Electromotive force

General physics

Electromotive Force and Internal Resistance — Physics Lesson

🎯 Learning Objectives

Learning Objectives

Define and use the electromotive force (e.m.f.) of a source as energy transferred per unit charge in driving charge around a complete circuit
Distinguish between e.m.f. and potential difference (p.d.) in terms of energy considerations
Understand the effects of the internal resistance of a source of e.m.f. on the terminal potential difference
Apply mathematical relationships to solve problems involving e.m.f., p.d., and internal resistance
Analyze real battery behavior using internal resistance concepts

🗣️ Language Objectives

Language Objectives

Use scientific terminology related to electrical energy and charge flow accurately
Distinguish between similar terms (e.m.f. vs potential difference) using precise language
Explain energy transfer processes in electrical circuits using appropriate scientific discourse
Describe the behavior of real batteries and power sources
Communicate problem-solving strategies for circuit analysis clearly

📝 Key Terms

Key Terms

English Term	Russian Translation	Kazakh Translation
Electromotive Force (e.m.f.)	Электродвижущая сила (ЭДС)	Электрқозғаушы күш (ЭҚК)
Potential Difference (p.d.)	Разность потенциалов	Потенциалдар айырмасы
Internal Resistance	Внутреннее сопротивление	Ішкі кедергі
Terminal Voltage	Напряжение на клеммах	Клеммадағы кернеу
Energy per Unit Charge	Энергия на единицу заряда	Заряд бірлігіне келетін энергия
Complete Circuit	Полная цепь	Толық тізбек
Load Resistance	Сопротивление нагрузки	Жүктеме кедергісі
Short Circuit	Короткое замыкание	Қысқа тұйықталу

🃏 Topic Flashcards

Interactive Flashcards

Practice with these flashcards to memorize key concepts about e.m.f., potential difference, and internal resistance.

📚 Glossary

Glossary

Electromotive Force (e.m.f.): The energy transferred per unit charge by a source in driving charge around a complete circuit. It represents the total energy per coulomb that the source can provide, measured in volts (V).
Translation
Russian: Электродвижущая сила (ЭДС) — энергия, передаваемая на единицу заряда источником при прохождении заряда по полной цепи. Представляет общую энергию на кулон, которую может обеспечить источник, измеряется в вольтах (В).
Kazakh: Электрқозғаушы күш (ЭҚК) — зарядты толық тізбек бойынша өткізуде көз арқылы заряд бірлігіне берілетін энергия. Көздің бере алатын кулон басына жалпы энергияны білдіреді, вольтпен (В) өлшенеді.
Potential Difference (p.d.): The energy transferred per unit charge between two points in a circuit. It represents the work done in moving charge from one point to another, measured in volts (V).
Translation
Russian: Разность потенциалов — энергия, передаваемая на единицу заряда между двумя точками в цепи. Представляет работу, совершаемую при перемещении заряда из одной точки в другую, измеряется в вольтах (В).
Kazakh: Потенциалдар айырмасы — тізбектегі екі нүкте арасында заряд бірлігіне берілетін энергия. Зарядты бір нүктеден екіншісіне жылжытуда атқарылатын жұмысты білдіреді, вольтпен (В) өлшенеді.
Internal Resistance: The resistance within a source of e.m.f. that opposes the flow of current through the source itself. It causes the terminal voltage to be less than the e.m.f. when current flows.
Translation
Russian: Внутреннее сопротивление — сопротивление внутри источника ЭДС, которое противодействует протеканию тока через сам источник. Оно приводит к тому, что напряжение на клеммах становится меньше ЭДС при протекании тока.
Kazakh: Ішкі кедергі — ЭҚК көзінің ішіндегі көз арқылы ток ағынына кедергі келтіретін кедергі. Ток ағып жатқанда клеммадағы кернеудің ЭҚК-тан кем болуына әкеледі.
Terminal Voltage: The potential difference measured across the terminals of a source when current is flowing. It equals the e.m.f. minus the voltage drop across the internal resistance.
Translation
Russian: Напряжение на клеммах — разность потенциалов, измеряемая на клеммах источника при протекании тока. Равно ЭДС минус падение напряжения на внутреннем сопротивлении.
Kazakh: Клеммадағы кернеу — ток ағып жатқанда көздің клеммаларында өлшенетін потенциалдар айырмасы. ЭҚК минус ішкі кедергідегі кернеу түсуіне тең.
Load Resistance: The external resistance connected to a source of e.m.f. that consumes electrical energy and converts it to other forms (heat, light, mechanical work, etc.).
Translation
Russian: Сопротивление нагрузки — внешнее сопротивление, подключенное к источнику ЭДС, которое потребляет электрическую энергию и преобразует ее в другие формы (тепло, свет, механическая работа и т.д.).
Kazakh: Жүктеме кедергісі — ЭҚК көзіне қосылған электр энергиясын тұтынып, оны басқа түрлерге (жылу, жарық, механикалық жұмыс және т.б.) түрлендіретін сыртқы кедергі.
Open Circuit: A circuit condition where no current flows, typically when the external circuit is disconnected. In this case, the terminal voltage equals the e.m.f.
Translation
Russian: Разомкнутая цепь — состояние цепи, при котором ток не течет, обычно когда внешняя цепь отключена. В этом случае напряжение на клеммах равно ЭДС.
Kazakh: Ашық тізбек — ток ағып жатпайтын тізбек жағдайы, әдетте сыртқы тізбек ажыратылған кезде. Бұл жағдайда клеммадағы кернеу ЭҚК-ға тең.

📖 Theory: Electromotive Force and Internal Resistance

Theory: E.M.F., Potential Difference, and Internal Resistance

Introduction to Electromotive Force (E.M.F.)

(e.m.f.) is a fundamental concept in understanding how of electrical energy work. Despite its name, e.m.f. is not actually a force, but rather a measure of per unit charge.

Kazakh Translation

Электрқозғаушы күш (ЭҚК) — электр энергиясының көздерінің қалай жұмыс істейтінін түсінудегі негізгі ұғым. Атауына қарамастан, ЭҚК шынында күш емес, керісінше заряд бірлігіне келетін энергияның өлшемі.

Battery with internal resistance diagram

A real battery showing e.m.f. (ε) and internal resistance (r)

Definition of E.M.F.

The e.m.f. of a source is defined as the energy transferred per unit charge in driving charge around a complete circuit.

Kazakh Translation

Көздің ЭҚК-ы зарядты толық тізбек бойынша өткізуде заряд бірлігіне берілетін энергия ретінде анықталады.

Mathematical Definition:

ε = W/Q

Where:

ε (epsilon) = e.m.f. (volts, V)
W = Work done or energy transferred (joules, J)
Q = Charge (coulombs, C)

Energy conversion inside a battery: Chemical energy → Electrical energy

Potential Difference vs E.M.F.

Key Distinctions:

Aspect	E.M.F. (ε)	Potential Difference (V)
Definition	Energy supplied per unit charge by the source	Energy transferred per unit charge between two points
Energy Role	Creates/supplies electrical energy	Uses/dissipates electrical energy
Measurement	Measured in open circuit (no current)	Measured across components when current flows
Independence	Independent of current drawn	Depends on current and resistance

Kazakh Translation

ЭҚК пен потенциалдар айырмасының арасындағы негізгі айырмашылықтар: ЭҚК көз арқылы заряд бірлігіне берілетін энергия, ал потенциалдар айырмасы екі нүкте арасында заряд бірлігіне берілетін энергия. ЭҚК электр энергиясын жасайды, потенциалдар айырмасы оны пайдаланады.

Internal Resistance

Real sources of e.m.f. have internal resistance (r) that opposes current flow through the source itself.

Kazakh Translation

ЭҚК-тың нақты көздерінде ішкі кедергі (r) болады, ол көздің өзі арқылы ток ағынына қарсы тұрады.

Complete circuit showing internal resistance (r) and load resistance (R)

Effects of Internal Resistance:

1. Terminal Voltage Equation:

V = ε — Ir

Where:

V = Terminal voltage (volts, V)
ε = E.M.F. of source (volts, V)
I = Current flowing (amperes, A)
r = Internal resistance (ohms, Ω)

2. Circuit Analysis:

For a complete circuit with load resistance R:

I = ε/(R + r)
V = ε × R/(R + r)

Graph showing how terminal voltage decreases with increasing current

Energy Considerations

Energy Distribution in a Circuit:

Total energy supplied by source: W_total = εIt
Energy dissipated in load: W_load = VIt = I²Rt
Energy lost in internal resistance: W_internal = I²rt

Energy Conservation: W_total = W_load + W_internal

Kazakh Translation

Тізбектегі энергия бөлінуі: жалпы энергия = жүктемедегі энергия + ішкі кедергідегі энергия. Энергия сақталу заңы орындалады.

Practice Questions

Question 1 (Easy):

A battery has an e.m.f. of 12V and internal resistance of 0.5Ω. What is the terminal voltage when no current flows?

Answer

When no current flows (I = 0), the terminal voltage equals the e.m.f.
Using V = ε — Ir
V = 12V — (0A)(0.5Ω) = 12V
The terminal voltage is 12V.

Question 2 (Medium):

The same battery from Question 1 is connected to a 3Ω resistor. Calculate:

The current in the circuit
The terminal voltage
The voltage across the 3Ω resistor

Answer

a) I = ε/(R + r) = 12V/(3Ω + 0.5Ω) = 12/3.5 = 3.43A
b) V = ε — Ir = 12V — (3.43A)(0.5Ω) = 12 — 1.71 = 10.29V
c) The voltage across the 3Ω resistor equals the terminal voltage = 10.29V

Question 3 (Medium):

Explain why the terminal voltage of a battery decreases when more current is drawn from it.

Answer

As more current flows, the voltage drop across the internal resistance (Ir) increases. Since V = ε — Ir, and ε is constant, the terminal voltage V decreases as current I increases. The internal resistance «steals» some of the battery’s voltage, leaving less available at the terminals.

Question 4 (Critical Thinking):

A student measures the terminal voltage of a battery as 9.0V when connected to a 2Ω resistor, and 8.5V when connected to a 1Ω resistor. Determine the e.m.f. and internal resistance of the battery. Analyze what this tells us about battery performance under different loads.

Answer

Setup equations:
For 2Ω load: 9.0 = ε — I₁r, where I₁ = ε/(2 + r)
For 1Ω load: 8.5 = ε — I₂r, where I₂ = ε/(1 + r)

Substituting:
9.0 = ε — εr/(2 + r) = ε(2)/(2 + r)
8.5 = ε — εr/(1 + r) = ε(1)/(1 + r)

Solving:
From first equation: ε = 9.0(2 + r)/2 = 9.0 + 4.5r
From second equation: ε = 8.5(1 + r) = 8.5 + 8.5r
Setting equal: 9.0 + 4.5r = 8.5 + 8.5r
Solving: 0.5 = 4r, so r = 0.125Ω
Therefore: ε = 9.0 + 4.5(0.125) = 9.56V

Analysis: Higher current loads cause greater voltage drop, demonstrating that internal resistance limits battery performance under heavy loads.

🧠 Memorization Exercises

Exercises on Memorizing Terms

Exercise 1: Fill in the Blanks

E.M.F. is the _______ transferred per unit _______ by a source.
The terminal voltage equals e.m.f. minus the voltage drop across the _______ resistance.
When no current flows, the terminal voltage equals the _______.
Potential difference represents energy _______ between two points.
The formula V = ε — Ir shows that terminal voltage _______ as current increases.

Answer

1. energy, charge
2. internal
3. e.m.f.
4. transferred (or dissipated)
5. decreases

Exercise 2: Equation Matching

Match each equation with its correct description:

Equations:

ε = W/Q
V = ε — Ir
I = ε/(R + r)
P = I²r

Descriptions:

Power lost in internal resistance
Current in complete circuit
Definition of e.m.f.
Terminal voltage equation

Answer

1-C: ε = W/Q → Definition of e.m.f.
2-D: V = ε — Ir → Terminal voltage equation
3-B: I = ε/(R + r) → Current in complete circuit
4-A: P = I²r → Power lost in internal resistance

Exercise 3: Concept Connections

Connect the concepts by filling in the cause-and-effect relationships:

Current increases → Internal voltage drop _______ → Terminal voltage _______
Load resistance decreases → Current _______ → Power dissipation _______
Internal resistance increases → Terminal voltage _______ → Efficiency _______
Open circuit → Current = _______ → Terminal voltage = _______

Answer

1. increases, decreases
2. increases, increases
3. decreases, decreases
4. zero, e.m.f.

📺 Educational Video

Video Tutorial: E.M.F. and Internal Resistance

Additional Resources:

🔬 Problem Solving Examples

Worked Examples

Example 1: Battery Testing

Problem: A battery has an open-circuit voltage of 6.0V. When connected to a 4Ω resistor, the terminal voltage drops to 5.6V. Calculate:

The internal resistance of the battery
The power delivered to the load
The power lost in internal resistance
The efficiency of the battery

🎤 Audio Solution

Detailed Solution with Pronunciation

Given: (pronounced: GIV-en)

Open-circuit voltage (e.m.f.) = 6.0V

Terminal voltage with load = 5.6V

Load resistance = 4Ω

Step 1: Find current

I = V/R = 5.6V/4Ω = 1.4A

Step 2: Find internal resistance

V = ε — Ir, so 5.6 = 6.0 — 1.4r

r = (6.0 — 5.6)/1.4 = 0.29Ω

Step 3: Power calculations

P_load = VI = 5.6 × 1.4 = 7.84W

P_internal = I²r = (1.4)² × 0.29 = 0.57W

Step 4: Efficiency

η = P_load/P_total = 7.84/8.41 = 93.2%

📝 Quick Solution

Brief Solution

Given: ε = 6.0V, V = 5.6V, R = 4Ω

1. Internal resistance:

I = V/R = 5.6/4 = 1.4A

r = (ε — V)/I = 0.4/1.4 = 0.29Ω

2. Load power: P = VI = 7.84W

3. Internal power: P = I²r = 0.57W

4. Efficiency: η = 93.2%

Example 2: Maximum Power Transfer

Problem: A 12V battery with internal resistance 2Ω is connected to a variable load resistor. Find the load resistance that gives maximum power transfer and calculate this maximum power.

🎤 Audio Solution

Detailed Solution with Pronunciation

Theory: (pronounced: THEE-oh-ree)

Maximum power transfer occurs when load resistance equals internal resistance.

Step 1: Condition for maximum power

R_load = r = 2Ω

Step 2: Calculate current

I = ε/(R + r) = 12V/(2Ω + 2Ω) = 3A

Step 3: Calculate maximum power

P_max = I²R = (3A)² × 2Ω = 18W

Verification: P = ε²R/(R + r)²

P_max = (12)² × 2/(2 + 2)² = 288/16 = 18W ✓

📝 Quick Solution

Brief Solution

Given: ε = 12V, r = 2Ω

Maximum Power Theorem:

R_load = r for maximum power transfer

Solution:

R_optimal = 2Ω

I = 12/(2+2) = 3A

P_max = 3² × 2 = 18W

Note: Efficiency = 50% at maximum power

🔬 Investigation Task

Interactive Simulation

Use this PhET simulation to investigate how internal resistance affects circuit behavior:

Investigation Questions:

How does terminal voltage change as you increase the load resistance?
What happens to the current when you decrease the load resistance to very small values?
How does the power delivered to the load vary with load resistance?
At what condition is maximum power transferred to the load?

Brief Answers

1. Terminal voltage increases as load resistance increases (less current, less internal voltage drop)
2. Current increases dramatically, approaching ε/r (short circuit current)
3. Power first increases then decreases with load resistance, maximum at R = r
4. Maximum power when load resistance equals internal resistance

👥 Group/Pair Activity

Collaborative Learning Activity

Work with your partner or group to complete this interactive battery analysis activity:

Discussion Points:

Why do car batteries have very low internal resistance?
How does battery age affect internal resistance?
Compare the internal resistance of different battery types (AA, car battery, phone battery)
When would you want high internal resistance in a source?

Practical Investigation Ideas:

Measure the internal resistance of different batteries using the voltage-divider method
Compare old vs new batteries of the same type
Test how temperature affects battery internal resistance
Investigate how different load resistances affect battery performance

✏️ Individual Assessment

Structured Questions — Individual Work

Question 1 (Analysis):

A solar cell has an e.m.f. of 0.6V and internal resistance of 0.8Ω when illuminated. It is connected to various load resistances.

Calculate the terminal voltage when connected to a 4Ω load.
Find the current delivered to a 0.2Ω load.
Determine the load resistance that gives maximum power transfer.
Calculate the efficiency when delivering maximum power.
Explain why solar cells are not operated at maximum power transfer in practice.

Answer

a) I = 0.6/(4 + 0.8) = 0.125A; V = 0.6 — 0.125×0.8 = 0.5V
b) I = 0.6/(0.2 + 0.8) = 0.6A
c) R_{max power} = r = 0.8Ω
d) At max power: η = R/(R+r) = 0.8/(0.8+0.8) = 50%
e) 50% efficiency is wasteful; practical systems operate at higher efficiency (70-90%) by using higher load resistance, sacrificing some power for better efficiency

Question 2 (Synthesis):

A student designs a battery testing system that measures both open-circuit voltage and short-circuit current to determine battery health.

Explain the theory behind this testing method.
Derive expressions for e.m.f. and internal resistance in terms of measured quantities.
A battery shows 9.2V open-circuit and 23A short-circuit current. Calculate its internal resistance.
Compare this battery to one with 9.0V open-circuit and 30A short-circuit current.
Discuss safety considerations for short-circuit testing.

Answer

a) Open-circuit gives ε directly; short-circuit gives maximum current I_sc = ε/r
b) ε = V_oc; r = ε/I_sc = V_oc/I_sc
c) r = 9.2V/23A = 0.4Ω
d) Battery 2: r = 9.0/30 = 0.3Ω. Lower internal resistance indicates better condition
e) Short-circuit tests generate high currents and heat; use brief pulses, current limiters, appropriate test equipment

Question 3 (Evaluation):

Compare the energy efficiency of three identical 1.5V batteries (each with 0.1Ω internal resistance) connected in different configurations to power a 0.8Ω load.

Calculate current, voltage, and power for series connection.
Calculate current, voltage, and power for parallel connection.
Calculate current, voltage, and power for single battery connection.
Determine which configuration is most efficient and explain why.
Analyze when each configuration would be preferred in practice.

Answer

a) Series: Total ε = 4.5V, r_total = 0.3Ω; I = 4.5/1.1 = 4.09A; V = 3.27V; P = 13.4W
b) Parallel: ε = 1.5V, r_total = 0.033Ω; I = 1.5/0.833 = 1.8A; V = 1.44V; P = 2.59W
c) Single: I = 1.5/0.9 = 1.67A; V = 1.33V; P = 2.22W
d) Parallel most efficient (96.3%) due to lowest internal resistance
e) Series for high voltage needs; parallel for high current/efficiency; single for low power applications

Question 4 (Critical Thinking):

An electric vehicle battery pack is designed with multiple cells. Each cell has e.m.f. = 3.7V and internal resistance = 0.05Ω.

Design a battery pack configuration to provide 300V with minimum internal resistance.
Calculate the internal resistance of your design.
If the motor draws 100A, find the power loss in internal resistance.
Propose methods to minimize internal resistance effects.
Analyze the trade-offs between voltage, current capacity, and efficiency.

Answer

a) Need 300V/3.7V ≈ 81 cells in series. For lower internal resistance, use multiple parallel strings
b) 81 cells in series: r = 81 × 0.05 = 4.05Ω
c) P_loss = I²r = (100)² × 4.05 = 40.5kW
d) Use multiple parallel strings, improve cell chemistry, active cooling, current control
e) Higher voltage reduces current for same power (lower I²r losses); parallel strings reduce internal resistance but increase complexity and cost

Question 5 (Application):

A power station generates electricity at 25kV and transmits it through power lines with total resistance 50Ω to supply 10MW to a city.

Calculate the current in the transmission lines.
Find the power loss in transmission.
Determine the voltage received by the city.
Calculate the transmission efficiency.
Explain how this relates to e.m.f. and internal resistance concepts.

Answer

a) I = P/V = 10×10⁶W/25×10³V = 400A
b) P_loss = I²R = (400)² × 50 = 8MW
c) V_drop = IR = 400 × 50 = 20kV; V_received = 25kV — 20kV = 5kV
d) η = (10MW)/(18MW) = 55.6%
e) Power station acts like source with e.m.f. = 25kV, transmission lines like internal resistance = 50Ω. High «internal resistance» causes significant voltage drop and power loss, similar to battery with high internal resistance

🔗 Additional Resources

Useful Links and References

📚 Study Materials:

🎥 Video Resources:

🧮 Practice Tools:

📖 Advanced Reading:

🤔 Lesson Reflection

Reflection Questions

Think about your learning today:

💡 Understanding:

Can you clearly distinguish between e.m.f. and potential difference in your own words?
How does the concept of internal resistance help explain real battery behavior?
What connections can you make between energy conservation and circuit analysis?
How has your understanding of «voltage» evolved after learning about e.m.f.?

🎯 Application:

How would you explain to someone why their phone battery voltage drops under heavy use?
What practical problems can you now solve using e.m.f. and internal resistance concepts?
How might this knowledge apply to renewable energy systems like solar panels?
Which problem-solving techniques did you find most effective?

🔄 Next Steps:

What aspects of circuit analysis would you like to explore further?
How confident do you feel about solving complex battery circuit problems?
What real-world applications of these concepts interest you most?
What questions do you still have about energy and charge flow?

📝 Self-Assessment Scale (1-5):

Rate your confidence in:

Defining e.m.f. as energy per unit charge: ___/5
Distinguishing e.m.f. from potential difference: ___/5
Understanding internal resistance effects: ___/5
Applying V = ε — Ir equation: ___/5
Analyzing real battery circuits: ___/5

🎯 Learning Goals Achieved:

☐ I can define e.m.f. as energy transferred per unit charge in driving charge around a complete circuit
☐ I can distinguish between e.m.f. and potential difference in terms of energy considerations
☐ I understand how internal resistance affects terminal potential difference
☐ I can solve problems involving e.m.f., p.d., and internal resistance
☐ I can explain real battery behavior using these concepts

🌟 Key Insights:

«The difference between e.m.f. and potential difference is like the difference between the total energy a source can provide versus the energy actually delivered to a specific component. Internal resistance acts like a ‘tax’ on the energy transfer, reducing the useful voltage available to external circuits.»