2. Conserving networks

Why a wire contributes equations

A resistor law relates its own voltage drop to its own current. It says nothing about which other component touches it. To build a circuit we need both local component laws and relations at connections.

We will use ideal wires: negligible resistance, no charge storage at a junction, and one potential everywhere along a connected wire. These approximations fit our steady lumped circuit. They are physical choices, not numerical settings.

By the end of this chapter you can write connection equations, explain ground, and diagnose a circuit whose voltage reference is missing.

Conservation at a connection

Suppose a junction connects $n$ terminals. Define each terminal current $i_j$ as entering its component, hence leaving the junction. Charge balance is

$$\frac{\mathrm{d}q_{\mathrm{junction}}}{\mathrm{d}t} =-\sum_{j=1}^{n}i_j.$$

Neglecting stored junction charge gives $\sum_j i_j=0$. Meanwhile the ideal wire imposes $\phi_1=\phi_2=\cdots=\phi_n$. These are two different kinds of relation: current conservation and potential compatibility. Kirchhoff’s junction and loop rules are introduced in OpenStax §10.3.

At the divider midpoint, the negative terminal of the upper resistor connects to the positive terminal of the lower resistor:

$$\phi_{1,-}=\phi_{2,+},\qquad i_{1,-}+i_{2,+}=0.$$

Each resistor also conserves charge internally, $i_{j,+}+i_{j,-}=0$. Combining these relations gives $i_{1,+}=i_{2,+}$. The familiar “same current in series” follows from conservation; we do not need to prescribe it again.

The entire divider as a network

text

                 top node a
           ┌────────●────────┐
           │                 │  I ↓
        +  │             [ R₁ = 1 kΩ ]
    Vs = 12 V                │
        −  │                 ● b: output
           │                 │  I ↓
           │             [ R₂ = 2 kΩ ]
           │                 │
           └────────●────────┘
                 node g: φg = 0 V

Conceptual circuit schematic. The source sets the potential difference from g to a. The resistors form the other branch from a to g, with output node b between them. The arrows define positive resistor current.

Source and resistor laws give

$$\phi_a-\phi_g=V_s,\qquad \phi_a-\phi_b=R_1I,\qquad \phi_b-\phi_g=R_2I.$$

Adding the two resistor drops telescopes to $\phi_a-\phi_g$. This is why the loop relation is $V_s-(R_1+R_2)I=0$ in this circuit. We have derived it from shared node potentials and component laws.

Ground fixes an origin

If $(\phi_a,\phi_b,\phi_g)=(12,8,0)$ V satisfies the three difference equations, then $(17,13,5)$ V also does. Adding the same constant to all node potentials leaves every difference unchanged. A voltmeter between two nodes cannot distinguish the two descriptions.

Ground selects $\phi_g=0$ and removes this undetermined offset. It does not say that the return wire carries no current, and it does not model a physical earth electrode. When interpreting a measured voltage, always identify the other terminal: “8 V” here means 8 V relative to node g.

Read the connection syntax

The next chapter will run the complete source. Its midpoint wire is simply

eqiora

connect upper.negative, lower.positive;

The connector declares voltage as across and current as through. These declarations tell the connection to equate voltages and sum signed currents. They do not establish a cause-and-effect order between components. The closing resistor and source equations determine the simultaneous solution.

Open the circuit connections. For a general introduction to equation independence, see algebraic relations and networks.

Try a missing reference

After running the next chapter, remove the ground instance and remove ground.terminal from the return connection. The differences still look plausible, but the node potentials have the arbitrary offset above. The resolution diagnostic identifies an unreferenced uniform shift.

Restore the physical reference. Picking a different linear solver cannot create the missing equation.

Exercises

1. Write the signed current equation where source.positive meets upper.positive. If the upper current is 4 mA, what is the source current?
2. Starting with the three difference equations, prove that a common potential shift leaves all three unchanged.
3. A third resistor carries current from the midpoint to ground. Does the upper resistor still carry the same current as the original lower one?
4. Is matching the number of unknowns and equations enough to guarantee a solution? Give an example using a prescribed voltage difference.

Answer sketches

1. $i_{s,+}+i_{1,+}=0$, hence $i_{s,+}=-4$ mA.
2. Substitute $\phi'_j=\phi_j+c$. Each difference contains $+c-c=0$.
3. No. The upper current splits: $I_1=I_2+I_{\mathrm{load}}$. Equal current was a consequence of the original connection topology.
4. No. Two repeated statements of the same difference add no independent constraint. Demanding that the same difference equal both 12 V and 13 V is inconsistent regardless of equation count.

Reference

Samuel J. Ling, William Moebs, and Jeff Sanny, University Physics Volume 2, OpenStax (2016), §10.3, Kirchhoff’s Rules.

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