Topic 11.4 Electric Power: calculate the power delivered or dissipated in circuit elements using P = IV and its resistive forms.

What this topic is asking

The College Board (Topic 11.4) wants you to calculate the electrical power delivered to or dissipated in a circuit element using $P = IV$ and its resistive forms, and to find the energy transferred over time. Power is where circuits do their work: heating, lighting and driving devices.

Power delivered to a circuit element

This follows directly from the definitions: each coulomb moving through a voltage $V$ gains or loses energy $qV$, and charge flows at rate $I$, so energy is transferred at rate $IV$. For a resistor the energy becomes heat; for a motor, mechanical work; for a charging capacitor, stored field energy. The expression $P = IV$ is completely general, holding for any circuit element regardless of whether it obeys Ohm's law, because it follows only from the definitions of current and voltage. For a non-ohmic device such as a filament bulb or a diode, you cannot use the resistive forms, but $P = IV$ still gives the instantaneous power if you read off the current and voltage at that operating point.

The resistive forms

For a resistor, substitute Ohm's law $V = IR$ into $P = IV$ to get three equivalent expressions:

All three give the same power; pick the one matching your known quantities. $P = I^2 R$ is handy when the current is shared (series), $P = V^2/R$ when the voltage is shared (parallel).

Energy over time

Power is the rate of energy transfer, so the energy delivered is the time integral:

This is the area under a power-versus-time graph. Electricity bills are charged in kilowatt-hours, an energy unit: $1$ kWh $= (1000\,\text{W})(3600\,\text{s}) = 3.6\times10^6$ J. When the current varies in time (as in a charging RC circuit), the power $P = I^2 R$ also varies, and the total heat dissipated is the genuine integral $E = \displaystyle\int I^2 R\,dt$ rather than a simple product. For a steady DC current, however, the power is constant and $E = Pt$ suffices.

In a multi-element circuit the powers add: the total power the source delivers equals the sum of the powers dissipated and stored in every element. This is energy conservation applied to power, and it is a quick way to check an answer. For a source of EMF $\varepsilon$ driving current $I$, the source supplies $\varepsilon I$; of this, $I^2 r$ is wasted internally and the remainder, $I^2 R$ across the external load, does the useful work.

Try this

Q1. A device draws $3.0$ A at $12$ V. Find its power. [1 point]

• Cue. $P = IV = (3.0)(12) = 36$ W.

Q2. A $5.0\,\Omega$ resistor carries $2.0$ A. Find the power it dissipates. [2 points]

• Cue. $P = I^2 R = (2.0)^2(5.0) = 20$ W.