Worked examples: dividing down and sizing up

The voltage-divider formula shows up in two directions. Sometimes you have resistors and a source and want to know the output. Sometimes you want a specific output and have to choose the resistors. One example of each.

Example 1 — forward problem

A 9 V battery drives a divider built from a 2 kΩ ($R_1$) in series with a 4 kΩ ($R_2$). Find $V_{out}$ across $R_2$ and the current drawn from the battery.

1. Apply the divider formula.
   $$V_{out} = 9 \cdot \frac{4\,k}{2\,k + 4\,k} = 9 \cdot \frac{4}{6} = 6\,\text{V}$$
2. Cross-check with series arithmetic. $I = 9 / 6\,k = 1.5\,\text{mA}$, so $V_{R_2} = 1.5 \cdot 4 = 6\,\text{V}$. The formula and the long way agree.
3. Power budget. $P = V \cdot I = 9 \cdot 1.5\,\text{mA} = 13.5\,\text{mW}$ — trivial, as expected for a kΩ-range divider.

Example 2 — reverse problem: design a 3.3 V rail

You have a 5 V supply and need to generate a 3.3 V reference. You've chosen $R_2$ = 10 kΩ for good loading margin. What's $R_1$?

1. Set up the ratio. We want $V_{out}/V_{in} = 3.3/5 = 0.66$. The divider formula gives $0.66 = R_2 / (R_1 + R_2)$.
2. Solve for R₁. $0.66 (R_1 + R_2) = R_2$, so $R_1 = R_2 \cdot (1/0.66 - 1) = R_2 \cdot 0.5151$. With $R_2 = 10\,k\Omega$,
   $$R_1 = 10\,k \cdot 0.5151 \approx 5.15\,k\Omega$$
3. Snap to an E12 standard value. The nearest common stock is $5.1\,k\Omega$. Recompute the actual output:
   $$V_{out} = 5 \cdot \frac{10}{5.1 + 10} = 5 \cdot 0.662 = 3.31\,\text{V}$$
   Close enough — a 0.3 % error for a reference that has ±5 % resistor tolerance anyway.
4. Check current. $I = 5 / (5.1 + 10)\,k = 0.331\,\text{mA}$ — about a third of a milliamp, nicely within budget for a battery-powered design.

Reverse problems always resolve into: pick one of the two resistors based on a current or loading constraint, then solve the formula for the other.