The second rule is about loops, closed paths around a circuit. With the junction rule, it solves any circuit, even ones with several batteries that series-parallel reduction cannot handle. The equations you get are a system, solved by elimination.
Why it works
Electric potential has one value at each point, like height on a hiking map. Walk around any closed loop and you come back to where you started, at the same potential, so all the rises and drops along the way add to zero. In energy terms: a charge going once around the loop gains from the batteries exactly what it gives to the resistors.
Key idea
Around any closed loop, $\sum \Delta V = 0$. Through a battery from $-$ to $+$: $+\varepsilon$ (from $+$ to $-$: $-\varepsilon$). Through a resistor in the direction of the current: $-IR$ (against it: $+IR$). Guess the current directions; a negative answer means that current flows the other way.
| You cross… | Going this way | Change in $V$ |
|---|---|---|
| a battery | from $-$ to $+$ | $+\varepsilon$ |
| a battery | from $+$ to $-$ | $-\varepsilon$ |
| a resistor | with the current | $-IR$ |
| a resistor | against the current | $+IR$ |
Worked examples
Worked example: One loop, two batteries
A loop contains $\varepsilon_1 = 12\,\text{V}$, $\varepsilon_2 = 3\,\text{V}$ facing the opposite way, $R_1 = 2\,\Omega$ and $R_2 = 4\,\Omega$. Find the current.
- Guess clockwise current $I$, the way $\varepsilon_1$ pushes. Walk around clockwise: $+12 - 2I - 3 - 4I = 0$.
- $6I = 9$, so $I = 1.5\,\text{A}$, clockwise. The bigger battery wins, and the smaller one is pushed backward through itself.
Answer: $1.5\,\text{A}$
Worked example: Two loops
In the circuit shown, $\varepsilon_1 = 10\,\text{V}$, $\varepsilon_2 = 9\,\text{V}$, $R_1 = 2\,\Omega$, $R_2 = 3\,\Omega$, $R_3 = 2\,\Omega$. Find $I_1$, $I_2$ and $I_3$.
- Junction (top): $I_3 = I_1 + I_2$.
- Left loop: $10 - 2I_1 - 2I_3 = 0$. Right loop: $9 - 3I_2 - 2I_3 = 0$.
- Substitute $I_3$: $4I_1 + 2I_2 = 10$ and $2I_1 + 5I_2 = 9$.
- Double the second and subtract the first: $8I_2 = 8$, so $I_2 = 1\,\text{A}$; then $I_1 = (10 - 2)/4 = 2\,\text{A}$ and $I_3 = 3\,\text{A}$.
- Check the left loop: $10 - 2(2) - 2(3) = 0$.
Answer: $I_1 = 2\,\text{A}$, $I_2 = 1\,\text{A}$, $I_3 = 3\,\text{A}$
Common mistake
Getting the sign wrong when crossing a resistor against the current. Current flows from the high end of a resistor to the low end, so walking against it you climb: $+IR$. Write the sign from the direction you walk compared with the current's arrow, not from which way the loop “looks”. Check: after solving, put the currents back into a loop you did not use.