Math Bridge: Rain Divider and RC Filter

A field team faces an unresolved physical question: How does wet resistance become a clean ADC code? They must answer it before changing rain sensor resistance in kilo-ohms on the real device. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is rain sensor resistance in kilo-ohms. The middle card applies this page's relationship. The green card is divider current. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Walk the arrows. Changing rain resistance moves voltage and ADC code. It does not move the fixed RC cutoff, so the signal path and noise path remain distinct but coupled parts of one measurement chain.

Derive the baseline in four named moves

1
Name the input. The chapter baseline for rain sensor resistance in kilo-ohms is 10.
2
Name the relationship. Vout=VinRs/(Rp+Rs); code=4095Vout/Vin; fc=1/(2πRC)
3
Substitute the chapter fixture. Set rain sensor resistance in kilo-ohms to 10. The page ledger gives divider current as 0.058 mA.
4
Read the result. Keep mA beside the value. Use it only inside the technical boundary on this page.

Predict, then change rain sensor resistance in kilo-ohms

Try Predict the direction of divider current. Move one control, calculate, then check your prediction.

Observe Changing rain resistance moves voltage and ADC code. It does not move the fixed RC cutoff, so the signal path and noise path remain distinct but coupled parts of one measurement chain. Reset the control to 10 and compare divider current.

Explain Only rain sensor resistance in kilo-ohms moves here. The other chapter fixtures remain fixed.

Check yourself

What should you do before trusting a moved-control result?

Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.

What does this small model leave out?

Answer: Only rain sensor resistance in kilo-ohms moves. Field effects named in the page's technical boundary stay fixed.

1. Give resistance a current path

The rain sensor's resistance is not yet a voltage. A 47 kΩ pull-up and the sensor form a series path from 3.3 V to ground.

Phoebe: The divider translates the wet plate into voltage. The filter decides which changes survive long enough for the ADC.

2. Derive the divider

Series currentI=Vin/(Rpullup+Rsensor).

Sensor voltageVout=I·Rsensor.

ADC codecode=(Vout/Vin)·4095 for this ideal 12-bit case.

3. Derive the low-pass

The 10 kΩ and 1 µF stage has τ=RC and cutoff fc=1/(2πRC). A sine at f is multiplied by 1/√(1+(f/fc)²).

τ=RC; fc=1/(2πRC); |H|=1/√(1+(f/fc)²)

4. Try the rain resistance

Vout=VinRs/(Rp+Rs); code=4095Vout/Vin; fc=1/(2πRC)

TryChange the wet plate's resistance while the chapter's pull-up and rain-filter values stay fixed.

ObserveAt the chapter's 10 kΩ heavy-rain point, the divider gives 0.579 V, 0.0579 mA, and ideal ADC code 718. The 10 ms filter cuts 20 kHz by about −62.0 dB, leaving 0.0398 mV from a 50 mV spike.

ExplainChanging rain resistance moves voltage and ADC code. It does not move the fixed RC cutoff, so the signal path and noise path remain distinct but coupled parts of one measurement chain.

Technical boundaries.

The divider assumes a purely resistive plate and ideal ADC input.

corrosion: Needs separate evidence
electrolysis: Needs separate evidence
surface contamination: Needs separate evidence
leakage: Needs separate evidence
cable capacitance: Needs separate evidence
ADC input loading: Needs separate evidence
reference tolerance: Needs separate evidence
quantisation: Needs separate evidence
wetting dynamics: Needs separate evidence
The RC formula treats one sine component, not an arbitrary transient spectrum: Needs separate evidence

Use field evidence or a deeper model before release.

5. Work the heavy-rain divider

I=3.3/(47 kΩ+10 kΩ)=0.0579 mA

Vout=I×10 kΩ=0.579 V; code≈718

6. Work the rain filter

τ=10 kΩ×1 µF=10.0 ms; fc=15.9 Hz

|H(20 kHz)|=0.000796 ⇒ −62.0 dB; 50 mV→0.0398 mV

Five time constants are 50.0 ms, inside the chapter's 100 ms response claim.

7. Check yourself

Why does lower sensor resistance lower Vout here?

Answer: The sensor is the lower divider leg, so it receives a smaller share of 3.3 V as its resistance falls.

What fixes the 15.9 Hz cutoff?

Answer: The 10 kΩ filter resistor and 1 µF capacitor, not the rain divider resistance in this simplified two-stage model.

Why check five time constants?

Answer: A first-order response is within about 0.7% of its final value after 5τ, giving a practical settling estimate.

Honesty boundary.

These are the chapter inputs, worked results, and named teaching assumptions.

3.3 V: Voltage or voltage-step value
47 kΩ pull-up: Resistance or impedance value
10 kΩ heavy-rain resistance: Resistance or impedance value
10 kΩ/1 µF filter: Resistance or impedance value
20 kHz disturbance: Frequency, sample rate, or event rate
50 mV spike: Voltage or voltage-step value
100 ms response claim: Time, interval, or service-life value

The 12-bit ADC code is explicitly an ideal teaching conversion.