC ( = 0)

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Example: Peak and RMS currents in C = 1.0 F, VRMS = 120 V for (a) f = 60 Hz and (b) 600 Khz

6

0

00 3

3

1 1 1(a) 2.7 k

2 2 60 Hz 1.0 10 F

2 2 120 V 170 V

170 V .063 A or 63 mA

2.7 10

120 V 44 mA

2.7 10

C

RMS

C

RMSRMS

C

XC fC

V V

VI

X

VI

X

5 6

00

1 1 1(b) 0.27

2 2 6.0 10 Hz 1.0 10 F

170 V 120 V 630 A 440 A

0.27 0.27

C

RMSRMS

C C

XC fC

V VI I

X X

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Filters:

High pass

Low pass

Replace Cs with Ls

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Series LCR Circuit:

D

At any time t, loop rule R L CV V V V

Continuity currents same in all elements at any time t

Consequence: 0 0 0 0 and RMS RMS RMSR L C RMS R L CV V V V V V V V

0 sinI I t everywhere in the series circuit. Because of their phase differences, the voltages add in a more complicated fashion.

2 Approaches: Complex variables

Graphical analysis, phasors

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Phasors: Represent voltages as vectors in a plane

0 sinI I t

t = 0

Length of each arrow = peak V gives phase w.r.t. I

Let this diagram rotate, angular velocity

0 sinRV I R t

0 sin2L LV I X t

0 sin2L CV I X t

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The vector sum of these voltages is the voltage across the whole circuit.

0 sinV V t

Source V is out of phase with I by

Define impedance, Z

0 0 or RMS RMSV I Z V I Z

Pythagoras

2 22 2 20 0 0 0 0 0 0

22 20 0 0

R L C L C

L C

V V V V I R I X I X

I R I X I X

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Phase angle 00 0

0 0

0

0

tan

or cos

L CL C L C

R

R

I X XV V X X

V I R R

V R

V Z

Power dissipated 2

cosRMSP I R

R Z

2 2 cos cosRMS RMS RMS RMSP I R I Z I V

cos is called the power factor

R alone: = 0, cos = 1

L or C alone: = ± 90°, cos = 0, no power dissipated

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Example: In series, R = 25.0 , L = 30 mH and C = 12.0 F. Driven by 90.0 VRMS at 500 Hz. (a) current in circuit, (b) voltmeter (RMS) reading across each element, (c) phase angle and (d) power dissipated.

6

2 2 22

2 2 500 Hz 0.0300 H = 94.2

1 1 126.5

2 2 500 Hz 12.0 10 F

25.0 94.2 26.5 72.2

L

C

L C

X L fL

XC fC

Z R X X

90.0 V(a) = = 1.25 A

72.2 RMS

RMS

VI

Z

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(b) RMS voltage across elements

1.25 A 25.0 31.2 V

1.25 A 94.2 118 V

1.25 A 26.5 33.1 V

R RMSRMS

L RMS LRMS

C RMS CRMS

V I R

V I X

V I X

Note that these do not add up to 90 V. They are not in phase. Instantaneous voltages do add up to source voltage at that instant.

(c) Phase angle25.0

cos 69.772.2

RArc

Z

(d) Power dissipated

cos 1.25 A 90.0 V cos 69.7 39.0 WRMS RMSP I V

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Parallel LCR Circuit 0 00

00

0 00 01

LL

R

CC

V VI

X L

VI

RV V

I V LX C

Phases differ by 90°

Here the voltage across each element is just the source voltage at any time t with no phase differences.

, , or 0 sinL R CV V t

The currents are not in phase but must obey the node rule at any point in time to preserve continuity.

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IR0

IC0

IL0

t

IL0 – IC0

IR0 I0

Pythagorean Theorem

220 0 0 0R L CI I I I

220 0 02

2

0 2

1 1 1

L C

L C

V V V

R X X

VR X X

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2

00 2

1 1 1 1

L C

VI

Z Z R X X

(for a parallel circuit}

Full expression for current at any time t

220 0 0 0 sinR L CI t I I I V t

IL0 – IC0

IR0 I0

1 1

tan1

L CX X

R

Note that here (parallel LCR) is actually a retardation rather than an advancement (series LCR).

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Resonance in LCR Circuits (revisited):

0

0

( ) sin

sin

D

D

I t I t

Vt

Z

D

Series 2

0 2

1 1

4

R

LC L LC

2

22 20 0

D 0

1

1 1Max. current when 0 or when

. . is a minimum

L C DD

DD

Z R I X I X R LC

LC LC

i e Z

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0

0

( ) sin

sin

D

D

I t I t

Vt

Z

D

Parallel 2

0 2

1 1

4

R

LC L LC

2 2

2 2

1 1 1 1 1 1

L C

CZ R X X R L

Parallel

Now Z is a maximum at resonance and the current goes through a minimum.

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Series

Parallel

Band pass filter

“notch” filters

Band block filter

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Example: 1040 kHz radio receiver. L = 4.0 mH, what C is needed

Use a series LCR circuit, Want maximum or resonance at 1040 kHz

3

0

2 23 3

12

1040 10 Hz 1

21 1

2 1040 10 Hz 0.004 H 2 1040 10 Hz

5.85 10 F or 5.85 pF

LC

CL