Energy Transferred Equation

Energy transfer in a circuit

When charge moves through an electrical component, energy is transferred. For example, a lamp transfers electrical energy to light and thermal energy, while a motor transfers it mainly to kinetic energy and thermal energy.

The total energy transferred depends on three things:

• the current, because a larger current means more charge passes each second;
• the potential difference, because a larger potential difference means more energy is transferred by each coulomb of charge;
• the time, because energy continues to be transferred while the current flows.

These relationships combine to give:

$$E=I\times V\times t$$

Here, $E$ is energy transferred in joules (J), $I$ is current in amperes (A), $V$ is the potential difference in volts (V), and $t$ is time in seconds (s). The potential difference must be measured across the component, and the current must be the current through it.

Where the equation comes from

Potential difference is energy transferred per unit charge, so:

$$E=Q\times V$$

Current is the rate of flow of charge. The charge that passes in time $t$ is therefore:

$$Q=I\times t$$

Substituting $I\times t$ for $Q$ gives:

$$E=(I\times t)\times V=I\times V\times t$$

This connection helps explain the equation rather than treating it as an isolated fact. If any one of $I$, $V$ or $t$ is doubled while the others remain constant, the energy transferred doubles.

Using the equation

Suppose a component has a potential difference of $12\ \text{V}$ across it and carries a current of $2.0\ \text{A}$ for $30\ \text{s}$. The energy transferred is:

$$E=2.0\times12\times30=720\ \text{J}$$

This means that the component transfers $720\ \text{J}$ of energy during those 30 seconds.

Time must be in seconds. For example, if a $0.50\ \text{A}$ current flows through a component with a potential difference of $6.0\ \text{V}$ for 2 minutes, first convert the time:

$$2\ \text{minutes}=2\times60=120\ \text{s}$$

Then:

$$E=0.50\times6.0\times120=360\ \text{J}$$

Finding a different quantity

The same equation can be rearranged:

$$I=\frac{E}{Vt}$$ $$V=\frac{E}{It}$$ $$t=\frac{E}{IV}$$

For instance, if $900\ \text{J}$ is transferred when the potential difference is $15\ \text{V}$ and the current is $2.0\ \text{A}$, the time is:

$$t=\frac{900}{2.0\times15}=30\ \text{s}$$

Use constant values, or suitable average values, for current and potential difference over the stated time.

Equation

$$E=IVt$$

• $E$: energy transferred in joules (J)
• $I$: current in amperes (A)
• $V$: potential difference in volts (V)
• $t$: time in seconds (s)

Use the current through the component and the potential difference across it.

Rearrangements

$$I=\frac{E}{Vt},\qquad V=\frac{E}{It},\qquad t=\frac{E}{IV}$$

Proportional relationships

With the other quantities constant:

• greater current means more energy transferred;
• greater potential difference means more energy transferred per unit charge;
• a longer time means more energy transferred.

Calculation check

Convert minutes or hours to seconds before substituting, then give energy in joules.