Today’s agendum: Electric Current.

2

Today’s agendum:

Electric Current.You must know the definition of current, and be able to use it in solving problems.

Current Density.You must understand the difference between current and current density, and be able to use current density in solving problems.

Ohm’s “Law” and Resistance.You must be able to use Ohm’s “Law” and electrical resistance in solving circuit problems.

Resistivity.You must understand the relationship between resistance and resistivity, and be able to calculate resistivity and associated quantities.

Temperature Dependence of Resistivity.You must be able to use the temperature coefficient of resistivity to solve problems involving changing temperatures.

3

Electric Current

Definition of Electric Current

The average current that passes any point in a conductor during a time t is defined as

where Q is the amount of charge passing the point.

One ampere of current is one coulomb per second:

av

QI

t

The instantaneous current isdQ

I = .dt

.1C

1A=1s

4

Here’s a really simple circuit:

+-

current

Don’t try that at home! (Why not?)

The current is in the direction of flow of positive charge… …opposite to the flow of electrons, which are usually the charge carriers.

Currents in battery-operated devices are often in the milliamp range: 1 mA = 10-3 A.

“m” for milli—another abbreviation to remember!

5

+ -

current electrons

An electron flowing from – to +

“Conventional” refers to our convention, which is always to consider the effect of + charges (for example, electric field direction is defined relative to + charges).

An electron flowing from – to + gives rise to the same “conventional current” as a proton flowing from + to -.

6

“Hey, that figure you just showed me is confusing.

+ -

current electrons

Good question.

“Hey, that figure you just showed me is confusing. Why don’t electrons flow like this?”

7

Chemical reactions (or whatever energy mechanism the battery uses) “force” electrons to the negative terminal. The battery won’t “let” electrons flow the wrong way inside it. So electrons pick the easiest path—through the external wires towards the + terminal.

Of course, real electrons don’t “want” anything.

Electrons “want” to get away from - and go to +.

+ -

current electrons

8

Note!

Current is a scalar quantity, and it has a sign associated with it.

In diagrams, assume that a current indicated by a symbol and an arrow is the conventional current.

I1

If your calculation produces a negative value for the current, that means the conventional current actually flows opposite to the direction indicated by the arrow.

9

Example: 3.8x1021 electrons pass through a point in a wire in 4 minutes. What was the average current?

av

Q NeI

t t

21 19

av

3.8 10 1.6 10I

4 60

avI 2.53A

Don’t worry, it gets “better” later.

“This is a piece of cake so far!”

10

Today’s agendum:




Resistivity.You must understand the relationship between resistance and resistivity, and be able to use calculate resistivity and associated quantities.


11

Current Density

When we study details of charge transport, we use the concept of current density.

Current density is the amount of charge that flows across a unit of area in a unit of time.

+

+

Current density: charge per area per time.

12

A current density J flowing through an infinitesimal area dA produces an infinitesimal current dI.

dA

J

dI J dA

The total current passing through A is just

surface

I J dA

Current density is a vector. Its direction is the direction of the velocity of positive charge carriers.

Current density: charge per area per time.

13

surface surface

II J dA J dA JA J

A

If J is constant and parallel to dA (like in a wire), then

A

J

Now let’s take a “microscopic” view of current.

Avvt

q

If n is the number of charges per volume, then the number of charges that pass through a surface A in a time t is

numbervolume n v t A

volume

14

The total amount of charge passing through A is the number of charges times the charge of each.

Avvt

qQ nqv t A

Divide by t to get the current…

QI nqv A

t

…and by A to get J:

J nqv .

15

To account for the vector nature of the current density,

J nqv

and if the charge carriers are electrons, q=-e so that

eJ n e v.

The – sign demonstrates that the velocity of the electrons is antiparallel to the conventional current direction.

16

Currents in Materials

Metals are conductors because they have “free” electrons, which are not bound to metal atoms.

In a cubic meter of a typical conductor there roughly 1028 free electrons, moving with typical speeds of 1,000,000 m/s.

But the electrons move in random directions, and there is no net flow of charge, until you apply an electric field...

17

-

E electron “drift” velocity

The voltage accelerates the electron, but only until the electron collides with a “scattering center.” Then the electron’s velocity is randomized and the acceleration begins again.Some predictions made by this theory are off by a factor or 10 or so, but it was the best we could do before quantum mechanics.

We will see later in this course that the electron would follow curved trajectories, but the idea here is still valid.

just one electron shown, for simplicity

inside a conductor

18

Even though the details of the model on the previous slide are wrong, it points us in the right direction, and works when you take quantum mechanics into account.

In particular, the velocity that should be used in

J n q v.

is not the charge carrier’s velocity (electrons in this example).

Instead, we should the use net velocity of the collection of electrons, the net velocity caused by the electric field.

This “net velocity” is like the terminal velocity of a parachutist; we call it the “drift velocity.”

dJ n q v .

19

It’s the drift velocity that we should use in our equations for current and current density in conductors:

dJ n q v

dI nqv A

d

Iv

nqA

20

Example: the 12-gauge copper wire in a home has a cross-sectional area of 3.31x10-6 m2 and carries a current of 10 A. The conduction electron density in copper is 8.49x1028 electrons/m3. Calculate the drift speed of the electrons.

d

Iv

nqA

d

Iv

neA

d 28 -3 19 6 2

10C/sv

(8.49 10 m )(1.60 10 C)(3.31 10 m )

4dv 2.22 10 m/s

21

Today’s agendum:






22

Resistance

The resistance of a material is a measure of how easily a charge flows through it.

Resistance: how much “push” is needed to get a given current to flow. V

RI

The unit of resistance is the ohm:1 V

1 .1 A

Resistances of kilo-ohms and mega-ohms are common:3 61 k 10 , 1 M =10 .

23

This is the symbol we use for a “resistor:”

All wires have resistance. Obviously, for efficiency in carrying a current, we want a wire having a low resistance. In idealized problems, we will consider wire resistance to be zero.Lamps, batteries, and other devices in circuits have resistance.

Every circuit component has resistance.

24

Resistors are often intentionally used in circuits. The picture shows a strip of five resistors (you tear off the paper and solder the resistors into circuits).

The little bands of color on the resistors have meaning, namely the amount of resistance.

25

Ohm’s “Law”

In some materials, the resistance is constant over a wide range of voltages.

For such materials, we write and call the equation “Ohm’s Law.”

V IR,

In fact, Ohm’s “Law” is not a “Law” in the same sense as Newton’s “Laws”…

Newton’s Laws demand; Ohm’s “Law” suggests.

… and in advanced classes you will write something other than V=IR when you write Ohm’s “Law.”

26

Today’s agendum:






27

This makes sense: a longer wire or higher-resistivity wire should have a greater resistance. A larger area means more “space” for electrons to get through, hence lower resistance.

It is also experimentally observed (and justified by quantum mechanics) that the resistance of a metal wire is well-described by

Resistivity

LR ,

A

where is a “constant” called the resistivity of the wire material, L is the wire length, and A its cross-sectional area.Notice that this relates R to a property of the material of which the wire is made.

28

L

The longer a wire, the “harder” it is to push electrons through it.

R = L / A,

The greater the resistivity, the “harder” it is to push electrons through it.

The greater the cross-sectional area, the “easier” it is to push electrons through it.

A

Resistivity is a useful tool in physics because it depends on the properties of the wire material, and not the geometry.

units of are m

29

Resistivities range from roughly 10-8 ·m for copper wire to 1015 ·m for hard rubber. That’s an incredible range of 23 orders of magnitude, and doesn’t even include superconductors (we might talk about them some time).

R = L / A

A = L / R

A = (d/2)2 geometry!

(d/2)2 = L / R

Example: Suppose you want to connect your stereo to remote speakers.(a) If each wire must be 20 m long, what diameter copper wire should you use to make the resistance 0.10 per wire.

30

(d/2)2 = L / R

d/2= ( L / R )½ don’t skip steps!

d = 2 ( L / R )½

d = 2 [ (1.68x10-8) (20) / (0.1) ]½

d = 0.0021 m = 2.1 mm

V = I R

(b) If the current to each speaker is 4.0 A, what is the voltage drop across each wire?

V = (4.0) (0.10) V = 0.4 V

31

Ohm’s “Law” Revisited

The equation for resistivity I introduced five slides back is a semi-empirical one. Here’s how we define resistivity:

E.

J

Our equation relating R and follows from the above equation.

We define conductivity as the inverse of the resistivity: 1 1

, or .

32

With the above definitions,

E J,

J E.

The “official” Ohm’s “Law”, valid for non-ohmic materials.

Cautions!

In this context: is not volume density! is not surface density!

33

Example: the 12-gauge copper wire in a home has a cross-sectional area of 3.31x10-6 m2 and carries a current of 10 A. Calculate the magnitude of the electric field in the wire.

IE J

A

8

6 2

(1.72 10 m) 10C/sE

(3.31 10 m )

2E 5.20 10 V/m

34

Today’s agendum:






35

Temperature Dependence of Resistivity

Many materials have resistivities that depend on temperature. We can model* this temperature dependence by an equation of the form 0 01 T T ,

where 0 is the resistivity at temperature T0, and is the temperature coefficient of resistivity.

*T0 is a reference temperature, often taken to be 0 °C or 20 °C. This approximation can be used if the temperature range is “not too great;” i.e. 100 °C or so.

36

Example: a carbon resistance thermometer in the shape of a cylinder 1 cm long and 4 mm in diameter is attached to a sample. The thermometer has a resistance of 0.030 . What is the temperature of the sample?

Look up resistivity of carbon, use it to calculate resistance. 5

0 3.519 10 m

This is the resistivity at 20 C.

0T 20 C L = 0.01 m r = 0.002 m

00 2

LR 0.028

r

This is the resistance at 20 C.

37

The result is very sensitive to significant figures in resistivity and .

-10.0005 C

RA(R)

L

00

1T(R) T 1

T(0.030) 122.6 C

38

Today’s agendum:

Emf, Terminal Voltage, and Internal Resistance.You must be able to incorporate all of these quantities in your circuit calculations.

Electric Power.You be able to calculate the electric power dissipated in circuit components, and incorporate electric power in work-energy problems.

39

Conservation of energy to implies that the voltage drop across circuit components in series is the sum of the individual voltage drops.

circuit components in series

C1 C2

+ -

C3

-Qa b

V2V1

Vab

V

Vab = V = V1 + V2 + V3

40

Therefore, the voltage drop across resistors in series is the sum of the individual voltage drops.

circuit components in series

Vab = V = V1 + V2 + V3

R3R2R1

+ -

V

V1 V3V2

a b

This is a consequence of conservation of energy. Use this in combination with Ohm’s “Law”, V=IR.

41

In Physics 104, whenever you work with currents in circuits, you should assume (unless told otherwise) “direct current.”

DC Currents

Current in a dc circuit flows in one direction, from + to -.We will not encounter ac circuits much in this course.

For any calculations involving household current, which is ac, assuming dc will be “close enough” to give you “a feel” for the physics.

If you need to learn about ac circuits, you’ll have courses devoted to them.

The mathematical analysis is more complex. We have other things to explore this semester.

42

We have been making calculations with voltages from batteries without asking detailed questions about the batteries. Now it’s time to look inside the batteries.

We introduce a new term – emf – in this section.

Any device which transforms a form of energy into electric energy is called a “source of emf.”

“emf” is an abbreviation for “electromotive force,” but emf does not really refer to force!

The emf of a source is the voltage it produces when no current is flowing.

emf, terminal voltage, and internal resistance

43

The voltage you measure across the terminals of a battery (or any source of emf) is less than the emf because of internal resistance.

Here’s a battery with an emf. All batteries have an “internal resistance:”

+ -a b The “battery” is

everything inside the green box.

Hook up a voltmeter to measure the emf:


everything inside the green box.Getting ready to connect the voltmeter (it’s not hooked up yet).

emf

emf is the zero-current potential difference

44

You can’t measure voltage without some (however small) current flowing, so you can’t measure emf directly.

Measuring the emf???



As soon as you connect the voltmeter, current flows.

You can only measure Vab.

I

emf

45

We model a battery as producing an emf, , and having an internal resistance r:



The terminal voltage, Vab, is the voltage you measure with current flowing. When a current I flows through the battery, Vab is related to the emf, , by

abV = ε ± I r .

r

Vab

46

Why the sign? If the battery is delivering current, the V it delivers is less than the emf, so the – sign is necessary.If the battery is being charged, you have to “force” the current through the battery, and the V to “force” the current through is greater than the emf, so the + sign is necessary.Your text writes and expects you to put the correct sign (+ or -) on the I. I’ll go along with your text, so our equation is

V = ε - I r

abV = ε - I r .

47

Useful facts:

When current passes through a battery in the direction from the - terminal toward the + terminal, the terminal voltage Vab of the battery is abV = ε - I r .

The sum of the potential changes around a circuit loop is zero. Potential decreases by IR when going through a resistor in the direction of the current and increases by when passing through an emf in the direction from the - to + terminal.

+-

I

V is ++

-

I

V is -

48

To model a battery, simply include an extra resistor to represent the internal resistance, and label the voltage source* as an emf instead of V (units are still volts):

+ -

r

*Remember, all sources of emf—not just batteries—have an internal resistance.

49

Example: a battery is known to have an emf of 9 volts. If a 1 ohm resistor is connected to the battery, the terminal voltage is measured to be 3 volts. What is the internal resistance of the battery?

+ -a b

the voltmeter’s resistance is so large that approximately zero current flows through the voltmeter

emf

terminal voltage Vabinternal resistance r

R=1 IBecause the voltmeter

draws “no” current, r and R are in series with a current I flowing through both.ε = Ir + IR

IR, the potential drop across the resistor, is just the potential difference Vab.

abV = IR

50

+ -a b

emf

R=1 I

ε = Ir + IR abV = IR

Ir = ε- IR

ε- IRr =

Iε

r = - RI

abV I =

R

ab

εRr = - R

V

ab

εr = R - 1

V

9

r = 1 - 1 = 3- 1 =23

A rather unrealistically large value for the internal resistance of a 9V battery.

51

By the way, the experiment described in the previous example is not a very good idea.

abV I =

R

3 I = = 3A

1

52

Today’s agendum:

Emf, Terminal Voltage, and Internal Resistance.You must be able to incorporate all of these quantities in your circuit calculations.

Electric Power.You be able to calculate the electric power dissipated in circuit components, and incorporate electric power in work-energy problems.

53

Last semester you defined power in terms of the work done by a force.

We’d better use the same definition this semester! So we will.

We focus here on the interpretation that power is energy transformed per time, instead of work by a force per time.

Electric Power

FF

dWP

dt

energy transformedP

time

54

However, we begin with the work aspect. We know the work done by the electric force in moving a charge q through a potential difference:

The instantaneous power, which is the work per time done by the electric force, is

i f i fdW dq V .

i f i fdW dq VP .

dt dt

The work done by the electric force in moving an infinitesimal charge dq through a potential difference is:

i f i f i fW U q V .

55

And one more thing… the negative sign means energy is being “lost.” So everybody writes

Let’s get lazy and drop the in front of the V, but keep in the back of our heads the understanding that we are talking about potential difference. Then

But wait! We defined I = dQ/dt. So

and understands that P<0 means energy out, and P>0 means energy in.

dW dqP V.

dt dt

P IV.

P IV

56

Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R.

Truth in Advertising II. Your power company doesn’t sell you power. It sells energy. Energy is power times time, so a kilowatt-hour (what you buy from your energy company) is an amount of energy.

Truth in Advertising I. The V in P=IV is a potential difference, or voltage drop. It is really a V.

57

What’s the meaning of this assumption about the current direction?

The current in your household wiring doesn’t flow in one direction, but because we haven’t talked about current other than a steady flow of charge, we’ll make the assumption. Our calculation will be a reasonable approximation to reality.

Example: an electric heater draws 15.0 A on a 120 V line. How much power does it use and how much does it cost per 30 day month if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh. For simplicity assume the current flows steadily in one direction.

58

An electric heater draws 15.0 A on a 120 V line. How much power does it use.

P IV

P 15 A 120 V 1800 W = 1.8 kW

How much does it cost per 30 day month if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh.

$17.00 cost

kWh

$0.105

day

h 3days 30kW8.1cost

59

How much energy is a kilowatt hour (kWh)?

So a kWh is a “funny” unit of energy. K (kilo) and h (hours) are lowercase, and W (James Watt) is uppercase.

1 kW 1 h 1000 W 3600 s

J1000 3600 s

s

6 = 3.6 10 J

60

How much energy did the electric heater use?

done by forceaverage

W Energy TransformedP

time time

averageEnergy Transformed P time

J 3 h used 3600 sEnergy Transformed 1800 30 days

s day h

Energy Transformed 583,200,000 Joules used

61

Energy Transformed 583,200,000 Joules used

That’s a ton of joules! Good bargain for $17. That’s about 34,000,000 joules per dollar (or 0.0000029¢/joule).OK, “used” is not an SI unit, but I stuck it in there to help me understand. And joules don’t come by the ton.

One last quibble. You know from energy conservation that you don’t “use up” energy. You just transform it from one form to another.

62

learn about lightning at

howstuffworksWhat kind of a problem is this?

You are given energy, potential difference, and time.

You need to calculate charge transferred, current, and average power. Equations for current and power are “obvious:”

Example: a typical lightning bolt can transfer 109 J of energy across a potential difference of perhaps 5x107 V during a time interval of 0.2 s. Estimate the total amount of charge transferred, the current and the average power over the 0.2 s.

avg

WP

t

avg

QI

t

http://www.howstuffworks.com/lightning.htm

63

We could calculate power right now, but let’s do this in the order requested. Besides, we can’t get current without Q, charge transferred.

We need to think in terms of energy transformations rather than work done by forces. The equation above tells us that potential energy stored in clouds can be transferred to the ground (at a different potential) by moving charge from cloud to ground. We are given energy transferred and potential difference, so we can calculate q.

“Could I think of the cloud-earth system as a giant capacitor which stores energy?”You could, except our capacitor equation U=QV/2 assumes the same charge on both plates; that’s untrue here.

U q V

64

Etransferred= Qtransferred Vif

Continuing with our energy transformation idea:

Qtransferred = Etransferred / Vif

Qtransferred = 109 J / 5x107 V

That’s a lot of charge (remember, typical charges are 10-6 C).

Qtransferred = 20 C

Once we have the charge transferred, the current is easy.

ΔQI =

Δt

20 CI = = 100 A

0.2 s

65

Average power is just the total energy transferred divided by the total time.

FF

WP =

t

transferredEP =

t

910 JP =

0.2 s

9P = 5×10 W

P = 5 GW

Holy ****, Batman. That’s the power output of five enormous power plants!

The numbers in this calculation differ substantially than the numbers in a homework problem(not necessarily assigned this semester). “This” lightning bolt carries relatively low current for a long time through a high potential difference, and transports a lot of energy.

In reality, there is no such thing as a universally-typical lightning bolt, so expect different results for different bolts.

66

(a) Rate of energy conversion.

R=4

+ -

I = 12 V

The total resistance in the circuit is 6 , so

V = I Rtotal

Energy is converted at the rate Pconverted=I=(2 A)(12 V)=24W.

Example: A 12 V battery with 2 internal resistance is connected to a 4 resistor. Calculate (a) the rate at which chemical energy is converted to electrical energy in the battery, (b) the power dissipated internally in the battery, and (c) the power output of the battery.

r=2

I = / Rtotal = 12 V / 6 = 2 A

67

(b) Power dissipated internally in the battery.

R=4

+ -

I=2A = 12 Vr=2

Pdissipated = I2r = (2 A)2 (2 ) = 8 W.

(c) Power output of the battery.

Poutput = Pconverted - Pdissipated = 24 W - 8 W = 16W.

68

(c) Power output of the battery (double-check).

R=4

+ -

I=2A = 12 Vr=2

I=2A

The output power is delivered to (and dissipated by) the resistor:

Poutput = Presistor = I2 R = (2 A)2 (4 ) = 16W.

69

VT

VHI

R

(a) Find the voltage at the point where the power wire enters your house.

VHT = IR

VT-VH = IR

VH = VT-IR

VH = (120 V) – (110 A) (0.03) = 116.7 V

Example: the electric utility company supplies your house with electricity from the main power lines at 120 V. The wire from the pole to your house has a resistance of 0.03 . Suppose your house is drawing 110 A of current…

70

(b) How much power is being dissipated in the wire from the pole to your house?

VT

VHI

R

P = IV = I2R = (V)2/R

P = I(VT-VH) = I2R = (VT-VH)2/R

P = (110 A) (120 V -116.7 V) = 363 W

or P = (110 A)2 (0.03) = 363 W

or P = (120 V – 116.7 V)2 / (0.03) = 363 W

Three different ways to solve; all will give the correct answer.

71

(c) How much power are you using inside your house?

VT

VHI

R

You need to understand that your household voltage represents the potential difference between the “incoming” and “outgoing” power lines, and the “outgoing” is at ground (0 V in this case)…except…

…because the “outgoing” power line is at 0 V, you can “accidentally” get this correct if you simply multiply the current by the voltage at the point where the power wire enters your house.

72

P = IV

(c) How much power are you using inside your house?

VT

VHI

R

P = (110 A) (116.7 V – 0 V)

P = 12840 W

You don’t want to use the P=I2R=V2/R equations because you don’t know the effective resistance of your house (although you could calculate it).

P = (110 A) (120 V) – (110 A)(3.3 V) is also a reasonable way to work this part.

73

Today’s agendum:

Resistors in Series and Parallel.You must be able to calculate currents and voltages in circuit components in series and in parallel.

Kirchoff’s Rules.You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit components that are not simply in series or in parallel.

74

Resistances in Circuits

There are “two” ways to connect circuit elements.

Series:

A B

Put your finger on the wire at A. If you can move along the wires to B without ever having a choice of which wire to follow, the circuit components are connected in series.

Truth in advertising: it is possible to have circuit elements that are connected neither in series nor in parallel. See problem 24.73 for an

example with capacitors.

75

Parallel:

A B

Put your finger on the wire at A. If in moving along the wires to B you ever have a choice of which wire to follow, the circuit components are connected in parallel.*

*Truth in advertising: actually, the circuit components are not connected in series, and may

be connected in parallel.

76

Are these resistors in series or parallel?

77

Here’s a circuit with three resistors and a battery:

R3R2R1

+ -

VI

Current flows…

…in the steady state, the same current flows through all resistors…

III

…there is a potential difference (voltage drop) across each resistor.

V1 V3V2

78

Applying conservation of energy allows us to calculate the equivalent resistance of the series resistors.

I am including the derivation in these notes, for the benefit of students who want to look at it.

79

An electric charge q is given a potential energy qV by the battery.

R3R2R1

+ -

VI

III

V1 V3V2

As it moves through the circuit, the charge loses potential energy qV1 as it passes through R1, etc.

The charge ends up where it started, so the total energy lost must equal the initial potential energy input:

qV = qV1 + qV2 + qV3 .

80

V = V1 + V2 + V3

V = IR1 + IR2 + IR3

Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in series.

R3R2R1

+ -

VI

III

V1 V3V2

qV = qV1 + qV2 + qV3

81

Req

+ -

VI

V

I

As above: V = IReq

From before: V = IR1 + IR2 + IR3

Combining: IReq = IR1 + IR2 + IR3

Req = R1 + R2 + R3

For resistors in series, the total resistance is the sum of the separate resistances.

82

We can generalize this to any number of resistors:

(resistors in series)a consequence of conservation of energy

R3R2R1

+ -

V

eq ii

R R

83

V

V

V

R3

R2

R1

+ -

VI

Current flows…

…different currents flows through different resistors…

…but the voltage drop across each resistor is the same.

I3

I1

I2

Here’s another circuit with three resistors and a battery.

84

Applying conservation of charge allows us to calculate the equivalent resistance of the parallel resistors.

I am including the derivation in these notes, for the benefit of students who want to look at it.

85

V

V

V

R3

R2

R1

+ -

VI

I3

I1

I2

A B

In the steady state, the current I “splits” into I1, I2, and I3 at point A.

I

I1, I2, and I3 “recombine” to make a current I at point B.

Therefore, the net current flowing out of A and into B is I = I1 + I2 + I3 .

1 2 31 2 3

V V VI = I = I =

R R R

Because the voltage drop across each resistor is V:

86

Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in parallel.

V

Req

+ -

VI

I

A B

IFrom above, I = I1 + I2 + I3, and

1 2 31 2 3

V V VI = I = I = .

R R R

So thateq 1 2 3

V V V V = + + .

R R R R

87

Dividing both sides by V gives

eq 1 2 3

1 1 1 1 = + + .

R R R R

We can generalize this to any number of resistors:

(resistors in parallel)a consequence of conservation of chargeieq i

1 1

R R

88

Summary:

Series A Beq i

i

R Rsame I, V’s add

Parallel A B

same V, I’s add

ieq i

1 1

R R

89

I’ll work this “conceptually.

A

B

Here’s the key to solving Physics problems: don’t bite off more than you can chew. Bite off little bite-sized chunks.

Example: calculate the equivalent resistance of the resistor “ladder” shown. All resistors have the same resistance R.

90

A

B

A hot dog. Where do you take the first bite?

91

A

B

Series

Not a “law” of physics, but sometimes helps with circuits: look for “bite-sized” chunks sticking out at one end.

92

A

B

Parallel

The new color indicates the value of the resistance is not R. In a real problem, you would calculate the “new color” resistor’s resistance.

Any more bite-sized chunks?

93

A

B

Series

94

A

B

Parallel

95

A

B

Series

96

A

B

All done!

97

3 8

8

10

6

1 9 V

The lecture notes are a “conceptual” solution. I may work the problem on the blackboard in lecture.

Example: For the circuit below, calculate the current drawn from the battery and the current in the 6 resistor.

98

In a few minutes, we will learn a general technique for solving circuit problems. For now, we break the circuit into manageable bits. “Bite-sized chunks.”

Replace the parallel combination (green) by its equivalent.

3 8

8

10

6

1 9 V

Do you see any bite-sized chunks that are simple series or parallel?

99

8 3

8

10

6

1 9 V

Replace the series combination (blue box) by its equivalent.

Any more “bite-sized chunks?” Remember that everything inside the green box is equivalent to a single resistor.

100

8 3

8

10

6

1 9 V

We are left with an equivalent circuit of 3 resistors in series, which is easy handle.

Next bite-sized chunk. Inside the blue box is “a” resistor. Replace the parallel combination (orange) by its equivalent.

101

Please make your selection...Example: two 100 light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. For which circuit will the bulbs be brighter?

1. parallel (left)

2. series (right)

102

(a) Series combination.

R2R1

+ -

IV = 24 V

Req = Ri

Req = R1 + R2

V = I Req

V = I (R1 + R2)

I = V / (R1 + R2) = 24 V / (100 + 100 ) = 0.12 A

Example: two 100 light bulbs are connected (a) in series and (b) in parallel to a 24 V battery. What is the current through each bulb and what is the equivalent resistance of each circuit?

103

The same current of 0.12 A flows through each bulb.

The equivalent resistance is

Req = R1 + R2

Req = 100 + 100 = 200 .

(b) Parallel combination.

V

V

R2

R1

+ -

V = 24 VI

I2

I1

I

ieq i

1 1 =

R R

eq 1 2

1 1 1 = +

R R R

104

V = I Req

1 2

IV=

1 1 +

R R

1 2

1 1I = V +

R R

1 1I = 24V +

100 Ω 100 Ω

200I = 24 = 0.48 A

10000

105

2

eq

1 1 1 200 Ω= + =

R 100 Ω 100 Ω 10000 Ω

The equivalent resistance is

eqR = 50 Ω

For which of the two circuits above would the bulb(s) be brighter…

106

To answer the question, we must calculate the power dissipated in the bulbs for each circuit. The more power “consumed,” the brighter the bulb.

In both circuits, the bulbs are identical and have identical currents passing through them. We pick either bulb for the calculation.

Series circuit: we know the resistance and current through each bulb, so we use:

P = I2R

P = (0.12 A)2 (100 )

P = 1.44 W

107

Parallel circuit: we know the resistance and voltage drop across each bulb, so we use:

P = V2 / R

P = (24 V)2 / ( 100 )

Compare: Pseries = 1.44 W

Pparallel = 5.76 W

The bulbs in parallel are brighter.

P = 5.76 W

108

This is what you see if you connect 40 W bulbs directly to a 120 V outlet. (DO NOT TRY AT HOME.)

Off On

109

Today’s agendum:

Resistors in Series and Parallel.You must be able to calculate currents and voltages in circuit components in series and in parallel.

Kirchhoff’s Rules.You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit components that are not simply in series or in parallel.

110

No, it is pronounced “KEERK-HOFF’s” rules. The ch sounds like “k,” not like “ch.” [Means church yard, cemetery.]Analyze this circuit for me, please. Find the currents I1, I2, and I3.

1 1 = 85 V

1 2 = 45 V

20

30

40 a

b cd

eg f

h

I3

I2

I1

Kirchhoff’s Rules

111

1 1 = 85 V

1 2 = 45 V

20

30

40 a

b cd

eg f

h

I3

I2

I1

I see two sets of resistors in series.This. And this.You know how to analyze those.

Further analysis is difficult. For example, series1 seems to be in parallel with the 30 resistor, but what about 2? You don’t know how to analyze that combination.

series1

series2

112

A new technique is needed to analyze this, and far more complex circuits.

Kirchhoff’s Rules

Kirchhoff’s Junction Rule: at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction. Also called Kirchhoff’s First Rule.*

Kirchhoff’s Loop Rule: the sum of the changes of potential around any closed path of a circuit must be zero. Also called Kirchhoff’s Second Rule.**

*This is just conservation of charge: charge in = charge out.**This is just conservation of energy: a charge ending up where it started out neither gains nor loses energy (Ei = Ef ).

113

Kirchhoff’s Rules

Starting Equations

at any junction I =0

around any closed loopV=0

simple… but there are details to worry about…

114

Solving Problems with Kirchhoff’s Rules

If this were Physics 103, you would have a procedure for circuit problems.

Procedure for Circuit Problems1. Draw the circuit.

2. Label + and – for each battery (the short side is -).3. Label the current in each branch of the circuit with a symbol and an arrow. You may choose whichever direction you wish for the arrow.

4. Apply Kirchoff’s Junction Rule at each junction. The direction of the current arrows tell you whether current is flowing in (+) or out (-).

Step 4 will probably give you fewer equations than variables. Proceed to step 5 go get additional equations.

115

5. Apply Kirchhoff’s Loop Rule for as many loops as necessary to get enough equations to solve for your unknowns. Follow each loop in one direction only—your choice.

5a. For a resistor, the sign of the potential difference is negative if your chosen loop direction is the same as the chosen current direction through that resistor; positive if opposite.

5b. For a battery, the sign of the potential difference is positive if your chosen loop direction moves from the negative terminal towards the positive; negative if opposite.

6. Collect equations, solve, and check results.

5a. Resistor: IV is -

+-5b. Battery:

loop

V is +

116

1 1 = 85 V

1 2 = 45 V

20

40 a

b cd

eg f

h

I3

I2

I1

Back to our circuit: we have 3 unknowns (I1, I2, and I3), so we will need 3 equations. We begin with the junctions.Junction a: I3 – I1 – I2 = 0 --eq. 1

Junction d: -I3 + I1 + I2 = 0

Junction d gave no new information, so we still need two more equations.

30

da

117

1 1 = 85 V

1 2 = 45 V

20

40 a

b cd

eg f

h

I3

I2

I1

There are three loops.

Loop 1. Loop 2. Loop 3.

Any two loops will produce independent equations. Using the third loop will provide no new information.

30

118

Reminders:

I

loop

V is -+-

loop

V is +

The “green” loop (a-h-d-c-b-a):

(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0

- 30 I1 + 45 - 41 I3 = 0 --eq. 2

The “blue” loop (a-b-c-d-e-f-g):

(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 041 I3 -130 + 21 I2 = 0 --eq. 3

Three equations, three unknowns; the rest is “algebra.”Make sure to use voltages in V and resistances in . Then currents will be

in A.

5

119

Collect our three equations:

I3 – I1 – I2 = 0

- 30 I1 + 45 - 41 I3 = 0

41 I3 -130 + 21 I2 = 0

Rearrange to get variables in “right” order:

– I1 – I2 + I3 = 0

- 30 I1 - 41 I3 + 45 = 0

21 I2 + 41 I3 -130 = 0

Use the middle equation to eliminate I1:

I1 = (41 I3 – 45)/(-30)

There are many valid sets of steps to solving a system of equations. Any that works is acceptable.

120

Two equations left to solve:

– (41 I3 – 45)/(-30) – I2 + I3 = 0

21 I2 + 41 I3 -130 = 0

Might as well work out the numbers:

1.37 I3 – 1.5 – I2 + I3 = 0

21 I2 + 41 I3 -130 = 0

– I2 + 2.37 I3 – 1.5 = 0

21 I2 + 41 I3 -130 = 0

Multiply the top equation by 21:

– 21 I2 + 49.8 I3 – 31.5 = 0

21 I2 + 41 I3 -130 = 0

121

Add the two equations to eliminate I2:

– 21 I2 + 49.8 I3 – 31.5 = 0

+ ( 21 I2 + 41 I3 -130 = 0 )

90.8 I3 – 161.5 = 0

Solve for I3: I3 = 161.5 / 90.8

I3 = 1.78

Go back to the “middle equation” two slides ago for I1:

I1 = (41 I3 – 45)/(-30)

I1 = - 1.37 I3 + 1.5

I1 = - (1.37) (1.78) + 1.5

I1 = - 0.94

122

Go back two slides to get an equation that gives I2:

– I2 + 2.37 I3 – 1.5 = 0

I2 = 2.37 I3 – 1.5

I2 = (2.37) (1.78) – 1.5

I2 = 2.72

Summarize answers so your lazy professor doesn’t have to go searching for them and get irritated (don’t forget to show units in your answer):

I2 = 2.72 A

I3 = 1.78 A

I1 = - 0.94 A

Are these currents correct? How could you tell? We’d better check our results.

123

I3 – I1 – I2 = 0

- 30 I1 + 45 - 41 I3 = 0

41 I3 -130 + 21 I2 = 0

I2 = 2.72 A

I3 = 1.78 A

I1 = - 0.94 A

1.78 – (-0.94) – 2.72 = 0

- 30 (-0.94) + 45 - 41 (1.78) = 0.22 ?

41 (1.78) -130 + 21 (2.72) = 0.10 ?

Are the last two indication of a mistake or just round off error? Recalculating while retaining 2 more digits gives I1=0.933, I2=2.714, I3=1.7806, and the last two results are 0.01 or less round off was the culprit.

124

Today’s agendum:

RC Circuits.You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.

Leftovers.Optional (not for test) material, if time permits.

125

RC Circuits

RC circuits contain both a resistor and a capacitor (duh).Until now we have assumed that

charge is instantly placed on a capacitor by an emf.The approximation resulting from this assumption is reasonable, provided the resistance between the emf and the capacitor being charged/discharged is small.

If the resistance between the emf and the capacitor is not small, then the charge on the capacitor does not change instantaneously.

Q

t

Q

t

126

Switch open, no current flows.

Charging a Capacitor

R

switch

C

t<0

Close switch, current flows.

t>0

I

Apply Kirchoff’s loop rule* (green loop) at the instant charge on C is q.

*Convention for capacitors is “like” batteries: negative if going across from + to -.

εq

- - IR=0C

This equation is deceptively complex because I depends on q and both depend on time.

127

When t=0, q=0 and I0=/R.

Limiting Cases

R

switch

C

When t is “large,” the capacitor is fully charged, the current “shuts off,” and Q=C.

Iε

q- - IR=0C

128

εq

- - IR=0C

Math:

ε qI = -

R RC

ε ε εdq q C q C -q= - = - =

dt R RC RC RC RC

ε

dq dt=

C -q RC

ε

dq dt= -

q-C RC

129

More math:0 ε q t

0

dq dt= -

q-C RC

0

ε tq

0

1ln q-C = - dt

RC

ε

ε q-C t

ln = --C RC

ε

ε

t-RCq-C

= e-C

ε εt

-RCq-C =-C e

130

Still more math: ε εt

-RCq=C -C e

ε

t-RCq=C 1-e

t-RCq t =Q 1-e

ε ε εt t t- - -RC RC RCdq C C

I t = = e = e = edt RC RC R

RC is the “time constant” of the circuit; it tells us “how fast” the capacitor charges and discharges.

131

Charging a capacitor; summary:

t-RCq t =Q 1-e ε t

-RCI t = e

R

Charging Capacitor

0

0.002

0.004

0.006

0.008

0.01

0 0.2 0.4 0.6 0.8 1

t (s)

q (

C)

Charging Capacitor

0

0.01

0.02

0.03

0.04

0.05

0 0.2 0.4 0.6 0.8 1

t (s)

I (A

)

Sample plots with =10 V, R=200 , and C=1000 F.RC=0.2 s

132

In a time t=RC, the capacitor charges to Q(1-e-1) or 63% of its capacity…

Charging Capacitor

0

0.002

0.004

0.006

0.008

0.01

0 0.2 0.4 0.6 0.8 1

t (s)

q (

C)

Charging Capacitor

0

0.01

0.02

0.03

0.04

0.05

0 0.2 0.4 0.6 0.8 1

t (s)

I (A

)

RC=0.2 s

…and the current drops to Imax(e-1) or 37% of its maximum.

=RC is called the time constant of the RC circuit

133

Capacitor charged, switch open, no current flows.

Discharging a Capacitor

R

switch

C

t<0

Close switch, current flows.

t>0

I

Apply Kirchoff’s loop rule* (green loop) at the instant charge on C is q.

*Convention for capacitors is “like” batteries: positive if going across from - to +.

q - IR=0

C

+Q

-Q-q

+q

134

q - IR=0

CMath:

qIR=

C

dq

I =dt

dq dt= -

q RC

dq q-R =

dt C

negative because charge decreases

135

More math:

q t t

Q 0 0

dq dt 1= - = - dt

q RC RC

tq

Q 0

1ln q = - dt

RC

q tln = -

Q RC

t-RCq(t)=Q e

t t- -RC RC

0

dq QI(t)=- = e =I e

dt RC

same equation as for charging

136

Disharging a capacitor; summary:

ε t-RCI t = e

R

Sample plots with =10 V, R=200 , and C=1000 F.RC=0.2 s

t-RCq(t)=Q e

Discharging Capacitor

0

0.002

0.004

0.006

0.008

0.01

0 0.2 0.4 0.6 0.8 1

t (s)

q (

C)


0

0.01

0.02

0.03

0.04

0.05

0 0.2 0.4 0.6 0.8 1

t (s)

I (A

)

137


0

0.01

0.02

0.03

0.04

0.05

0 0.2 0.4 0.6 0.8 1

t (s)

I (A

)


0

0.002

0.004

0.006

0.008

0.01

0 0.2 0.4 0.6 0.8 1

t (s)

q (

C)

In a time t=RC, the capacitor discharges to Qe-1 or 37% of its capacity…

RC=0.2 s

…and the current drops to Imax(e-1) or 37% of its maximum.

138

Today’s agendum:

RC Circuits.You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.

Leftovers.Optional (not for test) material, if time permits.

139

EMF’s in Series and in Parallel: Charging a Battery

If you put batteries in series the “right way,” their voltages add:

+ =6 V 3 V 9 V

+ =6 V 3 V 3 V

If you put batteries in series the “wrong way,” their voltages add algebraically:

magnitudes only

chosen loop direction -6 V +3 V -3 V algebraically, using chosen loop direction

Real quickly… some leftover circuits material…if time permits…

140

Why would you want to connect emf’s in series?

More voltage! Brighter flashlights, etc. Chemical reactions in batteries yield a fixed voltage. Without changing the chemical reaction (i.e., inventing a new battery type), the only way to change voltage is to connect batteries in series.

Why would you want to connect batteries in series the “wrong” way?

You could connect a source of emf – like the alternator in your car – so that it charges a battery.

Rechargeable batteries use an ac to dc converter as a source of emf for recharging.

141

Go to www.howstuffworks.com to see how batteries work.They even expose the secret of the 9 volt battery!

Click on the picture above only if you are mature enough to handle this graphic exposé.

http://www.howstuffworks.com/

http://www.howstuffworks.com/battery.htm

142

Go to www.howstuffworks.com to see how batteries work.They even expose the secret of the 9 volt battery!

Shocking!

Six 1.5 V batteries in series!

http://www.howstuffworks.com/

http://www.howstuffworks.com/battery.htm

143

Could you connect batteries (or sources of emf) in parallel?Sure!

3 V

3 V

You would still have a 3 V voltage drop across your resistor, but the two batteries in parallel would “last” longer than a single battery.

You could use Kirchhoff’s rules to analyze this circuit and show that Vab = 3 V.

a b

144

d. Energy storage in magnetic fieldsi) RL-circuit

0

0

0

dt

dI

R

LI

R

EMFdt

dILIREMF

EMFIREMF L

t

L

R

eR

EMFI 1

02 dt

dILIRIEMFI

2

00

2

1LIU

LIdIdU

dt

dILI

dt

dU

m

IU

m

m

m

AnL o2

nIB o

AB

LIUo

m 22

1 22

22

2B

B

A

Uu

o

mm

Specifically for a coil,

, and inside the coil (a long coil)

. This allows us to substitute in Um for the I in terms of B.

The energy density is the energy divided by the volume enclosed inside the coil.

145

ii) LC-circuitWe assume that there is no resistance in the circuit. Initially, the capacitor is charged with Qmax.

The total energy in the circuit is U = UC + Um. At t = 0, I = 0 and .C

QU

2

2max

When the switch is closed, the capacitor begins to discharge. The current increases to a maximum Imax at t = t’, at which time U

QLCdt

Qd

dt

QdL

C

Q

dt

dIL

C

Q

1

0

0

2

2

2

2

LC

tQQ cosmax

LC

t

LC

Q

dt

dQI sinmax

LC

1

146

iii) RLC-circuit

RIdt

dU 2

tL

R

LCeQQ

C

Q

dt

dQR

dt

QdL

L

Rt 2

12

2max

2

2

2

1cos

0

This is exactly the same equation we obtain for the damped harmonic

oscillator—a harmonic oscillator with a dissipative force, such as friction.

147

a. Alternating current & voltage

tItR

V

R

vi

tVv

sin sin

sin

maxmax

max

-1.5

-1

-0.5

0

0.5

1

1.5

0 100 200 300 400 500 600 700 800

time

cu

rre

nt

2

2

2 sin

sin

max

max

2max22

max

22max

2

VV

II

RItIRP

tRIRiP

rms

rms

i) reactance

Consider an inductor circuit.

0 sin

0

max

dt

diLtV

dt

diLv

2 sin

cos

sin

maxmax

0

max

tL

Vt

L

Vi

dttL

Vdi

L

t

LI

VX L

max

max

tXIv LL sinmax

Similarly for a capacitor, we define a capacitive reactance,

CX C

1

2 sin cos

sin

maxmax

max

tCVtCVdt

dQi

tCVQ

tXIv CC sinmax

RIV

XIV

XIV

R

LL

CC

max

max

max

148

ii) impedanceConsider an RLC circuit, with an alternating EMF.

22 sin

2 sin

2 sin

sin

sin

max

max

max

max

max

tXIv

tVv

tX

VI

tXIv

tVv

CC

R

LL

LL

ZIXXRIVVVV

VVVV

CLCLR

RCL

max22

max22

max

222max

c. TransformersConsider a circuit having two coils, a source EMF, and a load resistance as shown.

v1 = input voltage (primary)v2 = output voltage (secondary)N1 = number of turns (loops) on

primary coilN2 = number of turns on secondary

coilR = load resistance

dt

dNv

11 Riv

N

N

N

vN

dt

dNv 21

1

2

1

1222

eqRiRN

Ni

N

N

v

vR

N

Ni

v

vR

N

Niv 1

2

2

11

2

1

1

1

2

11

2

1

2

111

Today’s agendum: Electric Current.

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Transcript of Today’s agendum: Electric Current.