Worked examples: two loops, signs kept honest

KVL looks like bookkeeping once you settle on a walking direction. Two problems — one clean loop, one with opposing batteries — are enough to get the reflex.

Example 1 — single loop, one source

A 12 V battery drives a 3 Ω resistor (R₁) and a 5 Ω resistor (R₂) in a single series loop. Assume current $I$ flows clockwise. Find $I$ and the voltage drop across each resistor.

1. Pick a loop direction. Clockwise — same as the assumed current. Walk from the battery's − terminal, around the loop, back to the − terminal.
2. Write KVL. Walking − → + through the battery is $+12$; walking with the current through R₁ and R₂ gives two $-IR$ terms.
   $$+12 - I(3) - I(5) = 0$$
3. Solve for current. $12 = 8I \Rightarrow I = 1.5\,\text{A}$.
4. Branch voltage drops. $V_1 = IR_1 = 1.5 \cdot 3 = 4.5\,\text{V}$, $V_2 = IR_2 = 1.5 \cdot 5 = 7.5\,\text{V}$.
5. Sanity check. $V_1 + V_2 = 12\,\text{V} = \mathcal{E}$ — the drops sum back to the source, exactly as KVL promises.

This is the same answer you'd get from Ohm's Law on a 3 Ω + 5 Ω = 8 Ω series total. The point isn't a new answer — it's a method that keeps working when series reduction doesn't.

Example 2 — two batteries opposing each other

A single loop carries two batteries wired with opposite orientations: a 12 V source pushing current one way, a 6 V source pushing back. A 3 Ω resistor sits between them. Assume current $I$ flows clockwise, with the 12 V battery aiding that direction (− → + as you walk clockwise) and the 6 V battery fighting it (+ → − as you walk clockwise). Find $I$.

1. Walk the loop clockwise.
   $$+12 - I(3) - 6 = 0$$
   The 12 V is a lift (− → +), the resistor is a drop (walking with the current), the 6 V is a drop (+ → −).
2. Solve. $6 = 3I \Rightarrow I = 2\,\text{A}$. Clockwise is correct — positive answer means our assumed direction was right.
3. Net EMF check. The two batteries oppose, so the loop sees $\mathcal{E}_{net} = 12 - 6 = 6\,\text{V}$. $I = \mathcal{E}_{net}/R = 6/3 = 2\,\text{A}$ — the shortcut agrees with the full KVL walk.

Flip the 6 V battery around and its term becomes $+6$ instead of $-6$ — now both batteries push the same way and $I = 18/3 = 6\,\text{A}$. Changing one sign changes the answer by a factor of three. Signs are the whole job.