Potential Dividers

Every analogue quantity your phone measures passes through a divider on its way to being a number. Microcontrollers read voltages with an analogue-to-digital converter (ADC), and the standard recipe for reading anything — temperature, light, battery level, joystick position — is: put the sensor in a divider, feed the tap to the ADC pin. The ADC input resistance is enormous (megohms), so the unloaded formula holds to a fraction of a percent: the textbook idealisation is, for once, the engineering reality. An electric vehicle's battery-management system uses precision divider chains to scale hundreds of volts of pack voltage down to the few volts its monitoring chips can survive; the resistive touchscreen on ticket machines and factory panels is the divider made two-dimensional — pressing the screen taps a resistive sheet at the touch point, and the tap voltage is the coordinate. And in calibration laboratories the Kelvin–Varley divider still earns its keep, now verified against quantum voltage standards: matched resistors, dividing by ratio, exactly as equation (1) promised.

Think of the supply voltage as a staircase of electrical height that the current must descend on its way from the battery's positive terminal back to the negative. In a chain of two series resistors, the whole staircase is used up crossing them, and each resistor claims a share of the descent. What decides the shares? The current $I$ is the same through both (a series chain has no branches — the rule established in Resistors in Series & Parallel), so by Ohm's law each resistor drops $V = IR$: the bigger resistance takes the bigger drop, in exact proportion. Two equal resistors split the supply 50–50; a 9:1 ratio splits it 90–10. The junction between them sits at a definite electrical height, and a wire soldered there — the tap — exports that height as your output voltage.

Notice what the picture promises and what it does not. It promises a junction voltage set purely by the ratio $R_2 : (R_1 + R_2)$ — scale both resistors by ten and $V_{\text{out}}$ is untouched (though the current, and the wasted power, change tenfold). It does not promise that the junction will hold its voltage when you connect something there: a load gives the current a second route out of the junction, and the arithmetic changes. That is the loading story, treated honestly below because Cambridge asks it by name.

Everything follows from two facts you own from Resistors in Series & Parallel: series elements carry a common current, and series resistances add. The result, derived properly, then read physically:

where: $V_{\text{in}}$ — the supply across the whole chain; $R_1$ — the upper resistor (supply's positive terminal to tap); $R_2$ — the lower resistor (tap to 0 V), the one the output is measured across; $V_{\text{out}}$ — the tap-to-ground voltage, assuming nothing significant is connected to the tap. Read the fraction as "R₂'s share of the total ohms": $R_2 = R_1$ gives half the supply; $R_2 \gg R_1$ hands nearly everything to the output; $R_2 \to 0$ starves it toward zero. The output never exceeds the input — a divider only divides.

Two design remarks turn the formula into craft. First, only the ratio fixes the output, so the designer's second decision is the absolute size: small resistors waste power continuously (the chain conducts even when nothing uses the output — our 10 kΩ pair burns $V^2/R_{\text{total}} = 144/20000 = 7.2\ \mathrm{mW}$ forever), while very large resistors make the output fragile against loading, as the next section quantifies. Second, the potentiometer packages the whole idea in one component: a single resistive track with a sliding contact, so $R_1$ and $R_2$ are the track segments either side of the wiper, and turning the shaft sweeps $V_{\text{out}}$ smoothly from 0 to $V_{\text{in}}$. Wired with all three terminals it is a divider (controls voltage); wired with only two — wiper and one end — it degenerates into a variable series resistor, a rheostat (controls current). Exams test that distinction by drawing, so learn it by the terminal count (the hardware symbols live in Circuit Elements, Symbols & Diagrams).

Equation (1) describes a divider that nobody is using. Connect an actual device — a lamp, a motor, even a voltmeter — across $R_2$, and its resistance $R_L$ sits in parallel with $R_2$: the bottom leg becomes the smaller combination $R_2 \parallel R_L$, and the output sags:

where: $R_L$ — the load's resistance; $R_2 \parallel R_L$ — the parallel combination (always smaller than either, by the parallel rule from Resistors in Series & Parallel); $V_{\text{out}}'$ — the real, loaded output. Read the physics: the load opens a second exit from the junction, extra current flows through $R_1$, so $R_1$ drops more volts, leaving fewer for the output. The unloaded formula survives approximately when $R_L \gg R_2$ (rule of thumb: $R_L$ at least ten times $R_2$ keeps the sag under about 5%) — exactly why dividers feed high-resistance inputs and are never used to power anything hungry. The numbers make the lesson stick. Our 12 V, 10 kΩ/10 kΩ divider promises 6.0 V unloaded. Connect a 10 kΩ load: the bottom leg becomes $10\parallel10 = 5\ \mathrm{k\Omega}$, and $V_{\text{out}}' = 12 \times \dfrac{5}{15} = 4.0\ \mathrm{V}$ — a third of the promise gone. Connect a small motor of resistance 100 Ω instead: bottom leg $= \dfrac{10000 \times 100}{10100} = 99\ \Omega$, and $V_{\text{out}}' = 12 \times \dfrac{99}{10099} \approx 0.12\ \mathrm{V}$ — the "6 volt supply" collapses to a tenth of a volt. The divider did nothing wrong: you asked a ratio of resistances to behave like a battery, and it declined.

Now replace one resistor with a component whose resistance responds to the environment, and the divider becomes a transducer — a machine that turns a physical quantity into a voltage. Two components do almost all the syllabus work: the LDR (light-dependent resistor), whose resistance is high in darkness (hundreds of kilohms) and low in bright light (a few kilohms or less), and the NTC thermistor, whose resistance falls as temperature rises — the behaviour, and the semiconductor reason for it, come from Temperature Dependence, Thermistors & Superconductors-Lite. Put either into one leg and $V_{\text{out}}$ tracks light or temperature; feed it to a switching circuit and you have a night-light, a thermostat, a frost alarm. The design decision examiners test hardest: which leg does the sensor occupy? The output voltage is the bottom leg's share of the supply, so put the component whose share you want to grow with the sensed quantity in the bottom leg. For a light that must switch ON in darkness: place the LDR in the bottom leg. In darkness its resistance soars, its share of the supply soars, $V_{\text{out}}$ rises toward $V_{\text{in}}$, and the rising voltage triggers the lamp. Swap the LDR to the top leg and the behaviour inverts — darkness now starves the output, and your night-light becomes a day-light. No memorising required: ask "whose resistance grows?", and remember volts follow ohms.

One more design layer earns A-Level marks: choosing the fixed resistor's value. For the sharpest response, pick the fixed resistance comparable to the sensor's resistance in the middle of the range you care about — then the sensed change swings the ratio through its steepest region (a divider with wildly mismatched legs barely notices the sensor changing). And one boundary line, so the map stays honest: the potentiometer experiment — a null-balance instrument, not a readout — has its own module coming later in this chapter (The Potentiometer). Divider reads; potentiometer balances.

Board level — the LDR night-light (IGCSE Extended pattern)

A potential divider consists of a fixed 20 kΩ resistor connected from the +6.0 V supply rail to an output junction, and an LDR connected from the junction to 0 V. The LDR's resistance is 2.0 kΩ in daylight and 120 kΩ in darkness. (a) Calculate the output voltage in daylight and in darkness. (b) The lamp-driver circuit switches the lamp on when the output exceeds 3.0 V. Find the LDR resistance at which the lamp switches, and state whether the lamp is on in daylight or darkness.

Step 1. Identify the legs: fixed $R_1 = 20\ \mathrm{k\Omega}$ on top, LDR $= R_2$ on the bottom, output across the LDR. Equation (1) applies directly — the driver input draws negligible current (the exam's code phrase "switches when the output exceeds" signals a high-resistance input).

Step 2. Daylight: $V_{\text{out}} = 6.0 \times \dfrac{2.0}{20 + 2.0} = 6.0 \times \dfrac{2}{22} = \boxed{0.55\ \mathrm{V}}$.

Step 3. Darkness: $V_{\text{out}} = 6.0 \times \dfrac{120}{20 + 120} = 6.0 \times \dfrac{120}{140} = \boxed{5.1\ \mathrm{V}}$.

Step 4. Switching point, working backwards from $V_{\text{out}} = 3.0\ \mathrm{V}$: that is exactly half the supply, and the divider gives half when the legs are equal — so the LDR resistance is $\boxed{20\ \mathrm{k\Omega}}$ at switch-on. (Formally: $3.0 = 6.0\,\frac{R}{20+R} \Rightarrow 20 + R = 2R \Rightarrow R = 20\ \mathrm{k\Omega}$.)

Step 5. In daylight $V_{\text{out}} = 0.55\ \mathrm{V} < 3.0\ \mathrm{V}$: lamp off. In darkness $5.1\ \mathrm{V} > 3.0\ \mathrm{V}$: lamp on — a night-light, exactly as the bottom-leg placement predicted (dark → LDR resistance up → its share up).

Competitive level — thermostat design, then the load bites (CIE AS pattern, ~5 minutes)

A 9.0 V supply feeds a divider with an NTC thermistor as the top leg and a fixed resistor $R = 2.2\ \mathrm{k\Omega}$ as the bottom leg; the output is taken across the fixed resistor. The thermistor's resistance is 4.7 kΩ at 25 °C and 1.2 kΩ at 60 °C. (a) Calculate the output voltage at both temperatures. (b) The heater cut-off triggers when the output reaches 4.5 V. Find the thermistor resistance at the trigger point. (c) The output is now connected to a control circuit of input resistance 10 kΩ. Calculate the true output at 60 °C and comment on whether the trigger design still works.

Step 1. Part (a), at 25 °C: $V_{\text{out}} = 9.0 \times \dfrac{2.2}{4.7 + 2.2} = 9.0 \times \dfrac{2.2}{6.9} = \boxed{2.9\ \mathrm{V}}$. At 60 °C: $V_{\text{out}} = 9.0 \times \dfrac{2.2}{1.2 + 2.2} = 9.0 \times \dfrac{2.2}{3.4} = \boxed{5.8\ \mathrm{V}}$. Hotter → thermistor resistance falls → the fixed bottom leg's share grows → output rises. (Sensor in the top leg because we want heat to raise the cut-off signal.)

Step 2. Part (b), invert equation (1) for the unknown top leg: $4.5 = 9.0\,\dfrac{2.2}{R_{\text{th}} + 2.2}$ gives $R_{\text{th}} + 2.2 = \dfrac{9.0 \times 2.2}{4.5} = 4.4$, so $R_{\text{th}} = \boxed{2.2\ \mathrm{k\Omega}}$ — the trigger fires when the thermistor's resistance falls to match the fixed resistor (half-supply logic again: 4.5 V is half of 9.0 V).

Step 3. Part (c), load the bottom leg: $2.2 \parallel 10 = \dfrac{2.2 \times 10}{12.2} = 1.80\ \mathrm{k\Omega}$. At 60 °C: $V_{\text{out}}' = 9.0 \times \dfrac{1.80}{1.2 + 1.80} = 9.0 \times \dfrac{1.80}{3.00} = \boxed{5.4\ \mathrm{V}}$ — a sag of 0.4 V from the unloaded 5.8 V.

Step 4. Comment: the loaded output still crosses the 4.5 V threshold before 60 °C, so the thermostat works — but the trigger temperature has shifted (every output is now lower, so the thermistor must fall further, i.e. run hotter, to reach 4.5 V). A careful design recomputes the threshold with $R_2 \parallel R_L$, or chooses $R_L \gg R_2$. Writing that one sentence separates the A grade. Timing note: about five minutes. Parts (a)–(b) are ratio reads — spot the half-supply shortcut in (b) and skip the algebra. Part (c) is where the marks hide: the only new work is one parallel combination. Practise until "connected to a circuit of resistance X" triggers the reflex $R_2 \to R_2 \parallel R_L$.

Reading the wrong leg. Equation (1)'s $R_2$ is the resistor the output is measured across — the bottom leg, tap to ground. Substituting the first resistor named in the question is the most common divider error. Draw the chain, mark the tap, identify the bottom leg, then substitute.

Sensor-direction reversal. "Resistance of the LDR falls in light" plus "output across the LDR" means output falls in light. Double negations breed sign errors; resolve them with the mantra — volts follow ohms — not memorised outcomes.

Ignoring the load. Any question connecting a voltmeter, lamp, or "circuit of input resistance X" to the tap is a loading question in disguise; the unloaded answer will appear as a tempting wrong option. Recompute with $R_2 \parallel R_L$.

Potentiometer vs rheostat. Three terminals used = divider = controls voltage (output sweeps 0 to $V_{\text{in}}$). Two terminals used = rheostat = controls current (and the minimum is not zero). Cambridge draws both and asks which; count the wires leaving the component.

Expecting the divider to power things. A divider is a signal source, not a supply. If the load resistance is comparable to or below $R_2$, expect collapse — never pretend the unloaded formula still applies.

Quoting ratios without units discipline. Both legs must be in the same unit (both kΩ or both Ω). Mixing 2200 Ω with 10 kΩ inside one fraction is a silent factor-of-1000 error.

Constants and conventions. No constants beyond the circuit values; CIE expects 2–3 significant figures. IGCSE keeps dividers to qualitative-plus-formula; A-Level 9702 adds loading and sensor-threshold design (both worked above) — know which paper you are sitting.

Texts. CIE 9702 syllabus Topic 10.2 (potential dividers; sensor circuits with LDR and thermistor) — the definitive A-Level statement · Cambridge IGCSE 0625 §4.3.1 (variable potential dividers, Extended) · Duncan, Advanced Physics, circuits chapter for the classic loading treatment.

Provenance. All results are standard circuit theory. A verifier should recompute: unloaded 10 k/10 k on 12 V → 6.0 V, quiescent 0.6 mA and 7.2 mW; loaded by 10 kΩ → bottom leg 5 kΩ → 4.0 V; loaded by 100 Ω → 99 Ω → 0.118 V ≈ 0.12 V; LDR daylight $6 \times 2/22 = 0.545$ V, darkness $6 \times 120/140 = 5.14$ V, trigger at 20 kΩ; thermistor $9 \times 2.2/6.9 = 2.87$ V and $9 \times 2.2/3.4 = 5.82$ V, trigger $R_{\text{th}} = 2.2$ kΩ; ADC-loaded $2.2\parallel10 = 1.803$ kΩ → $9 \times 1.803/3.003 = 5.40$ V.

Prerequisites. Resistors in Series & Parallel · Temperature Dependence, Thermistors & Superconductors-Lite. Leads to. The Wheatstone Bridge · The Potentiometer (coming in E3) · Series and Parallel Networks (UG deep dive).