Physics - 8 - Electricity · 2020. 1. 3. · Electrical Current IB PHYSICS | UNIT 8 | ELECTRICITY...

ELECTRICITYIB PHYSICS | UNIT 8 | ELECTRICITY

Electrical CurrentIB PHYSICS | UNIT 8 | ELECTRICITY

8.1

Snap Circuits!

Challenges:1. Get two lightbulbs to light up

2. Build a circuit with two lightbulbs and a switch that turns on/off BOTH bulbs

3. Build a circuit with two lightbulbs and a switch that turns on/off ONE bulb

4. Get all five lightbulbs to light up

5. Build a circuit with five lightbulbs and a switch that turns on/off 2 bulbs

3 Volts 5 Volts 6 Volts

Quantifying Charge

Elementary Charge 1.60 × 10-19 C

Unit Coulombs C

Charge for one electron:

How many electrons per Coulomb?

Symbol 𝑞

1 C ×1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛

1.60 × 10−19 C= 𝟔. 𝟐𝟓 × 𝟏𝟎𝟏𝟖 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔

IB Physics Data Booklet

Current

The rate at which charges move through a conductor

Flow of Electrons

Symbol: Unit:

--

--

--

--

--

--

--

-

I Amperes [A]

Amperes?

Unit for Charge Coulomb [C]

Unit for Current Ampere [A]

A =C

s “Coulombs per second”

Try this…

How many electrons flow past a given point in a wire every second when the current is 3 A?

Elementary Charge = 1.6 × 10-19 C

3 A =3 C

1 s×

1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛

1.60 × 10−19 C= 𝟏. 𝟖𝟖 × 𝟏𝟎𝟏𝟗 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔

Current

Why do the electrons flow instead of protons or neutrons?

Outside of the atom

so they are more

easily transferred

Electron Drift

As electrons flow in a current they have a net velocity down the conductor called the _______________

*Click for animation*

drift speed

Electron Drift

I I

Area, A

vΔt

n, free electrons per unit volume

In a time Δt, there is a certain volume of electrons that flow through a wire

If we know the number of electrons per unit volume we can find charge

If we know the number of electrons per unit volume we can find charge

𝐼 =𝑛𝐴𝑣∆𝑡𝑞

∆𝑡= nAvq

𝐴𝑣∆𝑡 Volume

𝑛𝐴𝑣∆𝑡 # of electrons

𝑛𝐴𝑣∆𝑡𝑞 Total Charge

IB Physics Data Booklet

I → Current [A]

n → # of Electrons / m3

v → Drift Speed [m s-1]

q → Elementary Charge

A → Cross Sectional Area [m2]

[1.60 × 10-19 C]

Electron Drift – Try This

𝐼 = 𝑛𝐴𝑣𝑞Suppose the current in a 2.5 mm diameter copper wire is 1.5 A and there are 5.01026 free electrons per cubic meter. Find the drift velocity

Elementary Charge = 1.6 × 10-19 C𝐴 = 𝜋𝑟2 = 𝜋

0.0025

2

2

𝐴 = 4.9 × 10−6 m2

𝑣 =𝐼

𝑛𝐴𝑞=

1.5

(5.0 × 1026)(4.9 × 10−6)(1.60 × 10−19)

𝑣 = 0.0038 m s−1

𝐼 = 1.5 A

𝑛 = 5.0 × 1026 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑝𝑒𝑟 𝑚3

q = 1.60 × 10−19 C

Electron Drift – Try This

𝐼 = 𝑛𝐴𝑣𝑞The diagram shows a wire that has been 'pinched' such that the diameter at X is half the diameter at Y. There is a current of I flowing through Y and the drift velocity of the charge carriers at Y is v.

Which of the following are true of the current and drift velocity at X?

Drift velocity at X Current at X

a. 2v I

b. 4v I

c. 2v 2I

d. 4v 2I

*Current doesn’t change

𝑣 =𝐼

𝑛𝐴𝑞

* If r/2, then A/4 because A=πr2

Circuits

Conventional Current

When drawing a circuit, we always show the current flowing from positive to negative

Electron Flow

Conventional Current

*Ben Franklin defined

current as the flow of

positive charges

Kirchhoff’s First Law

The total current coming into a junction must equal the total current leaving the same junction

5 A

3 A

5 A

9 A

2 A 4 A

Kirchhoff’s First Law

Σ𝐼 = 0 (𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛)

5 A

3 A

5 A

9 A

2 A 4 A

Entering Junction →● Positive

Exiting Junction ●→ Negative

IB Physics Data Booklet

Try This

6 A

5 A

3 A

1 A

2 A1 A

7 A

4 A

9 A

Try This

2.0 A

0.3 A0.2 A

0.6 A

I2 = 1.1 A

I1 = 0.1 A

I3 = 2.0 A

Ohm’s Law and ResistanceIB PHYSICS | UNIT 8 | ELECTRICITY

8.2

Remember back…

What is potential energy?

Stored Energy

Voltage

Voltage is the Potential Energy ____________ between two locations

Symbol: Unit:

Which battery has more voltage?

Difference

Voltage = Potential Differencep.d.

V Volts [V]

Kirchhoff’s First Law

The total current coming into a junction must equal the total current leaving the same junction

5 A

3 A

5 A

9 A

2 A 4 A

(+5) + (-3) + (-2) = 0 (+5) + (-9) + (+4) = 0

Σ𝐼 = 0 (𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛)

Kirchhoff’s Second Law

The sum of the voltages (potential differences) provided must equal the voltages dissipated across components Σ𝑉 = 0 (𝑙𝑜𝑜𝑝)

8 V

12 V

1 V9 V

12 V

4 V 2 V

Kirchhoff’s Second Law

Σ𝑉 = 0 (𝑙𝑜𝑜𝑝)

8 V

12 V

1 V9 V

12 V

(12) + (-9) + (-1) + (-2) = 0(-8) + (12) + (-4) = 0

Negative to Positive → Positive

Positive to Negative → Negative Always Negative

Over Resistors:

Across Batteries

4 V 2 V

Resistance

How difficult it is for electrons to flow

Symbol: Unit:

Which one has more resistance for water flow?

R Ohms [Ω]

Conductors and Insulators

Conductors have a ___________ resistance

Insulators have a ___________ resistance

low

high

Resistance in a Circuit

There are many different components that act as resistors when placed in a circuit

Electrical Properties

Property What is it? Symbol Unit

Voltage Potential Difference VVolts[V]

CurrentThe rate at which the charges

move through wire IAmps

[A]

ResistanceHow hard it is for current to

flow through a conductor ROhms

[Ω]

How are they Related?

Voltage

Current

Resistance

Current

𝑉 ∝ 𝐼

𝑅 ∝ ൗ1 𝐼

How are they Related?

Ohm’s Law

×

Mathematical relationship between the electrical properties

V

I R

𝑉 = 𝐼 × 𝑅

𝐼 =𝑉

𝑅

𝑅 =𝑉

𝐼

IB Physics Data Booklet

Try this…

What is the voltage of a battery that produces a current of 1.5 amps through a 3 ohm resistor?

What resistance would produce a current of 5 amps from a 120-volt power source?

𝐼 = 1.5 A

𝑅 = 3 Ω𝑉 = 𝐼 × 𝑅 = 1.5 × 3 = 4.5 V

𝐼 = 5 A

𝑉 = 120 V 𝑅 =𝑉

𝐼=120

5= 24 Ω

Graphing Ohm’s Law

Linear Relationship means that our component is

________________

Cu

rren

t /

A

Potential Difference / V

Ohmic

Resistance

is constant

Graphing Ohm’s Law

Non-linear Relationship means that our component is ____________

Many/most electrical resistors don’t follow Ohm’s Law all of the time… For example, incandescent light bulbs have much more resistance as they heat up

Non-ohmic

Graphing Ohm’s Law

Find V and R for the resistors X and Y when the current is 2A

V 4 V

I 2 A

R 2 Ω

V 10 V

I 2 A

R 5 Ω

X

Y

Combining Resistors

R1 R2 R3R2

R3

R1

R1 R2

R1

R2

Series Parallel

Straw Competition

Whole Straw

2/3 of a Straw

1/3 of a Straw

2 Short Straws

3 Short Straws

Straw Competition

With the same full deep breath for each set up, time how quickly you can completely exhale all of your air

What it is What it represents Time to exhale (sec)

Whole Straw 3 resistors in ____________

2/3 of a Straw 2 resistors in ____________

1/3 of a Straw 1 resistor

2 Short Straws 2 resistors in ____________

3 Short Straws 3 resistors in ____________

When done collecting data, answer reflection questions on next slide

Time

Decreases(less resistance)

series

series

parallel

parallel

Straw Competition - Reflection

What does this model show about resistors in series?

What does this model show about resistors in parallel?

Adding resistors in series

increases overall resistance

Adding resistors in parallel

decreases overall resistance

Combining Resistors | Series

When combining resistors in series, the resistances are simply added up as if they were one large resistor

R1 R2 R3R1 R2

R1,2R1,2,3

𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅1 + 𝑅2 +⋯

Combining Resistors | Parallel

When combining resistors in parallel, the overall resistance decreases to produce a smaller equivalent resistance

R1,2R1,2,3

1

𝑅𝑡𝑜𝑡𝑎𝑙=

1

𝑅1+

1

𝑅2+⋯

R1

R2

R2

R3

R1

𝑅𝑡𝑜𝑡𝑎𝑙−1 = 𝑅1

−1 + 𝑅1−1 +⋯𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅1

−1 + 𝑅1−1 +⋯

−1

Combining Resistors – Try This

4 Ω

6 Ω

4 Ω 6 Ω 8 Ω

2.4 Ω 18 Ω

𝑅𝑇 =1

14 +

16

𝑅𝑇 = 4 + 6 + 8 = 18 Ω

= 4−1 + 6−1 −1 = 2.4 Ω

IB Physics Data Booklet

CircuitsIB PHYSICS | UNIT 8 | ELECTRICITY

8.3

Warm up

Which example has the least resistance? Assume that every resistor is the same.

Combining Resistors – Review

4 Ω

6 Ω

4 Ω 6 Ω 8 Ω

1

𝑅𝑡𝑜𝑡𝑎𝑙=

1

𝑅1+

1

𝑅2+⋯𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅1 + 𝑅2 +⋯

4 Ω + 6 Ω + 8 Ω = 𝟏𝟖 𝛀1

4 Ω+

1

6 Ω=

1

𝟐. 𝟒 𝛀

Equivalent Resistance

1 Ω

3 Ω

6 Ω

3 Ω

1 Ω 3 Ω

6 Ω

2 Ω

3−1 + 6−1 −1 = 2 Ω

1 + 2 + 3 = 6 Ω

Try This | Equivalent Resistance

10 Ω 8 Ω

7 Ω 2 Ω

18 Ω

9 Ω

6 Ω

7 + 2 = 9 Ω

10 + 8 = 18 Ω

9−1 + 18−1 −1 = 6 Ω

The Big Three

Ohm’s Law: If you know two of the three electrical properties: V, I, or R 𝑅 =

𝑉

𝐼

Kirchhoff’s Voltage Law Kirchhoff’s Current Law

Σ𝑉 = 0 (𝑙𝑜𝑜𝑝) Σ𝐼 = 0 (𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛)

• Draw a loop• The voltage provided must equal

the voltage dissipated• Useful if you have parallel

branches to solve for

• Calculate the current coming out of the battery (total current)

• If this splits into parallel branches, the total should still add up

Calculating Circuits - Series

R1 R2 R3

12 V

V I R

R1 2 V 2 A 1 Ω

R2 6 V 2 A 3 Ω

R3 4 V 2 A 2 Ω

Total 12 V 2 A 12 Ω

1 Ω 3 Ω 2 Ω

𝑅𝑇 = 1 + 3 + 2 = 6 Ω 𝐼𝑇 =𝑉

𝑅=12

6= 2 A 𝑉 = 𝐼 × 𝑅 =

Current is the same throughout

Calculating Circuits - Parallel

R2

12 V

R1V I R

R1 12 V 2 A 6 Ω

R2 12 V 4 A 3 Ω

Total 12 V 6 A 2 Ω

𝑅𝑇 = 6−1 + 3−1 −1 = 2 Ω

𝐼𝑇 =𝑉

𝑅=12

2= 6 A 𝐼 =

𝑉

𝑅=

6 Ω

3 Ω

Voltage loops to find that

voltage is the same across

each parallel branch

Think about it…

Are the circuits in this classroom parallel or series?

Both

Combination Circuits – Set up

Calculate V, I, and R for each resistor

V I R

R1 1 Ω

R2 3 Ω

R3 6 Ω

R4 3 Ω

Total 18 V

R36 Ω

3 Ω

1 Ω

R2

R1

3 Ω R4

18 V

Combination Circuits – Step 1

R36 Ω

3 Ω

1 Ω

R2

R1

3 Ω

Calculate the Equivalent Resistance

1 Ω

3 Ω

6 Ω

3 Ω

1 Ω 2 Ω 3 Ω

1 Ω + 2 Ω + 3 Ω = 𝟔 𝛀

1

3 Ω+

1

6 Ω=

1

𝟐 𝛀

R4

18 V

Combination Circuits – Step 2

R3

18 V

6 Ω

3 Ω

1 Ω

R2

R1

3 Ω R4

Use Ohm’s Law to find total current

V I R

R1 1 Ω

R2 3 Ω

R3 6 Ω

R4 3 Ω

Total 18 V 3 A 6 ΩV

I R×I =

V

R=18 V

6 Ω

3 A

Combination Circuits – Step 3

R3

18 V

6 Ω

3 Ω

1 Ω

R2

R1

3 Ω

Solve for resistors in series with battery

V I R

R1 3 V 3 A 1 Ω

R2 3 Ω

R3 6 Ω

R4 9 V 3 A 3 Ω

Total 18 V 3 A 6 Ω

3 A

3 A

3 A

V1 = I × R = 3 A × 1 Ω

V4 = I × R = 3 A × 3 Ω

R4

Combination Circuits – Step 4

Use Kirchhoff’s Voltage Law to study the loop

V I R

R1 3 V 3 A 1 Ω

R2 3 Ω

R3 6 V 1 A 6 Ω

R4 9 V 3 A 3 Ω

Total 18 V 3 A 6 Ω

R36 Ω

3 Ω

1 Ω

R2

R1

3 Ω R43 V 9 V

I3 =V

R=

18 V

6 V

6 V

6 Ω

Combination Circuits – Step 5

Use Kirchhoff’s Voltage Law to study the loop

V I R

R1 3 V 3 A 1 Ω

R2 6 V 2 A 3 Ω

R3 6 V 1 A 6 Ω

R4 9 V 3 A 3 Ω

Total 18 V 3 A 6 Ω

R36 Ω

3 Ω

1 Ω

R2

R1

3 Ω R43 V 9 V

I2 =V

R=

18 V

6 V

6 V

3 Ω

Combination Circuits – Check

Check current split on parallel branch

V I R

R1 3 V 3 A 1 Ω

R2 6 V 2 A 3 Ω

R3 6 V 1 A 6 Ω

R4 9 V 3 A 3 Ω

Total 18 V 3 A 6 Ω

R36 Ω

3 Ω

1 Ω

R2

R1

3 Ω R4

18 V

3 A

2 A

1 A

Combination Circuits – Try This #1

V I R

R1 4.8 V 1.6 A 3 Ω

R2 4.8 V 0.8 A 6 Ω

R3 7.2 V 2.4 A 3 Ω

Total 12 V 2.4 A 5 Ω

R26 Ω

3 Ω

3 Ω

R1

R3

12 V

2.4 A

1.6 A

0.8 A

1. Calculate total resistance

2. Calculate total current through battery

3. R3 has the same current as the battery

4. Voltage loops to find voltage in R1 and R2

5. Calculate missing currents

Combination Circuits – Try This #2

R15 Ω

R23 Ω

V I R

R1 7.5 V 1.5 A 5 Ω

R2 4.5 V 1.5 A 3 Ω

R3 12 V 1.5 A 8 Ω

Total 12 V 3 A 4 Ω

R3

12 V

8 Ω

1. Calculate total resistance

2. Calculate total current through battery

3. R3 is the only resistor in loop so it has full voltage

4. Current in top branch must be remaining 1.5 A

5. Calculate missing voltages

Measuring Circuits and ResistivityIB PHYSICS | UNIT 8 | ELECTRICITY

8.4

Warm up – What’s next??

V I R

R1 3 V 3 A 1 Ω

R2 9 V 3 A 3 Ω

R3 4 V 2 A 2 Ω

R4 8 V 2 A 4 Ω

Total 12 V 5 A 2.4 Ω

12 V

R11 Ω

R23 Ω

R32 Ω

R44 Ω

12 V

6 Ω

4 Ω

2 A

3 A • Each branch receives 12 V

• Current can be calculated

using total branch resistance

• Voltage can be calculated

from current and resistance

The Observer Effect

When taking any scientific measurement, there is always the possibility that the act of taking the measurement will

change what is being measured

The Observer Effect

When we measure voltage or current in a circuit, we want to make sure to minimize an effect that our tool has on

the circuit so that we get the most accurate results

Voltmeter

Ammeter

A

V

Ammeter

Hooked up in ____________ with the component being measured

Ideal Ammeter:

[R = ]A

series

0 Ω

What if Ammeter isn’t ideal?

A

10 V

What is the reading for the current flowing through the ammeter if it has an added resistance of 1 Ω?

R1 R2

5 Ω 5 Ω

1 Ω 𝑅𝑇 = 11 Ω

𝐼𝑇 =𝑉

𝑅=10

11= 𝟎. 𝟗 𝐀

Voltmeter

V

Hooked up in ____________ with the component being measured

Ideal Voltmeter:

[R = ]

parallel

∞ Ω

What if Voltmeter isn’t ideal?

10 V

What is the reading for the voltmeter across R1 if it has an added resistance of 20 Ω?

R2

5 Ω V I R

V 4.5 V 20 Ω

R1 4.5 V 5 Ω

R2 5.5 V 1.1 A 5 Ω

Total 10 V 1.1 A 9 Ω

20 Ω

V

R1

5 Ω

Try This

A

12 V

Calculate the resistance of this non-ideal meter:

AmmeterReading 1.2 A

R1 R2

3 Ω 6.5 Ω

V I R

R1 1.2 A 3 Ω

R2 1.2 A 6.5 Ω

A 1.2 A 0.5 Ω

Total 12 V 1.2 A 10 Ω

1.2 A

• Current is the same for all components

• Calculate total resistance from voltage and current

• Calculate resistance10 − 6.5 − 3 = 0.5 Ω

Try This

V

12 V

Calculate the resistance of this non-ideal meter:

VoltmeterReading 7 V

R1 R2

4 Ω 2.5 Ω

V I R

V 7 V 0.25 A 28 Ω

R1 7 V 1.75 A 4 Ω

R2 5 V 2 A 2.5 Ω

Total 12 V

• Use voltage loops to calculate voltage for R1 and R2

• Calculate current for R1 and R2

• Use current junction to find current through meter

• Calculate resistance of voltmeter

Remember Power?

Symbol: Unit:

New Equations:

P Watts [W]

𝑃 = 𝑉𝐼𝑃 = 𝐼2𝑅

𝑃 =𝑉2

𝑅

How much does is cost to

A blender runs on 5 amps of current on a 120 V. How much power is it drawing?

𝐼 = 5 A𝑉 = 120 V 𝑃 = 𝑉𝐼 = (120)(5)

= 𝟔𝟎𝟎𝐖

Different Devices… Different Power

Common Appliances Estimated WattsBlender 300-1000

Microwave 1000-2000

Waffle Iron 800-1500

Toaster 800-1500

Hair Dryer 1000-1875

TV 32" LED/LCD 50

TV 42" Plasma 240

Blu-Ray or DVD Player 15

Video Game Console(Xbox / PS4 / Wii)

40-140

IB Physics Data Booklet

Resistance

What factors affect the resistance of a wire?

•Cross-sectional Area

•Length

•Material

Resistance

𝑅 ∝1

𝐴𝑅 ∝ 𝐿

Imagine that you are testing the resistance of a straw while drinking a milkshake…

Calculating Resistance

R → Resistance

L → Length

A → Area

ρ → Resistivity

𝑅 = 𝜌𝐿

𝐴

𝐴 = 𝜋𝑟2

[Ω]

[m]

[m2]

[Ωm]

IB Physics Data Booklet

Resistivity

Resistivity ρ changes depending on the material used.

Resistivity – Try This #1

Calculate the resistance of a 1.8 m length of iron wire of with a diameter of 3 mm

𝑅 = 𝜌𝐿

𝐴

𝑅 = (12.0 × 10−8)(1.8)

(7.07 × 10−6)

L = 1.8 m

𝜌 = 12.0 × 10-8 Ωm

A =π(0.003/2)2 = 7.07 × 10-6 m2

𝑅 = 0.0306 Ω

Resistivity – Try This #2

A current of 4 A flowed through an 75 m length of metal alloy wire of area 2.4 mm2 when a p.d. of 12 V was applied across its ends. What was the resistivity of the alloy?

𝜌 =𝑅𝐴

𝐿

𝑅 =𝑉

𝐼=12

4= 3 Ω

𝐿 = 75 m

𝐴 = 2.4 mm2 ×1 m

1000 mm

2

𝐴 = 2.4 × 10−6 m2

𝜌 =(3)(2.4 × 10−6)

(75)

= 𝟗. 𝟔 × 𝟏𝟎−𝟖 𝛀𝐦

Voltage Dividersand BatteriesIB PHYSICS | UNIT 8 | ELECTRICITY

8.5

Warm Up – Mini Lab

Use the Voltmeter/Ammeter to measure the voltage, current, and resistance for this circuit.

6 Volts

V I R

A

B

Total

Types of Resistors

Types of Resistors

Light

Resistance

Light

Resistance

Types of Resistors

Heat

Resistance

Heat

Resistance

Types of Resistors

Potential Divider

Vout

R1

Vin

R2

R1

Vout

R1

Vout

Relationship between R1 and V?

Allows circuit designers to tune the voltage that is being delivered to the desired components

R2

Vout

R2

Vout

Relationship between R2 and V?

Potential Divider

Vout

10.0 V

Find the Output Voltage:

R1 R2

V I R

R1 1.2 V 0.04 A 30 Ω

R2 0.04 A 220 Ω

Total 10 V 0.04 A 250 Ω

30 Ω 220 Ω

1. Calculate total resistance and current2. Current is the same for each resistor3. Calculate voltage across R1

Assume Ideal Voltmeter

Potential Divider | Night Light

A light-dependent resistor (LDR) has R = 8 Ω in bright light and R = 140 Ω in low light. An electronic switch will turn on a light when its p.d. is above 7.0 V. What should the value of R2 be?

Electronic Switch

V I R

R1 7.0 V 0.05 A 140 Ω

R2 2.0 V 0.05 A 40 Ω

Total 9.0 V 0.05 A

*Night light should turn on in low light

1. Calculate current through R1

2. Current is the same throughout circuit (no current through switch)

3. Use voltage loop to find voltage across R2

4. Calculate resistance of R2

9.0 V

R1 R2

Potential Divider | Sprinkler System

A thermistor has a resistance of 2.5 Ω when it is in the heat of a fire and a resistance of 650 Ω in when at room temperature. An electronic switch will turn on a sprinkler system when its p.d. is above 7.0 V. What should the value of R1 be?

Electronic Switch

V I R

R1 7.0 V 0.8 A 8.75 Ω

R2 2.0 V 0.8 A 2.5 Ω

Total 9.0 V 0.8 A

9.0 V

R1 R2

1. Use voltage loop to find voltage across R2

2. Calculate current through R2

3. Current is the same throughout circuit (no current through switch)

4. Calculate resistance of R1

*Sprinkler should activate when hot

Batteries

Battery Shape Chemistry Nominal Voltage Rechargable?

AA, AAA, C, and D Alkaline or Zinc-carbon 1.5V No

9V Alkaline or Zinc-carbon 9V No

Coin cell Lithium 3V No

Silver Flat Pack Lithium Polymer (LiPo) 3.7V Yes

AA, AAA, C, D (Rechargeable)

NiMH or NiCd 1.2V Yes

Car battery Six-cell lead-acid 12.6V Yes

Primary Cells

Secondary CellsOne time use

Rechargable

Recharging?

Recharging Circuit

+V

Some batteries can reverse the chemical reaction that produces the potential difference by passing a current through the battery in the opposite direction as it would normally travel

Batteries | emf

We’ve been describing batteries so far as the voltage that they provide to the circuit, but that’s not the whole story…

Electromotive Force (emf)

The total energy transferred in the source per unit charge

passing through it

Symbol

ε

Unit

Volts [V]

ε r

Batteries | Internal Resistance

ε r

All batteries have some amount of internal resistance

Symbol

r

Unit

Ohms [Ω]

Batteries | Internal Resistance

10 Ω

0.5 Ω

1.2 A

What is the emf for a battery shown below?

ε

ε = 𝐼(𝑅 + 𝑟)

ε = 1.2(10 + 0.5)

ε = 𝟏𝟐. 𝟔 𝐕

IB Physics Data Booklet

Essentially the same as V = IR

Batteries | Internal Resistance

2.5 Ω

r

3 A

emf = 9 V

What is the internal resistance of this battery as shown below?

ε = 𝐼(𝑅 + 𝑟)

9 = 3(2.5 + 𝑟)

𝑟 = 𝟎. 𝟓 𝛀

Discharge Curves

Voltage isn’t constant across the battery’s lifetime…

Measuring Batteries