The Potential Divider

• Start with plain numbers. A 100 Ω and a 200 Ω resistor are in series across a 6.0 V supply.
• One loop, so one current: $I = \dfrac{6.0}{100 + 200} = 0.020$ A.
• Each resistor takes its own p.d., from $V = IR$: $0.020 \times 100 = 2.0$ V across the 100 Ω and $0.020 \times 200 = 4.0$ V across the 200 Ω.
• Check with Kirchhoff's second law: $2.0 + 4.0 = 6.0$ V. The supply voltage has been split into two parts.
• The current is the same through both, so $V = IR$ grows with R: the voltage splits in proportion to resistance. Bigger resistor, bigger part.

• Now add the extra wire. One output wire joins the junction between the two resistors; the other joins the bottom end of the pair.
• The p.d. between those two wires is the output $V_{\text{out}}$: simply the p.d. across R2. In the circuit above, taking the output across the 200 Ω gives 4.0 V; across the 100 Ω it would give 2.0 V.
• Nothing is connected across the output yet, so the output wires carry no current. The two resistors carry the same current as before.

• So a divider does one thing: it lets you take a chosen part of the supply voltage between two points, instead of all of it.

• The formula is just the two steps above, with letters instead of numbers.
• The pair carries $I = \dfrac{V_{\text{in}}}{R_1 + R_2}$, and the output is the p.d. across R2: $V_{\text{out}} = I R_2$. Put the first into the second:

• The fraction $\dfrac{R_2}{R_1 + R_2}$ is R2's share of the total resistance. So the output is R2's share of the input voltage.
• Check it on the first circuit: $\dfrac{200}{100 + 200} \times 6.0 = 4.0$ V. The same answer as $V = IR$ gave, in one line.

A divider is one ratio: the output is R2's share of the input. Every divider question starts from that one fraction.