Potential Dividers & Measuring Instruments

Nothing is connected to the output yet, so the two resistors are in series and carry the same current, I = Vᵢₙ/(R₁ + R₂). The output is whatever that current makes across the lower arm, Vₒᵤₜ = I R₂ = Vᵢₙ R₂/(R₁ + R₂). With 12.0 V across 4.7 kΩ above 2.2 kΩ, I = 1.74 mA and Vₒᵤₜ = 3.83 V. Notice what the result does not contain: any absolute resistance. Scale both arms by 100 and the output is unchanged while the current falls by 100. Two consequences follow at once. With positive resistors the divider can only attenuate — the output lies strictly between 0 and Vᵢₙ, so no choice of arms gives more than the supply. And the ratio inherits the resistors' tolerance: two 1% resistors leave this output uncertain by about 1%, roughly 0.04 V, before any load is mentioned.

Connect anything across the output and it sits in parallel with R₂, because both bridge the same pair of nodes. The lower arm is no longer R₂ but R₂R_L/(R₂ + R_L), which is smaller than either, so the lower arm's share of the chain shrinks and the output falls. Take the same 12.0 V ratio built high-resistance, R₁ = 470 kΩ over R₂ = 220 kΩ, and load it with 200 kΩ. The lower arm becomes 220 × 200/420 = 104.8 kΩ, and Vₒᵤₜ = 12.0 × 104.8/(470 + 104.8) = 2.19 V where the ratio promised 3.83 V. That drop is not a fault in the divider and it is not noise: it is a new circuit, and it is entirely predictable. A passive load can only pull the output down, never up, because a parallel resistor can only lower a resistance.

Recomputing the parallel combination for every load is wasteful. Collapse the divider once instead. Seen from its output terminals it behaves as a single source of V_oc = Vᵢₙ R₂/(R₁ + R₂) behind a single resistance Rₜₕ = R₁R₂/(R₁ + R₂), obtained by shorting the ideal supply so the two arms appear in parallel. Any load then forms a new divider with Rₜₕ alone: Vₒᵤₜ = V_oc R_L/(R_L + Rₜₕ). For 470 kΩ and 220 kΩ, Rₜₕ = 149.9 kΩ, so a 200 kΩ meter reads 3.826 × 200/349.9 = 2.19 V, matching the long calculation and showing at a glance why: the load is barely larger than Rₜₕ. The error fraction Rₜₕ/(R_L + Rₜₕ) is 42.8%. Since Rₜₕ ≤ (R₁ + R₂)/4, low error and low current pull against each other, and that tension is the whole design problem.

An ideal voltmeter has infinite resistance and takes no current; an ideal ammeter has zero resistance and drops no voltage. Real instruments miss both marks in ways you can quantify. A digital multimeter typically presents about 10 MΩ; an analogue meter is rated in ohms per volt, so 20 kΩ V⁻¹ on the 10 V range is 200 kΩ but only 50 kΩ on the 2.5 V range — changing range changes the circuit. An ammeter adds resistance in series: 9.0 V across 220 Ω should give 40.9 mA, but a 12 Ω ammeter reduces it to 38.8 mA, 5.2% low, while dropping 0.47 V of burden voltage. This also decides how you wire a resistance measurement. Put the voltmeter directly across R and the ammeter reads the sum of both branch currents, which spoils large R; put the ammeter in series with R and the voltmeter reads R plus the meter, which spoils small R.

Make the lower arm a sensor of resistance Rₛ and the divider becomes a transducer: Vₒᵤₜ = Vᵢₙ Rₛ/(R + Rₛ). What matters is not the output but how hard it moves, dVₒᵤₜ/dRₛ = Vᵢₙ R/(R + Rₛ)². Differentiating that with respect to the fixed arm gives Vᵢₙ(Rₛ − R)/(R + Rₛ)³, which vanishes at R = Rₛ, so the best fixed resistor equals the sensor's resistance at the operating point and the peak sensitivity is Vᵢₙ/(4Rₛ). For a 4.70 kΩ thermistor on a 5.00 V rail, choose 4.70 kΩ: the quiescent output is exactly 2.50 V and the sensitivity is 0.266 V kΩ⁻¹. Two cautions. That sensitivity is a tangent, so over a large swing the exact ratio must be used. And the sensor dissipates power — 2.50 V across 4.70 kΩ is 1.33 mW, enough to warm a small thermistor and corrupt the very quantity it measures.

Calibration fixes the two free numbers of a near-linear channel: a zero at a known low point and a span at a known high point. It cannot rescue a loading error that changes with the reading, so the load must be specified before the arms are chosen — a divider has no current to spare and is not a power supply. Supply drift is friendlier than it looks: the output is a fixed fraction of Vᵢₙ, so a converter that measures it against the same rail is ratiometric and the drift divides out. A sliding potentiometer shows the limits in one picture. With the wiper at fraction f the upper section is (1 − f)R_T and the lower is fR_T, so V_oc = fVᵢₙ and Rₜₕ = f(1 − f)R_T. The loaded output therefore bows below the straight line. The percentage error is worst at f = 0.5, where Rₜₕ reaches its maximum of R_T/4; the largest gap in volts sits further along, because the sag V_oc Rₜₕ/(R_L + Rₜₕ) goes as f²(1 − f) for a light load and so peaks at f = 2/3.