the last leg the ups and downs of circuits chapter 31

The Last LegThe Last LegThe Last LegThe Last LegThe Ups and Downs of CircuitsThe Ups and Downs of Circuits

Chapter 31Chapter 31

The End is Near!

• Examination #3 – Friday (April 15th)• Taxes Due – Friday (April 15th)• Watch for new WebAssigns• Final Exam is Wednesday, April

27th.• Grades will be submitted as quickly

as possible.

That’s Two Weeks. That’s Two Weeks. That’s Two Weeks.

So far we have considered

DC Circuits Resistors Capacitors Inductors

We looked at Steady State DC behaviors Transient DC behaviors.

We have not looked at sources that varied with time.

Example LR Circuit

i

0

equationcapacitor

theas form same

0

:0 drops voltageof sum

dt

dqR

C

qE

dt

diLiRE

Steady Source

Time Dependent Result:

R

L

eR

Ei LRt

constant time

)1( /

R

L

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10

Time

Vo

lts

Variable Emf Applied

emf

Sinusoidal

DC

Sinusoidal Stuff

)sin( tAemf

“Angle”

Phase Angle

Same Frequencywith

PHASE SHIFT

Different Frequencies

At t=0, the charged capacitor is connected to the inductor. What would you expect to happen??

The Math Solution:

LC

New Feature of Circuits with L and C

These circuits produce oscillations in the currents and voltages

Without a resistance, the oscillations would continue in an un-driven circuit.

With resistance, the current would eventually die out.

The Graph

Note – Power is delivered to our homes as an oscillating source (AC)

Producing AC Generator

x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x x

The Real World

A

The Flux:

tAR

emfi

tBAemf

t

BA

bulb

sin

sin

cos

AB

OUTPUT

)sin(0 tVVemf

WHAT IS AVERAGE VALUE OF THE EMF ??

Average value of anything:

Area under the curve = area under in the average box

T

T

dttfT

h

dttfTh

0

0

)(1

)(

T

h

Average Value

T

dttVT

V0

)(1

0sin1

0

0 T

dttVT

V

For AC:

So …

Average value of current will be zero. Power is proportional to i2R and is ONLY

dissipated in the resistor, The average value of i2 is NOT zero because

it is always POSITIVE

Average Value

0)(1

0

T

dttVT

V

2VVrms

RMS

2

2)(

2

2)

2(

2

1

)2

(1

0

02

0

20

0

20

0

20

220

VV

VdSin

VV

tT

dtT

SinT

TVV

dttT

SinT

VtSinVV

rms

rms

T

rms

T

rms

Usually Written as:

2

2

rmspeak

peakrms

VV

VV

Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit:

E

R

~

Power

tR

VRt

R

VRitP

tR

V

R

Vi

tVV

22

0

2

02

0

0

sin)sin()(

)sin(

)sin(

More Power - Details

R

VVV

RR

VP

R

VdSin

R

VP

tdtSinR

VP

dttSinTR

VP

tSinR

VtSin

R

VP

rms

T

T

200

20

20

2

0

22

0

0

22

0

0

22

0

22

022

0

22

1

2

1

2

1)(

2

1

)(1

2

)(1

Resistive Circuit

We apply an AC voltage to the circuit. Ohm’s Law Applies

Con

sid

er

this

cir

cuit

CURRENT ANDVOLTAGE IN PHASE

R

emfi

iRe

Alternating Current Circuits

is the angular frequency (angular speed) [radians per second].

Sometimes instead of we use the frequency f [cycles per second]

Frequency f [cycles per second, or Hertz (Hz)] f

V = VP sin (t -v ) I = IP sin (t -I )

An “AC” circuit is one in which the driving voltage andhence the current are sinusoidal in time.

v

V(t)

t

Vp

-Vp

v

V(t)

t

Vp

-Vp

V = VP sin (wt - v )Phase Term

Vp and Ip are the peak current and voltage. We also use the

“root-mean-square” values: Vrms = Vp / and Irms=Ip /

v and I are called phase differences (these determine whenV and I are zero). Usually we’re free to set v=0 (but not I).

2 2

Alternating Current Circuits

V = VP sin (t -v ) I = IP sin (t -I )

v

V(t)

t

Vp

-Vp

Vrms

I/

I(t)

t

Ip

-Ip

Irms

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.2

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.

2

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.

So V(t) = 170 sin(377t + v).Choose v=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

2

Resistors in AC Circuits

ER

~EMF (and also voltage across resistor): V = VP sin (t)Hence by Ohm’s law, I=V/R:

I = (VP /R) sin(t) = IP sin(t) (with IP=VP/R)

V and I“In-phase”

V

t

I

This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(C) the Capacitive Reactance

Capacitors in AC Circuits

E

~C Start from: q = C V [V=Vpsin(t)]

Take derivative: dq/dt = C dV/dtSo I = C dV/dt = C VP cos (t)

I = C VP sin (t + /2)

The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference).

V

t

I

V and I “out of phase” by 90º. I leads V by 90º.

I Leads V???What the **(&@ does that mean??

I

V

Current reaches it’s maximum at

an earlier time than the voltage!

1

2

I = C VP sin (t +/2)

Capacitor Example

E

~

CA 100 nF capacitor isconnected to an AC supply of peak voltage 170V and frequency 60 Hz.

What is the peak current?What is the phase of the current?

MX

f

C 65.2C

1

1077.3C

rad/sec 77.360227

Also, the current leads the voltage by 90o (phase difference).

Again this looks like IP=VP/R for aresistor (except for the phase change).

So we call XL = L the Inductive Reactance

Inductors in AC Circuits

LV = VP sin (t)Loop law: V +VL= 0 where VL = -L dI/dtHence: dI/dt = (VP/L) sin(t).Integrate: I = - (VP / L cos (t)

or I = [VP /(L)] sin (t - /2)

~

Here the current lags the voltage by 90o.

V

t

I

V and I “out of phase” by 90º. I lags V by 90º.

Phasor Diagrams

Vp

Ipt

Resistor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.

The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.

The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Resistor Capacitor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Vp Ip

t

Resistor Capacitor Inductor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

+ +i

+++

+

i

i

i

i

i

LC Circuit

time

U UB UE 12LI2 1

2q2

C

dU

dt ddt

(1

2LI2 1

2

q2

C) 0

LIdI

dtq

C

dq

dt0 L(

dq

dt)d2q

dt2q

C

dq

dt

Ld2q

dt 2

1

Cq 0

Analyzing the L-C Circuit

Total energy in the circuit:

Differentiate : N o change in energy

U UB UE 12LI2 1

2q2

C

dU

dt ddt

(1

2LI2 1

2

q2

C) 0

LIdI

dtq

C

dq

dt0 L(

dq

dt)d2q

dt2q

C

dq

dt

Ld2q

dt 2

1

Cq 0

Analyzing the L-C Circuit

Total energy in the circuit:

Differentiate : N o change in energy

U UB UE 12LI2 1

2q2

C

dU

dt ddt

(1

2LI2 1

2

q2

C) 0

LIdI

dtq

C

dq

dt0 L(

dq

dt)d2q

dt2q

C

dq

dt

Ld2q

dt 2

1

Cq 0

Analyzing the L-C Circuit

Total energy in the circuit:

Differentiate : N o change in energy

U UB UE 12LI2 1

2q2

C

dU

dt ddt

(1

2LI2 1

2

q2

C) 0

LIdI

dtq

C

dq

dt0 L(

dq

dt)d2q

dt2q

C

dq

dt

Ld2q

dt 2

1

Cq 0

Analyzing the L-C Circuit

Total energy in the circuit:

Differentiate : N o change in energy

The charge sloshes back andforth with frequency = (LC)-1/2

The charge sloshes back andforth with frequency = (LC)-1/2

tqq

qdt

qd

p

cos

022

2