Energy and Power in Circuit Devices

How a current transfers energy

When a current flows through a component, the power supply does work and transfers energy to the component and its surroundings. Work done is another name for the energy transferred, so both are measured in joules (J).

The change produced depends on the device. A current through a resistor heats it, increasing its internal energy; energy can then be transferred by heating to its surroundings. A filament lamp becomes hot and transfers energy to its surroundings by light and heating. An electric motor can do mechanical work, such as turning a fan and increasing the kinetic energy of moving parts, while also warming its surroundings.

Energy is conserved. The energy transferred from the supply is accounted for by changes in the devices and their surroundings. Charge is not used up as it passes through a component: it is energy that is transferred.

Potential difference tells us the energy transferred per coulomb

Current describes how quickly charge flows. Potential difference describes how much energy is transferred for each coulomb of charge passing through a component.

A potential difference of 6 V means that 6 J is transferred for every coulomb that passes through the component. The total energy therefore depends on both the potential difference and the amount of charge:

$$E = QV$$

Here, $E$ is energy transferred in joules (J), $Q$ is charge in coulombs (C), and $V$ is potential difference in volts (V).

For example, suppose 20 C passes through a component with a potential difference of 6 V across it. The energy transferred is $20 \times 6 = 120\text{ J}$. Passing twice as much charge through the same potential difference would transfer twice as much energy.

The relationship can also be rearranged: $Q = E/V$ or $V = E/Q$.

Power tells us how quickly energy is transferred

Two devices can transfer the same amount of energy but take different lengths of time. Power is the rate of energy transfer, not the total energy transferred:

$$P = \frac{E}{t}$$

Here, $P$ is power in watts (W), $E$ is energy transferred in joules (J), and $t$ is time in seconds (s). One watt means one joule transferred each second.

A device transferring 600 J in 30 s has a power of $600/30 = 20\text{ W}$. It transfers energy at an average rate of 20 J each second.

An appliance's power rating tells us its rate of energy transfer under its stated operating conditions. A 2000 W heater transfers 2000 J each second, whereas a 1000 W heater transfers 1000 J each second. Over the same operating time, the higher-power heater transfers twice as much energy. This describes the total transfer from the supply, not necessarily just the useful output.

To calculate total energy, rearrange the equation:

$$E = Pt$$

For example, a 20 W device operating for 60 s transfers $20 \times 60 = 1200\text{ J}$. A lower-power device operating for longer can therefore transfer more energy than a higher-power device operating briefly.

Connecting power to current and potential difference

Potential difference tells us the energy transferred per coulomb, while current tells us the number of coulombs passing each second. Multiplying them gives the energy transferred each second:

$$P = VI$$

Here, $V$ is potential difference in volts (V) and $I$ is current in amperes (A).

Suppose a motor has 12 V across it and a current of 2 A through it. Each coulomb transfers 12 J, and 2 C passes each second. Its electrical input power is therefore $12 \times 2 = 24\text{ W}$.

This equation can be rearranged to find an unknown current or potential difference: $I = P/V$ and $V = P/I$.

Calculating power using resistance

Resistance links potential difference and through . Substituting this into gives another useful equation:

$$P = I^2R$$

Here, $R$ is resistance in ohms (Ω). Use this form when the current and resistance are known.

For example, a 5 Ω resistor carrying a current of 2 A transfers energy at a rate of $2^2 \times 5 = 20\text{ W}$. The current must be squared: $2^2 = 4$.

For a fixed resistance, doubling the current makes the power four times as large. This square relationship explains why increasing current can greatly increase the rate of heating in a resistor.

Together, these equations connect the amount of charge passing, the energy transferred by each coulomb, and the time taken. Choose the relationship that matches the quantities you know.

Energy and work

Current through a component transfers energy from the supply to the device and its surroundings. Work done equals energy transferred.

• Resistor: heating and increased internal energy.
• Lamp: light and heating.
• Motor: mechanical work and heating.

Essential equations

$$P = \frac{E}{t}, \qquad E = Pt$$