Two Resistors in Series

Two resistors wired end to end sit on one path, and that single fact settles everything else. Charge has nowhere to go but forward, so the same current passes through both — it is not divided between them and it is not shared out according to size. Each resistor then takes its own voltage drop, $V_1 = I R_1$ and $V_2 = I R_2$, and the two drops must add up to whatever the supply provides. Divide that sum by the current they have in common and the resistances add: $R_t = R_1 + R_2$. This is not a rule to memorise. It is Ohm's law applied twice in a circuit that has only one loop.

Put a 47 Ω resistor in series with a 220 Ω resistor across a 12 V supply. The total is 267 Ω, so the current is $12/267 = 45\ \text{mA}$, and it is 45 mA at every point in the loop — before the first resistor, between them, and after the second. The 47 Ω part drops $0.045 \times 47 = 2.1\ \text{V}$; the 220 Ω part drops $0.045 \times 220 = 9.9\ \text{V}$; and 2.1 + 9.9 gives back the 12 V we started with. Adding the drops as a check costs nothing and catches most arithmetic errors on the spot.

The bookkeeping behind that check is Kirchhoff's voltage law, published by Gustav Kirchhoff in 1845 while he was still a student: go once around any closed loop and the voltage rises equal the voltage falls, because the loop returns you to the potential you started at. Every series result descends from it. The voltage divider is the same equation rearranged — each resistor claims the fraction $R_1/(R_t)$ of the supply — and that is how a potentiometer, a thermistor bridge and a sensor's biasing network all work. Solving this page backwards for one resistor is subtraction, $R_1 = R_t - R_2$, which is why the total must exceed the branch you already know.

The mistakes cluster in three places. The first is reaching for the wrong combination rule: series resistors add, parallel resistors combine as product over sum, and capacitors do exactly the reverse — series capacitors are the reciprocal case. Whenever you find yourself using product-over-sum on a series string, stop and ask which component you are holding. The second is assuming the larger resistor gets the larger current; it gets the larger voltage drop at the same current, and confusing those two makes a mess of any divider. The third only appears on AC: a coil or a capacitor in series with a resistor cannot be added arithmetically to it. Reactance is 90° out of phase with resistance, so a 30 Ω resistor in series with 40 Ω of reactance presents $\sqrt{30^2 + 40^2} = 50\ \Omega$, not 70. This page adds resistances, and resistances are what it will add — the quadrature sum belongs on the impedance pages.