Calculating Electrical Energy and Operating Costs

Welcome back. In the previous lesson, you calculated power using $P=VI$, $P=I^{2}R$, and $\frac{V^2}{R}$. Power tells you the rate at which energy is being used. This lesson adds time: it shows how to calculate the electrical energy an appliance uses and what that energy costs at a stated tariff.

This is a common exam pattern: identify the appliance power, convert the operating time, calculate kilowatt-hours, then multiply by the price per kilowatt-hour. Clear units at every stage will earn method marks and prevent the most common mistakes.

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From power to energy

A watt is a rate of energy use:

$$1\mathrm{W}=1J/s$$

So a $1000\mathrm{W}$ heater transfers $1000\mathrm{J}$ of energy every second while it runs. But an electricity retailer does not normally bill a household in joules. The billing unit is the kilowatt-hour, written $\mathrm{kWh}$.

A kilowatt-hour is an amount of energy:

$$1\mathrm{kWh}$$

means a load with a power of $1\mathrm{kW}$ operating for $1\mathrm{h}$.

The key energy relationship is:

$$E(\mathrm{kWh})=P(\mathrm{kW})\times t(\mathrm{h})$$

where:

Symbol	Meaning	Required unit for billing calculation
$E$	Electrical energy used	$\mathrm{kWh}$
$P$	Power of the appliance or load	$\mathrm{kW}$
$t$	Operating time	$\mathrm{h}$

The units show why this works:

$$\mathrm{kW}\times \mathrm{h}=\mathrm{kWh}$$

It is important not to confuse $\mathrm{kW}$ and $\mathrm{kWh}$:

• $\mathrm{kW}$ measures power: how quickly energy is being used.
• $\mathrm{kWh}$ measures energy: the total amount used over a period.

For example, a $2\mathrm{kW}$ heater running for $3\mathrm{h}$ uses:

$$E=2\times 3$$ $$E=6\mathrm{kWh}$$

A smaller $0.5\mathrm{kW}$ appliance running for $12\mathrm{h}$ uses exactly the same energy:

$$E=0.5\times 12$$ $$E=6\mathrm{kWh}$$

So a high-power appliance used briefly can consume the same energy as a low-power appliance used for much longer.

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Set up the units before you calculate

For billing questions, use kilowatts and hours. Convert before substituting into the formula.

Power conversion

$$1\mathrm{kW}=1000\mathrm{W}$$

To change watts to kilowatts, divide by $1000$:

$$P(\mathrm{kW})=\frac{P(\mathrm{W})}{1000}$$

Examples:

$$2000\mathrm{W}=2\mathrm{kW}$$ $$750\mathrm{W}=0.75\mathrm{kW}$$ $$60\mathrm{W}=0.060\mathrm{kW}$$

Time conversion

$$1\mathrm{h}=60\min$$

To change minutes to hours, divide by $60$:

$$t(\mathrm{h})=\frac{t(\min )}{60}$$

Examples:

$$30\min =0.5\mathrm{h}$$ $$45\min =0.75\mathrm{h}$$ $$90\min =1.5\mathrm{h}$$

For days:

$$1\mathrm{day}=24\mathrm{h}$$ $$3\mathrm{days}=3\times 24=72\mathrm{h}$$

There is another valid energy pairing used in science:

$$E(\mathrm{J})=P(\mathrm{W})\times t(\mathrm{s})$$

That gives energy in joules. For an electricity-cost question, however, use $\mathrm{kW}$ and $\mathrm{h}$, so the answer is directly in $\mathrm{kWh}$.

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From energy used to operating cost

A tariff tells you the price of one unit of energy. In New Zealand, it is commonly stated in cents per kilowatt-hour, written $c/kWh$.

The cost formula is:

$$C(\mathrm{c})=E(\mathrm{kWh})\times r(c/kWh)$$

where $r$ is the tariff.

The units cancel neatly:

$$\mathrm{kWh}\times \frac{\mathrm{c}}{\mathrm{kWh}}=\mathrm{c}$$

If the tariff is given in dollars per kilowatt-hour instead, your result will be in dollars:

$$C(\mathrm{NZD})=E(\mathrm{kWh})\times r(NZD/kWh)$$

A useful exam rule is:

• tariff in cents per kWh: calculate cost in cents, then divide by $100$ if the answer is required in dollars;
• tariff in dollars per kWh: calculate cost directly in dollars.

$$100\mathrm{c}=1\mathrm{NZD}$$

Worked example 1: energy and cost for one appliance

Question: A $2.4\mathrm{kW}$ heater operates for $45\min$. Electricity costs $32c/kWh$. Calculate:

1. energy used;
2. operating cost.

Step 1: List the known values

$$P=2.4\mathrm{kW}$$ $$t=45\min$$ $$r=32c/kWh$$

Step 2: Convert time to hours

$$t=\frac{45}{60}$$ $$t=0.75\mathrm{h}$$

Step 3: Calculate energy

$$E=Pt$$ $$E=2.4\times 0.75$$ $$E=1.8\mathrm{kWh}$$

Step 4: Calculate cost

$$C=E\times r$$ $$C=1.8\times 32$$ $$C=57.6\mathrm{c}$$ $$\boxed{C=57.6\mathrm{c}}$$

Or, in dollars:

$$C=\frac{57.6}{100}$$ $$\boxed{C=0.576\mathrm{NZD}}$$

Rounded to the nearest cent:

$$\boxed{C=0.58\mathrm{NZD}}$$

A quick check: the heater uses less than $2\mathrm{kWh}$, and each kWh costs $32\mathrm{c}$, so the cost must be less than $64\mathrm{c}$. The answer is sensible.

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A reliable exam method

For almost every energy-cost question, use this order.

1. Write the required quantities.
   Usually energy in $\mathrm{kWh}$, cost in cents, or cost in dollars.

2. List the known values with units.
   Include power, time, tariff, number of appliances, and number of days if given.

3. Calculate total power if there is more than one identical load.
   For example, four $60\mathrm{W}$ lamps have a total power of:

   $$P=4\times 60$$ $$P=240\mathrm{W}$$

4. Convert power to $\mathrm{kW}$.

5. Convert operating time to $\mathrm{h}$.

6. Calculate energy using $E=Pt$.

7. Calculate cost using $C=E\times r$.

8. Round money sensibly.
   Usually, round the final answer to the nearest cent, unless the question specifies otherwise.

This layout also protects you from a frequent error: multiplying watt-hours by cents per kilowatt-hour. The numerical answer may look believable, but the units do not match. Convert watts to kilowatts first.

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Worked example 2: daily use over a month

Question: Four $60\mathrm{W}$ lamps operate for $4\mathrm{h}$ per day over a $30$-day month. Electricity costs $28c/kWh$. Calculate the energy used and the operating cost.

Step 1: Find the total lamp power

There are four lamps:

$$P=4\times 60$$ $$P=240\mathrm{W}$$

Convert to kilowatts:

$$P=\frac{240}{1000}$$ $$P=0.240\mathrm{kW}$$

Step 2: Find total operating time

$$t=4\times 30$$ $$t=120\mathrm{h}$$

Step 3: Calculate energy

$$E=Pt$$ $$E=0.240\times 120$$ $$E=28.8\mathrm{kWh}$$

Step 4: Calculate cost

$$C=E\times r$$ $$C=28.8\times 28$$ $$C=806.4\mathrm{c}$$

Convert cents to dollars:

$$C=\frac{806.4}{100}$$ $$\boxed{C=8.06\mathrm{NZD}}$$

The total energy is $28.8\mathrm{kWh}$, and the variable energy cost is $\mathrm{NZD}8.06$.

The table shows an efficient way to manage a multi-appliance question. For each item, calculate:

$$E(\mathrm{kWh})=P(\mathrm{kW})\times t(\mathrm{h})$$

Then add the energy values:

$$E_{\mathrm{total}}=E_1+E_2+E_3+\cdots$$

Finally, multiply the total by the tariff.

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Worked example 3: calculate power first, then energy and cost

Some questions do not give appliance power directly. They may give voltage and current, which connects directly to the previous lesson.

Question: A $230\mathrm{V}$ appliance draws $8.0\mathrm{A}$ for $2.5\mathrm{h}$. The tariff is $29.0c/kWh$. Calculate the operating cost.

Step 1: Calculate power

Voltage and current are known, so use:

$$P=VI$$ $$P=230\times 8.0$$ $$P=1840\mathrm{W}$$

Convert power to kilowatts:

$$P=\frac{1840}{1000}$$ $$P=1.84\mathrm{kW}$$

Step 2: Calculate energy

$$E=Pt$$ $$E=1.84\times 2.5$$ $$E=4.6\mathrm{kWh}$$

Step 3: Calculate cost

$$C=E\times r$$ $$C=4.6\times 29.0$$ $$C=133.4\mathrm{c}$$ $$C=\frac{133.4}{100}$$ $$\boxed{C=1.33\mathrm{NZD}}$$

Keep the power calculation, energy calculation, and cost calculation on separate lines. Even if a later arithmetic slip occurs, this working can still earn method marks.

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What is included in an electricity bill?

For most exam questions, “operating cost” means the variable energy charge:

$$\text{energy cost}=\text{kWh used}\times \text{tariff}$$

A real electricity bill may also include a fixed daily charge. The Electricity Authority notes that a fixed charge is paid for maintaining the connection and network access, even if no energy is used.

Only include a fixed charge if the question states one. If it does, calculate it separately:

$$\text{fixed charge}=\text{daily charge}\times \text{number of days}$$

Then add it to the energy cost.

If a question gives different tariffs for different time periods, such as peak and off-peak rates, calculate each period’s energy cost separately and add the costs. Do not apply one tariff to all energy unless the question tells you to.

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Common errors to avoid

Using watts directly with a kWh tariff

This is incorrect:

$$500\mathrm{W}\times 2\mathrm{h}=1000\mathrm{kWh}$$

The correct calculation converts $500\mathrm{W}$ first:

$$500\mathrm{W}=0.5\mathrm{kW}$$ $$E=0.5\times 2=1\mathrm{kWh}$$

Leaving minutes as hours

This is incorrect:

$$E=2\mathrm{kW}\times 30\min$$

Convert first:

$$30\min =0.5\mathrm{h}$$ $$E=2\times 0.5=1\mathrm{kWh}$$

Forgetting the number of appliances

If there are five identical lamps, heaters, or chargers, multiply the power of one item by five before calculating energy.

Confusing daily use with monthly use

If an appliance operates $3\mathrm{h}$ per day for $30$ days:

$$t=3\times 30=90\mathrm{h}$$

Do not use $3\mathrm{h}$ as the time for the whole month.

Rounding too early

Keep extra calculator digits until the final cost. For money, then round to the nearest cent unless instructed otherwise.

Treating a stated appliance rating as a guaranteed real-life average

In an exam, use the stated rating and stated operating time unless told otherwise. In real installations, a thermostat-controlled heater, refrigerator, or heat pump may switch on and off, so its actual energy use can be less than its maximum-rated power multiplied by the full elapsed time.

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Key takeaways

For electricity billing calculations, use:

$$E(\mathrm{kWh})=P(\mathrm{kW})\times t(\mathrm{h})$$

Then calculate the energy charge:

$$C(\mathrm{c})=E(\mathrm{kWh})\times r(c/kWh)$$

Remember these conversions:

$$1\mathrm{kW}=1000\mathrm{W}$$ $$1\mathrm{h}=60\min$$ $$100\mathrm{c}=1\mathrm{NZD}$$

The safe exam sequence is: total the appliance power if needed, convert to $\mathrm{kW}$, convert time to hours, calculate $\mathrm{kWh}$, then apply the tariff. Check whether the final cost is reasonable for the appliance power and duration.

Next, you will calculate efficiency from input and output power or energy, using the same careful unit handling and percentage calculations.