10.3 Potential dividers
- Syllabus
- 9702–2028–2029
- Topic
- 10.3
- Level
- AS
Learning objectives
10.3.1The principle of a potential divider circuit• understand the principle of a potential divider circuit10.3.2The principle of the potentiometer as a means of comparing potential• recall and use the principle of the potentiometer as a means of comparing potential differences10.3.3The use of a galvanometer in null methods• understand the use of a galvanometer in null methods10.3.4The use of thermistors and light-dependent resistors in potential dividers• explain the use of thermistors and light-dependent resistors in potential dividers to provide a potential difference that is dependent on temperature and light intensity
A potential divider selects a fraction of the supply voltage
Two resistors R₁ and R₂ in series carry the same current. If the output is measured across R₂ and the supply is V_s, then the output is the fraction of total series resistance contributed by R₂.
I=Vs/(R1+R2),soVout=IR2=VsR2/(R1+R2)
With R₁ = 2.0 kΩ and R₂ = 3.0 kΩ across 10 V, V_out = 10 × 3/(2 + 3) = 6.0 V across R₂. Across R₁ it is 4.0 V, and the two outputs sum to the supply.
A device of resistance R_L connected across R₂ is a parallel load. Replace R₂ by R_b = R₂R_L/(R₂ + R_L), then use V_out = V_sR_b/(R₁ + R_b). For R_L = 3.0 kΩ above, R_b = 1.5 kΩ and V_out falls from 6.0 V to 4.29 V.
A divider gives half the supply only when the two effective series resistances are equal. Always identify the two output terminals and include any load that changes the effective resistance.
A potentiometer compares p.d.s by their null balance lengths
A steady current in a uniform wire of constant cross-sectional area produces resistance R ∝ l and therefore p.d. V = IR ∝ l. The potential gradient k = V_wire/L is constant, so a balance length l represents p.d. E = kl.
Connect the test p.d. with polarity opposing the wire's p.d., slide the contact and locate the point where the galvanometer reads zero. Keep the same wire current, wire and temperature while comparing the two balance lengths.
E1=kl1,E2=kl2,henceE1/E2=l1/l2
A 1.20 V reference balances at 60.0 cm. An unknown balances at 45.0 cm without changing the wire current. E_x/1.20 = 45.0/60.0, so E_x = 0.900 V.
Balance length is not itself a voltage. The ratio method fails if the potential gradient changes between readings, and the available wire p.d. must be large enough for the unknown to reach a balance point.
A galvanometer locates equality without loading the test source
Atnull:Ig=0⇔Vg=0⇔thegalvanometerterminalsareatequalpotential
A sensitive centre-zero galvanometer is a detector, not the measuring scale. Away from balance, its deflection direction shows which terminal is at higher potential; moving the contact until the direction changes and then narrowing the interval locates zero.
At balance no current is drawn through the galvanometer branch or from the test source. The test source therefore has no internal voltage loss in that branch, so a potentiometer can compare its e.m.f. without the loading caused by an ordinary finite-resistance voltmeter.
On a potentiometer wire, a contact left of balance gives one deflection and a contact right of balance gives the opposite deflection. The zero point between them is the length whose wire p.d. exactly opposes the test p.d.
A null reading does not mean every voltage or current in the apparatus is zero. The driver current still flows in the potentiometer wire; only the detector branch has zero current and equal endpoint potential.
Sensor placement sets the direction of a divider's voltage response
| Sensor | Increasing stimulus | Resistance response |
|---|---|---|
| NTC thermistor | temperature increases | R_NTC decreases |
| LDR | light intensity increases | R_LDR decreases |
| Output measured across | When sensor resistance decreases | Why |
|---|---|---|
| sensor | V_out decreases | V_out = V_sR_sensor/(R_fixed + R_sensor) |
| fixed resistor | V_out increases | the fixed resistor receives a larger fraction of V_s |
Explain any response in four links: stimulus change → sensor resistance change → selected divider fraction change → output p.d. change. Reversing the sensor and fixed-resistor positions reverses the output trend for the same stimulus.
An NTC thermistor is the output resistor below a 20 kΩ fixed resistor across 60 V. At 20 kΩ, V_out = 30 V. Cooling raises the NTC resistance by 50% to 30 kΩ, so V_out = 60 × 30/(20 + 30) = 36 V. Heating would lower its resistance and lower this sensor-output voltage.
The simple ratio assumes negligible output loading and an approximately fixed terminal supply p.d. A finite load changes the selected effective resistance; appreciable source internal resistance can also make terminal p.d. change as total current changes.