How Two Simple Rules Can Solve Any Tangled Circuit

Many circuits — two batteries feeding one lamp, or a bridge of four resistors — cannot be reduced by simple series and parallel steps. Two conservation rules named after Kirchhoff solve all of them, once we also account for the hidden resistance inside every cell.

This part covers emf and internal resistance, combinations of cells, Kirchhoff's rules, and the Wheatstone bridge.

**The emf $\epsilon$ is the potential difference across a cell when no current flows, while the terminal potential difference is lower when a current $I$ is drawn, because some voltage is lost across the cell's internal resistance $r$: $V=\epsilon -Ir$.

Relations:**

$$V=\epsilon -Ir,I=\frac{\epsilon }{R+r}$$

- Supplying current: $V<\epsilon$
- Open circuit, $I=0$: $V=\epsilon$
- Being charged: $V=\epsilon +Ir$

Worked example. A cell of emf $1.5$ V and internal resistance $0.5\Omega$ drives a $2.5\Omega$ resistor:

$$I=\frac{1.5}{2.5+0.5}=0.5\text{ A},V=1.5-0.5\times 0.5=1.25\text{ V}$$

Shorting the cell with $R=0$ gives the maximum current, $\frac{\epsilon }{r}=3$ A.

An everyday example. Car headlights dim for a moment when the engine is started, because the starter motor draws a large current and the battery's terminal voltage drops across its internal resistance.

The substance. Emf is not a force — it is the work done by the cell per unit charge, measured in volts.

**For cells in series, the emfs add with their signs and the internal resistances add; for two cells in parallel, ${\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2}$ and $r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$.

Series:**

$${\epsilon }_{eq}={\epsilon }_1+{\epsilon }_2,r_{eq}=r_1+r_2$$

If one cell is reversed, its emf is subtracted, but its internal resistance still adds.

Parallel, with like terminals joined:

$${\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2},r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$$

Worked example — series. Two cells of $1.5$ V and $0.5\Omega$ each drive a $4\Omega$ load:

$$I=\frac{3}{4+1}=0.6\text{ A}$$

Worked example — parallel. Cells of $2$ V, $1\Omega$ and $3$ V, $2\Omega$:

$${\epsilon }_{eq}=\frac{2\times 2+3\times 1}{1+2}=\frac{7}{3}\text{ V},r_{eq}=\frac{2}{3}\Omega$$

For $n$ identical cells in parallel, ${\epsilon }_{eq}=\epsilon$ and $r_{eq}=\frac{r}{n}$.

An everyday example. A TV remote holds two AA cells in series to double the voltage, while a home inverter bank can join batteries in parallel to supply larger currents.

The substance. Series combinations raise the voltage, while parallel combinations of identical cells lower the internal resistance and so can deliver more current.

The junction rule says the total current entering a junction equals the total current leaving it, from charge conservation; the loop rule says the algebraic sum of potential changes around any closed loop is zero, from energy conservation.

The rules: $\sum I_{in}=\sum I_{out}$ at a junction, and $\sum \Delta V=0$ around a loop.

Sign conventions:

- Crossing a resistor in the direction of current: $-IR$
- Crossing a cell from its negative to its positive terminal: $+\epsilon$

Worked example. Three branches join the same two junctions. Branch 1 has a $6$ V cell and $2\Omega$; branch 2 has a $4$ V cell and $1\Omega$; the middle branch has $4\Omega$ only. Let $I_1$ and $I_2$ flow up through branches 1 and 2, so $I_1+I_2$ flows down the middle.

$$\text{Loop 1: }6-2I_1-4(I_1+I_2)=0\Rightarrow 6I_1+4I_2=6$$

$$\text{Loop 2: }4-I_2-4(I_1+I_2)=0\Rightarrow 4I_1+5I_2=4$$

Solving gives $I_1=1$ A and $I_2=0$. The middle resistor has $4$ V across it, exactly matching the $4$ V cell, so no current flows in that branch.

An everyday example. Water pipes splitting and rejoining in a building obey the junction rule — the water entering a junction must equal the water leaving it.

The substance. A negative answer for a current just means it flows opposite to the direction you assumed — its size is still correct.

**A Wheatstone bridge of four resistors $P$, $Q$, $R$ and $S$ is balanced, with no current through the galvanometer between its two middle points, when $\frac{P}{Q}=\frac{R}{S}$ — so an unknown resistance can be found from three known ones.

Set-up.** $P$ and $Q$ form one path from A to C through B; $R$ and $S$ form another through D; a galvanometer joins B and D, and a cell joins A and C.

Derivation. At balance no current flows through the galvanometer, so B and D are at the same potential. The current $I_1$ in $P$ therefore also flows through $Q$, and $I_2$ in $R$ also flows through $S$:

- Loop ABDA: $I_{1}P=I_{2}R$
- Loop BCDB: $I_{1}Q=I_{2}S$

Dividing the two equations:

$$\frac{P}{Q}=\frac{R}{S}$$

Worked example. With $P=10\Omega$, $Q=30\Omega$ and $R=15\Omega$, balance needs

$$S=\frac{QR}{P}=\frac{30\times 15}{10}=45\Omega$$

Meter bridge. A uniform wire forms two arms; balance at $l$ cm from one end gives $\frac{R}{S}=\frac{l}{100-l}$.

An everyday example. A traditional two-pan weighing balance is level only when both sides match — likewise, the galvanometer reads zero only when the bridge's two ratios match.

The substance. At balance, the galvanometer's own resistance does not matter, because no current flows through it.