Electrical Energy and Power

3.4 Electrical Energy and Power
PHYSICS • Form 5 • Chapter 3: Electricity

3.4 Electrical Energy and Power

Electrical appliances in our daily lives convert electrical energy into other forms of energy such as heat, light, and mechanical energy. Understanding the mathematical relationships between electrical energy, power, voltage, current, and resistance allows consumers to determine energy usage, calculate electrical consumption costs, and choose energy-efficient appliances to reduce household expenditure.

Learning Objectives

  • State the relationship between electrical energy (\(E\)), voltage (\(V\)), current (\(I\)), and time (\(t\)).
  • Define electrical power (\(P\)) and formulate its relationship with voltage (\(V\)), current (\(I\)), and resistance (\(R\)).
  • Solve numerical problems involving electrical energy and power in daily life.
  • Compare the power and energy consumption rates of various electrical devices, such as CFL and LED lamps.
  • Calculate the cost of electrical energy consumption based on kilowatt-hours (\(\text{kWh}\) or units).
  • Identify and suggest steps to reduce household electrical energy usage.

Summary of Formulae for Electrical Energy and Power

See how voltage, current, resistance and time connect to electrical energy, power and electricity cost.

Formula Tree

\(E = VIt\) Electrical energy
\(P = \dfrac{E}{t} = VI\) Electrical power
\(P = I^2R\) Using \(V = IR\)
\(P = \dfrac{V^2}{R}\) Using \(I = \dfrac{V}{R}\)

Electricity Cost Conversion

From joules to billing units

\(1\text{ unit} = 1\text{ kWh}\)

\(\text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (h)}\)

\(\text{Cost} = \text{Energy (units)} \times \text{Cost per unit}\)

Power rating and working voltage

An appliance labelled \(240\text{ V}, 700\text{ W}\) uses \(700\text{ J}\) of energy every second when supplied with \(240\text{ V}\).

Short Explanation

Electrical Energy

Electrical energy is the work done by electric charges moving through a potential difference. It is given by \(E = VIt\), where \(V\) is potential difference, \(I\) is current and \(t\) is time.

Electrical Power

Electrical power is the rate at which electrical energy is used. \(P = \dfrac{E}{t} = VI\). Using Ohm's law, it can also be written as \(P = I^2R\) or \(P = \dfrac{V^2}{R}\).

Working Voltage

Working voltage is the potential difference required for an electrical appliance to operate under normal conditions. At this voltage, the appliance consumes its stated power rating.

Electricity Billing Unit

Household electricity is billed in kilowatt-hours. \(1\text{ unit} = 1\text{ kWh}\). Energy used is calculated as power in kilowatts multiplied by time in hours.

Problem-solving framework

Step 1: Identify the problem → Step 2: Identify the information given → Step 3: Identify the formula that can be used → Step 4: Solve the problem numerically
Quantity Symbol SI unit
Electrical energy \(E\) joule, \(\text{J}\)
Electrical power \(P\) watt, \(\text{W}\) or \(\text{J s}^{-1}\)
Potential difference \(V\) volt, \(\text{V}\)
Current \(I\) ampere, \(\text{A}\)
Resistance \(R\) ohm, \(\Omega\)
Time \(t\) second, \(\text{s}\)
Lamp type Power Energy used per day Cost for 30 days
Compact Fluorescent Lamp (CFL) \(40\text{ W}\) \(0.04\text{ kW} \times 12\text{ h} = 0.48\text{ units}\) \(\text{RM}3.139\)
Light Emitting Diode (LED) \(12\text{ W}\) \(0.012\text{ kW} \times 12\text{ h} = 0.144\text{ units}\) \(\text{RM}0.942\)

Steps in Reducing Household Electrical Energy Usage

Close all windows and doors when using the air conditioner and keep the filter clean. Switch off appliances when not in use. Use only full loads of laundry in the washing machine. Use energy-saving lamps.

Try to Answer First

Answer in your mind, then press “Check Answer”.

1

An electric heater has an output power of \(80\text{ W}\). How much electrical energy does it supply in \(10\text{ seconds}\)?

Check Answer
Answer: \(E = Pt = 80\text{ W} \times 10\text{ s} = 800\text{ J}\).
2

A fluorescent lamp is rated \(240\text{ V}, 32\text{ W}\). What is its resistance under normal operating conditions?

Check Answer
Answer: \(R = \dfrac{V^2}{P} = \dfrac{240^2}{32} = \dfrac{57600}{32} = 1800\ \Omega\).
3

If electricity costs \(30\text{ sen}\) per unit, what is the cost of operating a \(1\text{ kW}\) vacuum cleaner for half an hour?

Check Answer
Answer: Energy used \(= 1\text{ kW} \times 0.5\text{ h} = 0.5\text{ units}\). Cost \(= 0.5\text{ units} \times 30\text{ sen} = 15\text{ sen}\) (or \(\text{RM}0.15\)).

Common Mistakes

  • !Using time in minutes instead of seconds when calculating energy in joules using \(E = VIt\) or \(E = Pt\).
  • !Using power in watts (\(\text{W}\)) directly instead of converting it to kilowatts (\(\text{kW}\)) when calculating electrical energy consumption in units (\(\text{kWh}\)).
  • !Using seconds instead of hours when calculating energy in kilowatt-hours (\(\text{kWh}\)).
  • !Assuming an appliance always consumes its rated power even when the supplied voltage differs from its stated working voltage.

Concept Misunderstandings

Misunderstanding

1 kilowatt-hour (\(\text{kWh}\)) is a unit of electrical power.

Correct Concept

Kilowatt-hour (\(\text{kWh}\)) is a unit of electrical energy, not power. It represents the energy consumed by a \(1\text{ kW}\) appliance running for \(1\text{ hour}\) (\(1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J}\)).

Misunderstanding

A higher power rating always means an appliance is more energy-saving.

Correct Concept

A higher power rating means the appliance consumes more electrical energy per second. For the same output brightness or task, an appliance with a lower power rating (such as a \(12\text{ W}\) LED compared to a \(40\text{ W}\) CFL) consumes less energy and has higher efficiency.

Summary

  • Electrical energy: \(E = VIt\) (S.I. unit: Joule, \(\text{J}\)).
  • Electrical power: \(P = \dfrac{E}{t} = VI = I^2R = \dfrac{V^2}{R}\) (S.I. unit: Watt, \(\text{W}\) or \(\text{J s}^{-1}\)).
  • Working voltage: The potential difference required for an appliance to operate normally at its stated power rating.
  • Electrical billing unit: \(1\text{ unit} = 1\text{ kWh} = \text{Power in kW} \times \text{Time in hours}\).
  • Energy conservation: Use high-efficiency appliances (e.g., LED), full washing machine loads, clean air-conditioner filters, closed rooms during air conditioning, and switch off unused devices.

Short Activity

Objective: Test your understanding of electrical energy, power, working voltage and electricity cost.

A. Objective Quiz

1 What is the electrical power of an appliance that operates with a potential difference of \(240\text{ V}\) and carries a current of \(0.5\text{ A}\)?

2 Which of the following defines 1 unit of electricity used in household billing?

B. Fill in the Blanks

3 The potential difference required for an electrical appliance to operate under normal conditions is known as the voltage.

4 The S.I. unit for electrical power is the , which is equivalent to \(\text{J s}^{-1}\).

C. Matching / Drag and Drop

Drag each term to the correct formula or description. If using a phone, tap the answer first, then tap the matching box.

Concept / Term
Electrical energy (\(E\))
Electrical power (\(P\)) in terms of \(V\) and \(R\)
Electrical power (\(P\)) in terms of \(I\) and \(R\)
1 Joule
Formula / Description
\(VIt\)Correct ✓Try againEmpty
\(\dfrac{V^2}{R}\)Correct ✓Try againEmpty
\(I^2R\)Correct ✓Try againEmpty
Energy used when \(1\text{ A}\) flows under \(1\text{ V}\) for \(1\text{ s}\)Correct ✓Try againEmpty
 

Keywords

Electrical energy (\(E\)) Electrical power (\(P\)) Working voltage Joule (\(\text{J}\)) Watt (\(\text{W}\)) Kilowatt-hour (\(\text{kWh}\))

Electrical Energy and Power

3.4 Electrical Energy and Power
PHYSICS • Form 5 • Chapter 3: Electricity

3.4 Electrical Energy and Power

Electrical appliances in our daily lives convert electrical energy into other forms of energy such as heat, light, and mechanical energy. Understanding the mathematical relationships between electrical energy, power, voltage, current, and resistance allows consumers to determine energy usage, calculate electrical consumption costs, and choose energy-efficient appliances to reduce household expenditure.

Learning Objectives

  • State the relationship between electrical energy (\(E\)), voltage (\(V\)), current (\(I\)), and time (\(t\)).
  • Define electrical power (\(P\)) and formulate its relationship with voltage (\(V\)), current (\(I\)), and resistance (\(R\)).
  • Solve numerical problems involving electrical energy and power in daily life.
  • Compare the power and energy consumption rates of various electrical devices, such as CFL and LED lamps.
  • Calculate the cost of electrical energy consumption based on kilowatt-hours (\(\text{kWh}\) or units).
  • Identify and suggest steps to reduce household electrical energy usage.

Summary of Formulae for Electrical Energy and Power

See how voltage, current, resistance and time connect to electrical energy, power and electricity cost.

Formula Tree

\(E = VIt\) Electrical energy
\(P = \dfrac{E}{t} = VI\) Electrical power
\(P = I^2R\) Using \(V = IR\)
\(P = \dfrac{V^2}{R}\) Using \(I = \dfrac{V}{R}\)

Electricity Cost Conversion

From joules to billing units

\(1\text{ unit} = 1\text{ kWh}\)

\(\text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (h)}\)

\(\text{Cost} = \text{Energy (units)} \times \text{Cost per unit}\)

Power rating and working voltage

An appliance labelled \(240\text{ V}, 700\text{ W}\) uses \(700\text{ J}\) of energy every second when supplied with \(240\text{ V}\).

Short Explanation

Electrical Energy

Electrical energy is the work done by electric charges moving through a potential difference. It is given by \(E = VIt\), where \(V\) is potential difference, \(I\) is current and \(t\) is time.

Electrical Power

Electrical power is the rate at which electrical energy is used. \(P = \dfrac{E}{t} = VI\). Using Ohm's law, it can also be written as \(P = I^2R\) or \(P = \dfrac{V^2}{R}\).

Working Voltage

Working voltage is the potential difference required for an electrical appliance to operate under normal conditions. At this voltage, the appliance consumes its stated power rating.

Electricity Billing Unit

Household electricity is billed in kilowatt-hours. \(1\text{ unit} = 1\text{ kWh}\). Energy used is calculated as power in kilowatts multiplied by time in hours.

Problem-solving framework

Step 1: Identify the problem → Step 2: Identify the information given → Step 3: Identify the formula that can be used → Step 4: Solve the problem numerically
Quantity Symbol SI unit
Electrical energy \(E\) joule, \(\text{J}\)
Electrical power \(P\) watt, \(\text{W}\) or \(\text{J s}^{-1}\)
Potential difference \(V\) volt, \(\text{V}\)
Current \(I\) ampere, \(\text{A}\)
Resistance \(R\) ohm, \(\Omega\)
Time \(t\) second, \(\text{s}\)
Lamp type Power Energy used per day Cost for 30 days
Compact Fluorescent Lamp (CFL) \(40\text{ W}\) \(0.04\text{ kW} \times 12\text{ h} = 0.48\text{ units}\) \(\text{RM}3.139\)
Light Emitting Diode (LED) \(12\text{ W}\) \(0.012\text{ kW} \times 12\text{ h} = 0.144\text{ units}\) \(\text{RM}0.942\)

Steps in Reducing Household Electrical Energy Usage

Close all windows and doors when using the air conditioner and keep the filter clean. Switch off appliances when not in use. Use only full loads of laundry in the washing machine. Use energy-saving lamps.

Try to Answer First

Answer in your mind, then press “Check Answer”.

1

An electric heater has an output power of \(80\text{ W}\). How much electrical energy does it supply in \(10\text{ seconds}\)?

Check Answer
Answer: \(E = Pt = 80\text{ W} \times 10\text{ s} = 800\text{ J}\).
2

A fluorescent lamp is rated \(240\text{ V}, 32\text{ W}\). What is its resistance under normal operating conditions?

Check Answer
Answer: \(R = \dfrac{V^2}{P} = \dfrac{240^2}{32} = \dfrac{57600}{32} = 1800\ \Omega\).
3

If electricity costs \(30\text{ sen}\) per unit, what is the cost of operating a \(1\text{ kW}\) vacuum cleaner for half an hour?

Check Answer
Answer: Energy used \(= 1\text{ kW} \times 0.5\text{ h} = 0.5\text{ units}\). Cost \(= 0.5\text{ units} \times 30\text{ sen} = 15\text{ sen}\) (or \(\text{RM}0.15\)).

Common Mistakes

  • !Using time in minutes instead of seconds when calculating energy in joules using \(E = VIt\) or \(E = Pt\).
  • !Using power in watts (\(\text{W}\)) directly instead of converting it to kilowatts (\(\text{kW}\)) when calculating electrical energy consumption in units (\(\text{kWh}\)).
  • !Using seconds instead of hours when calculating energy in kilowatt-hours (\(\text{kWh}\)).
  • !Assuming an appliance always consumes its rated power even when the supplied voltage differs from its stated working voltage.

Concept Misunderstandings

Misunderstanding

1 kilowatt-hour (\(\text{kWh}\)) is a unit of electrical power.

Correct Concept

Kilowatt-hour (\(\text{kWh}\)) is a unit of electrical energy, not power. It represents the energy consumed by a \(1\text{ kW}\) appliance running for \(1\text{ hour}\) (\(1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J}\)).

Misunderstanding

A higher power rating always means an appliance is more energy-saving.

Correct Concept

A higher power rating means the appliance consumes more electrical energy per second. For the same output brightness or task, an appliance with a lower power rating (such as a \(12\text{ W}\) LED compared to a \(40\text{ W}\) CFL) consumes less energy and has higher efficiency.

Summary

  • Electrical energy: \(E = VIt\) (S.I. unit: Joule, \(\text{J}\)).
  • Electrical power: \(P = \dfrac{E}{t} = VI = I^2R = \dfrac{V^2}{R}\) (S.I. unit: Watt, \(\text{W}\) or \(\text{J s}^{-1}\)).
  • Working voltage: The potential difference required for an appliance to operate normally at its stated power rating.
  • Electrical billing unit: \(1\text{ unit} = 1\text{ kWh} = \text{Power in kW} \times \text{Time in hours}\).
  • Energy conservation: Use high-efficiency appliances (e.g., LED), full washing machine loads, clean air-conditioner filters, closed rooms during air conditioning, and switch off unused devices.

Short Activity

Objective: Test your understanding of electrical energy, power, working voltage and electricity cost.

A. Objective Quiz

1 What is the electrical power of an appliance that operates with a potential difference of \(240\text{ V}\) and carries a current of \(0.5\text{ A}\)?

2 Which of the following defines 1 unit of electricity used in household billing?

B. Fill in the Blanks

3 The potential difference required for an electrical appliance to operate under normal conditions is known as the voltage.

4 The S.I. unit for electrical power is the , which is equivalent to \(\text{J s}^{-1}\).

C. Matching / Drag and Drop

Drag each term to the correct formula or description. If using a phone, tap the answer first, then tap the matching box.

Concept / Term
Electrical energy (\(E\))
Electrical power (\(P\)) in terms of \(V\) and \(R\)
Electrical power (\(P\)) in terms of \(I\) and \(R\)
1 Joule
Formula / Description
\(VIt\)Correct ✓Try againEmpty
\(\dfrac{V^2}{R}\)Correct ✓Try againEmpty
\(I^2R\)Correct ✓Try againEmpty
Energy used when \(1\text{ A}\) flows under \(1\text{ V}\) for \(1\text{ s}\)Correct ✓Try againEmpty
 

Keywords

Electrical energy (\(E\)) Electrical power (\(P\)) Working voltage Joule (\(\text{J}\)) Watt (\(\text{W}\)) Kilowatt-hour (\(\text{kWh}\))