Resistance

General physics

Electrical Power — Physics Lesson

🎯 Learning Objectives

Learning Objectives

Recall the relationship between electrical power, voltage, and current (P = VI)
Apply the power formula P = I²R in electrical circuits
Use the power formula P = V²/R for circuit calculations
Solve problems involving electrical power in various circuit configurations
Analyze energy transfer in electrical components

🗣️ Language Objectives

Language Objectives

Use scientific terminology related to electrical power accurately
Explain power calculations using appropriate mathematical language
Describe energy transfer processes in electrical circuits
Communicate problem-solving strategies clearly
Write conclusions using scientific discourse

📝 Key Terms

Key Terms

English Term	Russian Translation	Kazakh Translation
Power	Мощность	Қуат
Voltage	Напряжение	Кернеу
Current	Ток	Ток
Resistance	Сопротивление	Кедергі
Energy	Энергия	Энергия
Circuit	Цепь	Тізбек
Watt	Ватт	Ватт

🃏 Topic Flashcards

Interactive Flashcards

Practice with these flashcards to memorize key electrical power concepts and formulas.

📚 Glossary

Glossary

Electrical Power: The rate at which electrical energy is transferred or converted to other forms of energy, measured in watts (W).
Translation
Russian: Электрическая мощность — скорость передачи или преобразования электрической энергии в другие формы энергии, измеряется в ваттах (Вт).
Kazakh: Электрлік қуат — электр энергиясының басқа энергия түрлеріне берілуі немесе түрлендірілуінің жылдамдығы, ватпен (Вт) өлшенеді.
Voltage (V): The electrical potential difference between two points in a circuit, measured in volts (V).
Translation
Russian: Напряжение — разность электрических потенциалов между двумя точками в цепи, измеряется в вольтах (В).
Kazakh: Кернеу — тізбектегі екі нүкте арасындағы электрлік потенциалдар айырмасы, вольтпен (В) өлшенеді.
Current (I): The flow of electric charge through a conductor, measured in amperes (A).
Translation
Russian: Ток — поток электрического заряда через проводник, измеряется в амперах (А).
Kazakh: Ток — өткізгіш арқылы электр зарядының ағыны, амперпен (А) өлшенеді.
Resistance (R): The opposition to the flow of electric current in a material, measured in ohms (Ω).
Translation
Russian: Сопротивление — противодействие протеканию электрического тока в материале, измеряется в омах (Ом).
Kazakh: Кедергі — материалдағы электр тогының ағуына қарсы тұру, оммен (Ом) өлшенеді.

📖 Theory: Electrical Power

Theory: Electrical Power

Introduction to Electrical Power

Electrical power is the rate at which electrical energy is transferred or converted to other forms of energy. It is a fundamental concept in understanding how electrical circuits work and how much energy is consumed by electrical devices.

Kazakh Translation

Электрлік қуат — электр энергиясының басқа энергия түрлеріне берілуі немесе түрлендірілуінің жылдамдығы. Бұл электрлік тізбектердің қалай жұмыс істейтінін және электрлік құрылғылардың қанша энергия тұтынатынын түсінудегі негізгі ұғым.

The Three Power Formulas

1. P = VI (Power = Voltage × Current)

This is the most fundamental formula for electrical power. It states that power is equal to the product of voltage and current.

Kazakh Translation

Бұл электрлік қуаттың ең негізгі формуласы. Ол қуат кернеу мен токтың көбейтіндісіне тең екенін айтады.

Formula: P = V × I

Where:

P = Power (watts, W)
V = Voltage (volts, V)
I = Current (amperes, A)

2. P = I²R (Power = Current² × Resistance)

This formula is useful when you know the current and resistance in a circuit. It’s derived from Ohm’s law (V = IR) and P = VI.

Kazakh Translation

Бұл формула тізбектегі ток пен кедергіді білгенде пайдалы. Ол Ом заңынан (V = IR) және P = VI формуласынан алынған.

Formula: P = I² × R

Derivation: P = VI, and V = IR, so P = (IR) × I = I²R

3. P = V²/R (Power = Voltage² / Resistance)

This formula is useful when you know the voltage applied across a resistor and its resistance value.

Kazakh Translation

Бұл формула резисторға қолданылған кернеу және оның кедергі мәнін білгенде пайдалы.

Formula: P = V²/R

Derivation: P = VI, and I = V/R (from Ohm’s law), so P = V × (V/R) = V²/R

Practice Questions

Question 1 (Easy):

A light bulb operates at 12V and draws a current of 2A. Calculate the power consumed by the bulb.

Answer

Using P = VI
P = 12V × 2A = 24W
The bulb consumes 24 watts of power.

Question 2 (Medium):

A resistor has a resistance of 50Ω and carries a current of 0.5A. Calculate the power dissipated by the resistor.

Answer

Using P = I²R
P = (0.5A)² × 50Ω = 0.25 × 50 = 12.5W
The resistor dissipates 12.5 watts of power.

Question 3 (Medium):

A heating element has a resistance of 20Ω and is connected to a 240V supply. Calculate the power rating of the heating element.

Answer

Using P = V²/R
P = (240V)²/20Ω = 57,600/20 = 2,880W = 2.88kW
The heating element has a power rating of 2.88 kilowatts.

Question 4 (Critical Thinking):

A circuit contains three identical resistors, each with resistance R, connected in series to a voltage source V. If one resistor is removed, how does the total power consumed by the circuit change? Explain your reasoning using the appropriate power formulas.

Answer

Analysis:
Initial situation: 3 resistors in series, total resistance = 3R
Current: I₁ = V/(3R)
Power: P₁ = V²/(3R)

After removing one resistor: 2 resistors in series, total resistance = 2R
Current: I₂ = V/(2R)
Power: P₂ = V²/(2R)

Comparing: P₂/P₁ = [V²/(2R)]/[V²/(3R)] = 3/2 = 1.5

Conclusion: The power increases by 50% when one resistor is removed because the total resistance decreases, allowing more current to flow and increasing the power consumption.

🧠 Memorization Exercises

Exercises on Memorizing Terms

Exercise 1: Fill in the Blanks

Electrical _______ is measured in watts.
The formula P = VI shows that power equals _______ multiplied by _______.
When current is squared and multiplied by resistance, we get the formula P = _______.
The unit of electrical resistance is the _______.
Voltage divided by resistance gives us the _______.

Answer

1. power
2. voltage, current
3. I²R
4. ohm (Ω)
5. current

Exercise 2: Match the Formulas

Match each scenario with the most appropriate power formula:

You know voltage and current → _______
You know current and resistance → _______
You know voltage and resistance → _______

Options: P = VI, P = I²R, P = V²/R

Answer

1. P = VI
2. P = I²R
3. P = V²/R

📺 Educational Video

Video Tutorial: Electrical Power

Additional Resources:

🔬 Problem Solving Examples

Worked Examples

Example 1: Light Bulb Power Calculation

Problem: A 60W light bulb is connected to a 120V household supply. Calculate:

The current through the bulb
The resistance of the bulb filament

🎤 Audio Solution

Detailed Solution with Pronunciation

Step 1: Finding current (pronounced: CURR-ent)

We use P = VI, so I = P/V

I = 60W ÷ 120V = 0.5A

The current is zero point five amperes.

Step 2: Finding resistance (pronounced: ree-ZIS-tanse)

We use V = IR, so R = V/I

R = 120V ÷ 0.5A = 240Ω

The resistance is two hundred forty ohms.

📝 Quick Solution

Brief Solution

Given: P = 60W, V = 120V

Find: I and R

Current: I = P/V = 60/120 = 0.5A

Resistance: R = V/I = 120/0.5 = 240Ω

Example 2: Heating Element Power

Problem: A heating element with resistance 25Ω is connected to a 230V supply. Calculate the power dissipated and the current drawn.

🎤 Audio Solution

Detailed Solution with Pronunciation

Step 1: Finding power using P = V²/R

P = (230V)² ÷ 25Ω

P = 52,900 ÷ 25 = 2,116W

The power is two thousand one hundred sixteen watts.

Step 2: Finding current using I = V/R

I = 230V ÷ 25Ω = 9.2A

The current is nine point two amperes.

📝 Quick Solution

Brief Solution

Given: R = 25Ω, V = 230V

Find: P and I

Power: P = V²/R = (230)²/25 = 2,116W

Current: I = V/R = 230/25 = 9.2A

🔬 Investigation Task

Interactive Simulation

Use this PhET simulation to investigate how changing voltage, current, and resistance affects electrical power:

Investigation Questions:

What happens to power when you double the voltage while keeping resistance constant?
How does power change when you double the current while keeping voltage constant?
If you increase resistance while keeping voltage constant, what happens to power?

Brief Answers

1. Power increases by a factor of 4 (P = V²/R, so doubling V quadruples P)
2. Power doubles (P = VI, so doubling I doubles P)
3. Power decreases (P = V²/R, so increasing R decreases P)

👥 Group/Pair Activity

Collaborative Learning Activity

Work with your partner or group to complete this interactive quiz about electrical power:

Discussion Points:

Which power formula is most useful for different types of problems?
How can you verify your answers using alternative formulas?
What safety considerations are important when dealing with electrical power?

✏️ Individual Assessment

Structured Questions — Individual Work

Question 1 (Analysis):

A circuit contains two resistors: R₁ = 10Ω and R₂ = 20Ω connected in parallel to a 12V battery.

Calculate the total resistance of the circuit.
Determine the current through each resistor.
Find the power dissipated by each resistor.
Calculate the total power supplied by the battery.
Verify your answer using P = V²/R_total.

Answer

a) 1/R_total = 1/10 + 1/20 = 3/20, so R_total = 6.67Ω
b) I₁ = V/R₁ = 12/10 = 1.2A; I₂ = V/R₂ = 12/20 = 0.6A
c) P₁ = V²/R₁ = 144/10 = 14.4W; P₂ = V²/R₂ = 144/20 = 7.2W
d) P_total = P₁ + P₂ = 14.4 + 7.2 = 21.6W
e) P_total = V²/R_total = 144/6.67 = 21.6W ✓

Question 2 (Synthesis):

Design a heating system for a small room that requires 3kW of power. You have access to 240V mains supply and heating elements with the following resistances: 20Ω, 30Ω, 40Ω, and 60Ω.

Determine all possible combinations of resistors that would give approximately 3kW.
Which combination would be most energy efficient and why?
Calculate the monthly electricity cost if the system runs 8 hours per day and electricity costs $0.15 per kWh.

Answer

a) P = V²/R, so R = V²/P = 240²/3000 = 19.2Ω
Possible combinations: 20Ω alone (2.88kW), or parallel combinations close to 19.2Ω
b) Single 20Ω resistor is most efficient (fewer connections, less heat loss)
c) Monthly cost = 2.88kW × 8h × 30days × $0.15 = $103.68

Question 3 (Critical Analysis):

A student claims that connecting identical light bulbs in series will reduce the total power consumption compared to connecting them in parallel. Analyze this statement using mathematical evidence.

Answer

Analysis:
For n identical bulbs (resistance R each):
Series: R_total = nR, I = V/(nR), P_total = V²/(nR)
Parallel: R_total = R/n, I = nV/R, P_total = nV²/R
Comparison: P_parallel/P_series = n²
Conclusion: The student is correct. Series connection reduces power by factor of n², where n is the number of bulbs.

Question 4 (Problem Solving):

An electric kettle rated at 2.5kW takes 4 minutes to boil 1 liter of water. If the voltage drops to 80% of its rated value, calculate:

The new power consumption
The new time required to boil the same amount of water
The additional energy cost per boiling cycle

Answer

a) P_new = P × (0.8)² = 2.5 × 0.64 = 1.6kW
b) Energy needed is same, so t_new = 4 × (2.5/1.6) = 6.25 minutes
c) Additional energy = 1.6kW × (6.25-4)/60 h = 0.06kWh extra

Question 5 (Evaluation):

Evaluate the environmental and economic benefits of replacing a 100W incandescent bulb with a 20W LED bulb that provides the same brightness. Consider:

Daily energy savings (assuming 6 hours use per day)
Annual cost savings (electricity at $0.12 per kWh)
CO₂ emission reduction (0.5 kg CO₂ per kWh)
Discuss the long-term implications

Answer

a) Daily savings = (100-20) × 6/1000 = 0.48 kWh
b) Annual savings = 0.48 × 365 × $0.12 = $21.02
c) CO₂ reduction = 0.48 × 365 × 0.5 = 87.6 kg per year
d) Long-term: Reduced energy demand, lower emissions, economic savings offset higher initial cost

🔗 Additional Resources

Useful Links and References

📚 Study Materials:

🎥 Video Resources:

🧮 Practice Tools:

🤔 Lesson Reflection

Reflection Questions

Think about your learning today:

💡 Understanding:

Which of the three power formulas do you find most useful and why?
What connections can you make between electrical power and everyday appliances?
How has your understanding of energy transfer in circuits changed?

🎯 Application:

How would you explain the relationship between power, voltage, and current to a friend?
What real-world problems could you solve using today’s formulas?
Which activities helped you learn most effectively?

🔄 Next Steps:

What aspects of electrical power would you like to explore further?
How confident do you feel about solving power problems?
What questions do you still have about this topic?

📝 Self-Assessment Scale (1-5):

Rate your confidence in:

Using P = VI formula: ___/5
Using P = I²R formula: ___/5
Using P = V²/R formula: ___/5
Choosing the appropriate formula: ___/5
Solving complex power problems: ___/5