4. Open the components

Why did the 8 V output fall?

The previous chapter assumed the voltmeter drew negligible current. Now connect a 2 kΩ load between the midpoint and ground. The upper resistor must carry the load current as well as the lower-resistor current. Predicting an unchanged 8 V would silently reuse the old connection equations.

We will derive the new voltage, inspect the library definitions, and decide which part of the model actually needs to change.

Derive the loaded divider

Let $v$ be the midpoint voltage. Ohm’s law gives three currents:

$$I_1=\frac{V_s-v}{R_1},\qquad I_2=\frac{v}{R_2},\qquad I_L=\frac{v}{R_L}.$$

Junction conservation requires $I_1=I_2+I_L$. Substitution gives

$$V_s=v\left(1+\frac{R_1}{R_2}+\frac{R_1}{R_L}\right),\qquad v=\frac{V_s}{1+R_1/R_2+R_1/R_L}.$$

Equivalently, the two lower branches form $R_p=(1/R_2+1/R_L)^{-1}$, followed by $v=V_sR_p/(R_1+R_p)$. The equivalence follows from shared voltage and summed currents, as in OpenStax §10.2.

For $R_1=1$ kΩ and $R_2=R_L=2$ kΩ, $R_p=1$ kΩ. Thus $v=6$ V, $I_1=6$ mA, and $I_2=I_L=3$ mA. The powers are 0.036 W in the upper resistor, 0.018 W in each lower branch, and −0.072 W in the source.

What the import actually says

Here is the full source imported as electrical in the packaged divider:

Eqiora.Electrical.Basic/src/basic.eqi

public connector Pin {
  across voltage: kg * m ^ 2 / (s ^ 3 * A);
  through current: A;
}

public component IdealVoltageSource(
  parameter voltage: kg * m ^ 2 / (s ^ 3 * A),
  port positive: Pin,
  port negative: Pin
) {
  relation law {
    positive.voltage - negative.voltage - voltage = 0;
    positive.current + negative.current = 0;
  }
}

public component Resistor(
  parameter resistance: kg * m ^ 2 / (s ^ 3 * A ^ 2),
  port positive: Pin,
  port negative: Pin
) {
  relation law {
    positive.voltage - negative.voltage - resistance * positive.current = 0;
    positive.current + negative.current = 0;
  }
}

public component Ground(
  port terminal: Pin
) {
  relation law {
    terminal.voltage = 0;
  }
}

Read the definition in its package.

The resistor owns two statements: its voltage-current law and internal charge conservation. Neither mentions the load or the surrounding junctions. Therefore we can reuse precisely this resistor definition for the new load.

The source owns a prescribed voltage difference and internal current balance. It does not prescribe how much current must flow. The network determines that. Ground owns only the zero-potential reference.

Add the physical branch

In the complete direct divider.eqi from chapter 3, add an instance:

eqiora

instance load: Resistor(resistance = 2[kOhm]);

Replace the midpoint and return connections with:

eqiora

connect upper.negative, lower.positive, load.positive;
connect lower.negative, load.negative, source.negative, ground.terminal;

Keep the top connection unchanged. Add an observation for the new branch:

eqiora

observable load_power: W = (load.positive.voltage - load.negative.voltage) * load.positive.current;

Add ("load_power", "W") to the direct runner’s observation list. Run again and compare all powers, including the load, with the derived values. The old sum of just three powers should now be −0.018 W: it omits the load’s absorption.

In the packaged source, the corresponding instance uses electrical.Resistor. This name change only locates the component definition; the new junction equations have exactly the same role.

A parameter edit or a different law?

Changing a constant resistance from 1 kΩ to 2 kΩ preserves the form $v=Ri$. Connecting another resistor changes topology. Both can use the same component. By contrast, asking resistance to vary with temperature changes the law and introduces another physical quantity.

As a modeling exercise, consider

$$R(T)=R_0[1+\alpha(T-T_0)].$$

Here $\alpha$ has units K⁻¹, so the bracket is dimensionless. This is an assumed local linear approximation, not a universal law for all resistors. It requires a temperature and an appropriate range in which $R(T)$ remains positive. Predicting $T$ in turn requires the thermal energy balance introduced in chapter 1.

This is the useful question to ask when opening a component: which assumption does its equation make, and does my experiment still satisfy it? See constitutive laws for a wider discussion, or follow the heat into heat transfer.

Exercises

1. Find the load resistance that lowers the original 8 V output to 4 V.
2. Show that the unloaded prediction returns as $R_L$ grows without bound.
3. A loaded calculation gives the correct midpoint but the three-power sum fails. Explain why this alone need not indicate a solver error.
4. At a fixed current, a resistor’s resistance rises by 10%. What happens to its power? What if voltage, rather than current, is held fixed?

Answer sketches

1. $4=12/(1+1/2+1000/R_L)$, so $R_L=2000/3$ Ω, about 667 Ω.
2. $R_1/R_L$ tends to zero, leaving $V_sR_2/(R_1+R_2)$.
3. The load is a fourth component that absorbs power. Audit all branches before changing numerical tolerances.
4. At fixed current, $p=I^2R$ rises by 10%. At fixed voltage, $p=V^2/R$ becomes $1/1.1$ of its original value. The surrounding circuit determines what is actually held fixed.

Reference

Samuel J. Ling, William Moebs, and Jeff Sanny, University Physics Volume 2, OpenStax (2016), §10.2, Resistors in Series and Parallel.

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