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Electrical Energy and Power: Formulas & Calculations

CSEC Physics: Electricity Fundamentals

Essential Understanding: Electrical energy powers our modern world, and understanding how to calculate it is crucial. Energy is the capacity to do work, while power is the rate at which energy is transferred. Master these concepts to calculate electricity costs, design circuits, and solve real-world problems.

🔑 Key Skill: Calculating Energy ($E = VIt$) & Power ($P = VI$)

📈 Exam Focus: kWh calculations & cost of electricity

🎯 Problem Solving: Power ratings & efficiency

Core Concepts & Formulas

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Electrical Energy

Definition: The ability of an electric current to do work, measured in joules (J).

Main Formula: \[ E = V \times I \times t \]

$ E $: Energy in joules (J)
$ V $: Potential difference in volts (V)
$ I $: Current in amperes (A)
$ t $: Time in seconds (s)

Alternative Formulas: $ E = I^2 R t $ or $ E = \frac{V^2}{R} t $

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Electrical Power

Definition: The rate at which electrical energy is transferred, measured in watts (W).

Main Formula: \[ P = V \times I \]

$ P $: Power in watts (W)
$ V $: Potential difference in volts (V)
$ I $: Current in amperes (A)

Alternative Formulas: $ P = I^2 R $ or $ P = \frac{V^2}{R} $

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Cost of Electricity

Definition: The monetary cost of electrical energy consumed, calculated using kilowatt-hours.

Key Unit: 1 kilowatt-hour (kWh) = 3.6 × 10⁶ J

Cost Formula: \[ \text{Cost} = \text{Power (kW)} \times \text{Time (h)} \times \text{Rate (\$/kWh)} \]

Utility companies bill customers based on kWh consumed.

The Complete Formula Network

All electrical energy and power formulas are interconnected through Ohm’s Law ($V = IR$):

\[ P = VI = I^2R = \frac{V^2}{R} \]

\[ E = Pt = VIt = I^2Rt = \frac{V^2}{R}t \]

Choose the formula based on the given variables in the problem.

Interactive Circuit Simulator

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Energy & Power Explorer

Objective: Adjust voltage and resistance to see how it affects current, power, and energy consumption over time.

Voltage (V): 12 V Range: 1-24V

Resistance (R): 6 Ω Range: 1-20Ω

Time: 5 s Range: 1-60s

Current (I)

2.00 A

$I = V/R$

Power (P)

24.00 W

$P = VI$

Energy (E)

120.00 J

$E = Pt$

Observations:

With V=12V and R=6Ω, the current is 2A. The power dissipated is 24W, and over 5 seconds, 120J of energy is used.

Units & Conversions

Analysis: The chart compares power ratings of common household appliances. Note how heating devices (kettle, iron) typically have higher power ratings than electronics (TV, laptop). Higher power means more energy consumed per second.

Joule to Kilowatt-hour

\[ 1 \text{ kWh} = 3.6 \times 10^6 \text{ J} \]

Utility companies use kWh because joules are too small for household energy measurement.

Watt to Kilowatt

\[ 1 \text{ kW} = 1000 \text{ W} \]

Convert watts to kilowatts before calculating energy costs.

Energy from Power

\[ E (\text{kWh}) = P (\text{kW}) \times t (\text{h}) \]

For cost calculations, ensure all units are consistent (kW and hours).

Unit	Symbol	Equivalent	Usage
Joule	J	1 N·m	Basic unit of energy
Watt	W	1 J/s	Unit of power
Kilowatt-hour	kWh	3.6 × 10⁶ J	Electricity billing
Kilowatt	kW	1000 W	Appliance power ratings

Electricity Cost Calculator

Calculate how much it costs to run an electrical appliance:

Power Rating (W)

Hours used per day

Days used per month

Electricity Rate ($/kWh)

Click “Calculate” to see the monthly cost

Worked Examples & Past Paper Questions

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Example 1: Basic Energy Calculation (CSEC 2019)

Question: A 12V battery supplies a current of 2A to a resistor for 5 minutes. Calculate the energy supplied.

Identify known values: $V = 12\text{V}$, $I = 2\text{A}$, $t = 5 \text{ minutes} = 300 \text{ s}$

Select appropriate formula: $E = VIt$

Substitute values: $E = 12 \times 2 \times 300$

Calculate: $E = 7200 \text{ J}$

Answer: The energy supplied is 7200 joules.

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Example 2: Cost Calculation (CSEC 2021)

Question: A 2kW electric heater is used for 3 hours each day. If electricity costs $0.30 per kWh, calculate the cost of running the heater for 30 days.

Calculate daily energy consumption: $E_{\text{daily}} = P \times t = 2\text{kW} \times 3\text{h} = 6 \text{ kWh}$

Calculate monthly energy consumption: $E_{\text{monthly}} = 6 \text{ kWh/day} \times 30 \text{ days} = 180 \text{ kWh}$

Calculate cost: $\text{Cost} = 180 \text{ kWh} \times \$0.30/\text{kWh} = \$54.00$

Answer: The cost for 30 days is $54.00.

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Example 3: Finding Resistance (CSEC 2018)

Question: An electric lamp is rated 60W, 240V. Calculate the resistance of the lamp when operating normally.

Identify known values: $P = 60\text{W}$, $V = 240\text{V}$

Select appropriate formula: $P = \frac{V^2}{R}$ (since we know P and V)

Rearrange formula: $R = \frac{V^2}{P}$

Substitute values: $R = \frac{240^2}{60} = \frac{57600}{60}$

Calculate: $R = 960 \Omega$

Answer: The resistance of the lamp is 960Ω.

Key Examination Insights

Common Mistakes

Forgetting to convert time to seconds when using $E = VIt$ (energy in joules).
Mixing up watts and kilowatts in cost calculations.
Using the wrong formula variant (e.g., using $P = VI$ when you only know $V$ and $R$).
Not converting minutes to seconds or hours appropriately.

Success Strategies

Always write down all known variables before selecting a formula.
For cost questions: Power in kW × Time in hours = Energy in kWh.
Remember that a “rating” on an appliance (e.g., “60W, 240V”) gives the power at that voltage.
Check units at each step of your calculation.

CSEC Practice Arena

Test Your Understanding

Which formula is used to calculate electrical energy?

$E = VIt$

$P = VI$

$V = IR$

$I = Q/t$

Explanation: The fundamental formula for electrical energy is $E = VIt$, where V is voltage, I is current, and t is time in seconds. This gives energy in joules.

A 100W light bulb is left on for 8 hours. How much energy does it consume in kWh?

0.08 kWh

0.8 kWh

8 kWh

800 kWh

Solution: First convert 100W to kW: 100W = 0.1kW. Then $E = P \times t = 0.1\text{kW} \times 8\text{h} = 0.8 \text{kWh}$.

A television rated at 240V, 120W operates for 5 hours. If electricity costs $0.25 per kWh, what is the cost of operation?

$0.15

$0.25

$0.15

$1.50

Solution: Energy used = $120\text{W} = 0.12\text{kW} \times 5\text{h} = 0.6 \text{kWh}$. Cost = $0.6 \times 0.25 = \$0.15$.

Which appliance would cost the MOST to run for 1 hour?

60W light bulb

2kW electric kettle

150W computer

1000W microwave

Explanation: The electric kettle has the highest power rating (2kW = 2000W), so it consumes the most energy per hour, resulting in the highest cost.

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CSEC Examination Mastery Tip

Power vs Energy: Many students confuse these concepts. Remember:

Power (Watts): The rate at which energy is used. It’s like speed (how fast you’re using energy).
Energy (Joules or kWh): The total amount of energy used. It’s like distance (how much energy you’ve used total).
Relationship: Energy = Power × Time (just like Distance = Speed × Time).

In cost calculation questions, always convert power to kilowatts and time to hours before multiplying.