Http:// (nz014.jpg)This relates (sort of) to a demo I’ll do later

http://www.nearingzero.net (nz014.jpg) This relates (sort of) to a demo I’ll do later.

http://www.nearingzero.net/

Today’s lecture is brought to you by…

PhysicsMan

Sometimes he appears in a homework problem!

Not to be confused with…

Electro-Man (http://www.thinkgeek.com).

Today’s agenda:

Potential Changes Around a Circuit.You must be able to calculate potential changes around a closed loop.

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

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

Examples.

In lecture 7 (the first capacitors lecture) I suggested using conservation of energy to show that the voltage drop across circuit components in series is the sum of the individual voltage drops:

circuit components in series

C1 C2

+ -

C3

a b

V2V1

Vab

V

Vab = V = V1 + V2 + V3

V3

In general, the voltage drop across resistors in series (or other circuit components) is the sum of the individual voltage drops.

circuit components in series

Vab = V = V1 + V2 + V3

R3R2R1

+ -

V

V1 V3V2

a b

You may use this in tomorrow’s homework. It “is”* on your starting equations sheet, and is a consequence of conservation of energy. Use this in combination with Ohm’s Law, V=IR.

I “derived” this in

lecture 7.

Here’s what your text means by Vab:

Vab=Va-Vb=Vba

* V = 0 around closed loop

R3R2R1

+ -

V

V1 V3V2

a b

Start at point a (or any other point) and follow the current in a clockwise path around the circuit and back to point a…

- V1 - V2 - V3 + V = 0

- IR1 - IR2 - IR3 + V = 0

I

Example: add the potential changes around the loop shown.

- V1 - V2 - V3 + V = 0- V1 - V2 - V3 + V = 0- V1 - V2 - V3 + V = 0- V1 - V2 - V3 + V = 0- V1 - V2 - V3 + V = 0

For tomorrow’s homework, your path around the circuit should go in the same direction as your guessed current.

R3R2R1

+ -

VA

V1 V3V2

a b

- V1 - V2 - V3 + VA - VB = 0

+-

VB

Again, start at point a and follow the current in a clockwise path around the circuit and back to point a…

- IR1 - IR2 - IR3 + VA - VB = 0

For tomorrow’s homework, your path around the circuit should go in the same direction as your guessed current.

Example: add the potential changes around the loop shown.

I

- V1 - V2 - V3 + VA - VB = 0- V1 - V2 - V3 + VA - VB = 0- V1 - V2 - V3 + VA - VB = 0- V1 - V2 - V3 + VA - VB = 0- V1 - V2 - V3 + VA - VB = 0- V1 - V2 - V3 + VA - VB = 0

5

+ -

9 V

a

b

To be worked at the blackboard in lecture.

10

Example: calculate I, Vab, and Vba for the circuit shown.

I

5

+ -

9 V

a

b

10

Graph the potential rises and drops in this circuit.


I = 0.6 A

= 9V

5 I = 3V

10 I = 6V

a

b

5

+ -

9 V

a

b

10 +-

6 V


To be worked at the blackboard in lecture.

I

5

+ -

9 V

a

b

10 +-

6 V


What if you guess the wrong current direction?

I

In Physics 2135, 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.

Today’s agenda:




Examples.

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 is not a force!

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

emf, terminal voltage, and internal resistance

http://www.energizer.com

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

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)

Homework hint: an ideal voltmeter would be able to measure . For example, problem 25.32.

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



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

abV = ε ± I r .

r

Vab

Not recommended for use by children under 6. Do not continue use if you experience dizziness, shortness of breath, or trouble sleeping. Do not operate heavy machinery

after using.

An extinct starting

equation.

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.

If the internal resistance is negligible, simply don’t include it!

If you are asked to calculate the terminal voltage, it is just Vab = Va – Vb, calculated using the techniques I am showing you today.

(Terminal voltage is usually expressed as a positive number, so it is better to take the absolute value of Vab.)

a b

Summary of procedures for tomorrow’s homework:

Potential increases by when current passes through an emf in the direction from the - to + terminal.

Draw the current in a circuit so that it flows from – to + through the battery. The sum of the potential changes around a circuit loop is zero.

+-

I

V is +

Potential decreases by IR when current goes through a resistor.

I

loop

V is -

Treat a battery internal resistance like any other resistor. If I flows through a battery + to -, potential decreases by .

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=0

IR, the potential drop across the resistor R, is also the potential difference Vab.

abV = IR

+ -a b

emf

R=1 I

ε- Ir - IR= 0 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.

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

abV I =

R

3 I = = 3A

1

I may do a demo on this some time.

Today’s agenda:




Examples.

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

The above equation doesn’t appear on your equation sheet, but it should appear in your brain.

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

Using the above, 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 .

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

Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R.I can’t believe it, but I got soft and put P = I2R = V2/R on your starting equation sheet.

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.

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.Skip to slide 34 for now.

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.

3 h $0.105cos t 1.8 kW 30 days

day kWh

cos t $17.00

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

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

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.

I

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

R=4

+ -

= 12 V

Calculate (a) the rate at which chemical energy is converted to electrical energy in the battery.

R=4

+ -

I = 12 Vr=2

V 0 around any closed circuit loop

Start at negative terminal of battery…

- I R2 - I R4 = 0

I = / (R2 + R4) = 12 V / 6 = 2 A

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

Calculate (b) the power dissipated internally in the battery.

R=4

+ -

I=2A = 12 Vr=2

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

Calculate (c) the power output of the battery.

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

R=4

+ -

I=2A = 12 Vr=2

Calculate (c) the power output of the battery (double-check).

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

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

*Assume zero internal resistance unless the problem suggests otherwise.

Example: a 3 volt and 6 volt battery are connected in series, along with a 6 ohm resistor. The batteries* are connected the “wrong” way (+ to + or - to -). What is the power dissipated in the resistor? In the 3 volt battery?

6

+ -

6 V

a+-

3 VI

Starting at point a...

+ 6 – 3 – 6 I = 0

I = (6 – 3) / 6 = 0.5 A

What is the power dissipated in the resistor?

6

+ -

6 V

a+-

3 VI = 0.5 A

PR = I2R = (0.5)2 (6) = 1.5 W

What is the power dissipated in the 3 volt battery?

P3V = IV = (0.5) (3) = 1.5 W

P6V = IV = (0.5) (6) = 3 W = PR + P3VNote:

Documents

Http:// (nz014.jpg)This relates (sort of) to a demo I’ll do later