physics - 8 - electricity · 2020. 1. 3. · electrical current ib physics | unit 8 | electricity...
TRANSCRIPT
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