4.2.5 - Electrical power, energy and cost

Power as Energy Transfer Rate

Power is a rate. A component with a large power transfers energy quickly; it does not necessarily transfer a large total amount of energy unless it is operating for long enough.

In an electrical component, the p.d. tells you the energy transferred per unit charge, and the current tells you the charge passing per unit time.

• $V$ is the potential difference across the component, in volts.
• $I$ is the current through the component, in amperes.
• $P$ is the power transferred by the component, in watts.

If a charge $Q$ passes through a p.d. $V$, the energy transferred is $W = VQ$. If the charge passes in time $t$, then $I = Q / t$. So:

The unit check is useful:

That is why a p.d. of $12\,\mathrm{V}$ and a current of $0.50\,\mathrm{A}$ gives a power of $6.0\,\mathrm{W}$.

For a practical measurement of electrical power, measure the current through the component with an ammeter in series, and measure the p.d. across the same component with a voltmeter in parallel. The word same matters: using the total supply p.d. with the current through only one branch is a common circuit error.

Worked example: power of a lamp

A lamp has a p.d. of $5.8\,\mathrm{V}$ across it and a current of $0.42\,\mathrm{A}$ through it.

Use $P = VI$:

$P = 5.8\times 0.42 = 2.436\,\mathrm{W}$

The measurements are both given to 2 significant figures, so:

$P = 2.4\,\mathrm{W}$

This means the lamp transfers $2.4\,\mathrm{J}$ of energy each second.

[DIAGRAM: asset_slug: 4_02_5_electrical_power_and_energy_cost__diagram_01; description: Electrical power is calculated using the p.d. across and current through the same load.; caption: Measure current through and p.d. across the same load to determine power.]

Resistor Power Equations

For a resistor, the p.d., current and resistance are linked by $V = IR$. Combining this with $P = VI$ gives two more useful forms.

Use these with quantities for the same resistor or load:

• use $P = VI$ when $V$ and $I$ are known
• use $P = I^{2}R$ when $I$ and $R$ are known
• use $P = V^{2} / R$ when $V$ and $R$ are known

The two resistance forms can seem to contradict each other. They do not. They describe different controlled conditions.

• If the current is fixed, $P = I^{2}R$, so a larger resistance transfers more power.
• If the p.d. is fixed, $P = V^{2} / R$, so a larger resistance transfers less power.

The control condition decides which comparison is valid.

Worked example: choosing the equation

A $22\,\Omega$ resistor has a current of $0.15\,\mathrm{A}$ through it.

The known quantities are $I$ and $R$, so use $P = I^{2}R$:

$P = (0.15)^{2}\times 22$

$P = 0.0225\times 22 = 0.495\,\mathrm{W}$

To 2 significant figures:

$P = 0.50\,\mathrm{W}$

As a check, the p.d. across the resistor would be:

Then $P = VI = 3.3\times 0.15 = 0.495\,\mathrm{W}$, the same result.

Energy Transferred by a Circuit

Power tells you the rate of energy transfer. To find the total energy transferred, multiply by the time.

Here $W$ is the energy transferred or work done, in joules. This is a different use of the letter $W$ from the unit symbol W for watt. In written work, use context and units to keep them separate.

The equation $W = VIt$ follows from $W = VQ$ and $Q = It$. It is useful when a circuit question gives p.d., current and time directly.

Worked example: energy transferred by a heater

A low-voltage heater has a p.d. of $12\,\mathrm{V}$ across it and a current of $1.8\,\mathrm{A}$ through it. It is switched on for $3.0\,\mathrm{min}$.

First convert the time:

Then use $W = VIt$:

$W = 12\times 1.8\times 180$

$W = 3888\,\mathrm{J}$

To 2 significant figures:

$W = 3.9\times 10^{3}\,\mathrm{J}$

You can also calculate the power first:

Both routes are the same physics.

The product $W=Pt$ assumes constant power, or that $P$ is the mean power over the stated interval. If a device cycles between settings, calculate $P_1t_1+P_2t_2+\cdots$. More generally, the area under a power–time graph is the energy transferred.