1. Voltage, current, and power

A small circuit that warms its surroundings

A 12 V source drives two series resistors. The first resistance is 1 kΩ and the second is 2 kΩ. A current of 4 mA will emerge from the calculation in chapter 3. Before finding that current, ask a different question: if these values are right, how much power goes into each component?

By the end of this chapter you can define terminal directions, distinguish current from energy transfer, and audit the signs of a complete power balance.

Choose directions before calculating

For a two-terminal component, call the terminals positive and negative. These are names that define a measurement direction; they do not promise that one potential will always exceed the other. Define

$$v=\phi_+-\phi_-,\qquad i=\text{current entering the positive terminal}.$$

The potential $\phi$ is measured in volts, while $i$ is measured in amperes, or coulombs of charge per second. A negative current means charge transport opposite to the chosen arrow. It does not invalidate the model.

text

                     current i →
       positive terminal ● ── [ component ] ── ● negative terminal
           potential φ+                         potential φ−

       component voltage v = φ+ − φ−

Conceptual terminal diagram. The current arrow enters the terminal used first in the voltage difference. This convention makes positive power mean energy entering the component.

Derive the power sign

If charge $\mathrm{d}q$ enters at $\phi_+$ and leaves at $\phi_-$, the energy transferred into the component is $(\phi_+-\phi_-)\mathrm{d}q$. Dividing by elapsed time gives

$$p=v\frac{\mathrm{d}q}{\mathrm{d}t}=vi.$$

Thus $p>0$ means absorption and $p<0$ means delivery. The dimensions provide a quick check: $\mathrm{V\,A}=(\mathrm{J/C})(\mathrm{C/s})=\mathrm{W}$. This energy-per-charge argument and its resistor specialization are developed in OpenStax §9.5.

For a constant positive resistance, Ohm’s law gives $v=Ri$, hence

$$p=Ri^2=\frac{v^2}{R}\geq0.$$

The resistor absorbs power for either direction of current. Its constitutive law converts electrical energy into heat; it does not consume charge.

Predict the divider’s energy budget

Choose downward current through both resistors. At $I=0.004\,\mathrm{A}$, the upper resistor drops 4 V and the lower drops 8 V. Their absorbed powers are

$$p_1=(0.004)^2(1000)=0.016\,\mathrm{W},\qquad p_2=(0.004)^2(2000)=0.032\,\mathrm{W}.$$

Current exits the source’s positive terminal, so its into-positive-terminal current is $-0.004\,\mathrm{A}$. Its absorbed power is

$$p_s=(12)(-0.004)=-0.048\,\mathrm{W}.$$

The sum $p_s+p_1+p_2=0$ accounts for every component. After ten seconds at these constant values, the resistors have received 0.16 J and 0.32 J. The source has supplied 0.48 J. These are predictions from the stated values, ready to compare with the divider run.

Do not add a positive “power supplied” to positive absorbed powers and expect zero. Either use signed absorbed power for every component, or equate the positive amount supplied to the positive amount absorbed.

What a warm resistor tells us

The electrical calculation determines heat generation, but does not determine temperature. To predict temperature, we need thermal storage and a path for heat to leave. The simplest energy balance would read

$$C_{\mathrm{th}}\frac{\mathrm{d}T}{\mathrm{d}t} =p-\dot Q_{\mathrm{out}},$$

where $C_{\mathrm{th}}$ is thermal capacity in J/K and $\dot Q_{\mathrm{out}}$ is outward heat flow in W. This is the bridge to heat transfer. Treating $R$ as constant is a physical approximation; if heating materially changes it, the electrical law must reflect that dependence.

Exercises

1. Reverse both the voltage and current measurement directions. Show that the computed power is unchanged. What happens if you reverse only one?
2. Double the supply voltage while retaining both resistances. Predict the factor by which resistor power changes.
3. A calculation reports absorbed powers $0.016$, $0.032$, and $+0.048$ W for the two resistors and the source. Identify the likely sign mistake.
4. Can you determine the resistor’s temperature from 0.016 W alone? Name two missing physical inputs.

Answer sketches

1. $(-v)(-i)=vi$. Reversing only one makes the reported sign opposite to the absorbed-power convention.
2. The current doubles, so each resistor’s power quadruples. The source delivers 0.192 W in total.
3. The source current was probably counted as outgoing while the resistor currents were counted as incoming. With one absorbed-power convention its power is $-0.048$ W.
4. You need, for example, thermal capacity for a transient and a heat-transfer law plus ambient temperature for cooling. Power alone supplies neither.

Reference

Samuel J. Ling, William Moebs, and Jeff Sanny, University Physics Volume 2, OpenStax (2016), §9.5, Electrical Energy and Power.

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