Worked examples: three-resistor chain and a bulb-string break

Two problems, one in steady state, one in failure mode. Series circuits are small enough that you can solve them in your head once the pattern clicks.

Example 1 — three-resistor chain

A 12 V battery drives three resistors in series: 2 Ω, 4 Ω, and 6 Ω. Find the current in the loop and the voltage drop across each resistor.

1. Given. $V = 12\,\text{V}$, $R_1 = 2\,\Omega$, $R_2 = 4\,\Omega$, $R_3 = 6\,\Omega$.
2. Add resistors. $R_{total} = 2 + 4 + 6 = 12\,\Omega$.
3. Current from Ohm's Law.
   $$I = \frac{12\,\text{V}}{12\,\Omega} = 1\,\text{A}$$
4. Voltage drops (V = IR applied to each). $V_1 = 1 \cdot 2 = 2\,\text{V}$, $V_2 = 1 \cdot 4 = 4\,\text{V}$, $V_3 = 1 \cdot 6 = 6\,\text{V}$.
5. Check. $2 + 4 + 6 = 12\,\text{V}$ — the drops sum to the source voltage, as KVL promises.

Note the pattern: the biggest resistor (6 Ω) drops the biggest voltage (6 V) because they're all sharing the same 1 A. Drop is proportional to resistance whenever current is common.

Example 2 — bulb-string break (why series is fragile)

Ten identical 5 Ω festive bulbs run in series off a 50 V supply. Normally they all glow equally. One bulb fails open (burnt filament). What happens to the current and to the remaining nine bulbs?

1. Before the break. $R_{total} = 10 \cdot 5 = 50\,\Omega$. Current $I = 50\,\text{V}/50\,\Omega = 1\,\text{A}$. Each bulb dissipates $P = I^2 R = 1^2 \cdot 5 = 5\,\text{W}$ — glowing steadily.
2. The break. An open filament means $R_{total} = \infty$. No complete loop exists for current to flow.
3. After. $I = V/R_{total} = 50/\infty = 0\,\text{A}$. Current in a series loop is zero the instant any link opens.
4. Consequence for the other nine bulbs. $P = I^2 R = 0$ — they stop glowing too. You can't see which one failed; you check them one by one (or buy the kind with a bypass-diode shunt).