Physics 9702 · Paper 1

Adjusting the Divider

A fixed pair of resistors gives one fixed output. Real circuits usually need an output you can change.
Two ways do it: make one resistor variable, or use a single resistor with a sliding contact.

Changing one resistor

Worked example

The range of an adjustable divider

A 10 V supply drives a 10 Ω resistor in series with a variable resistor R2 that runs from 0 to 40 Ω. The output is across R2. Find the range of the output.

At $R_2 = 0$ : $V_{\text{out}} = 0$ V.
At $R_2 = 40$ : $V_{\text{out}} = \dfrac{40}{10 + 40} \times 10 = 8.0$ V.

Answer

0 \text{ to } 8.0 \text{ V}

Note the top value: 8.0 V, not 10 V. The fixed 10 Ω always keeps a share. The output can only reach the full supply if the other resistance can reach zero.

Your turn— tap to reveal the worked answer (9702-style)

A divider runs from a 9.0 V supply. The output, across a 470 Ω resistor, must be 2.0 V. Find the other resistance.

Answer: the 470 Ω takes 2.0 V, so the other resistor takes 7.0 V. Same current, so resistances are in the same ratio as the voltages: $R_1 = 470 \times \dfrac{7.0}{2.0} = 1600 \; \Omega$ (2 s.f.). (9702-style)

Common mistake

Putting the wrong resistor on top of the fraction. The numerator is always the resistor the output is measured across. Ask “where is the voltmeter?” before you write anything.

The sliding-contact form

One resistor with a sliding contact works like two resistors: the contact splits it into a top part and a bottom part.

One resistor, three connections. The contact splits it into a top part and a bottom part. Slide the contact to the bottom and Vout is 0. Slide it to the top and Vout is the full Vin.

This form has a real advantage: the output runs over the full range, from 0 up to the whole supply. Either part can shrink to zero.
Compare a lamp controlled by a plain variable resistor in series: that lamp can never be fully off. It always keeps some share of the voltage. A lamp on the sliding divider can be turned fully off: its share can go all the way down to 0 V.

Why the formula fails when something is connected across R2

The formula assumes nothing is connected across R2 except a perfect voltmeter. Connect a real component there (a lamp, a motor, anything that takes current) and it is in parallel with R2, lowering the bottom resistance, so the real output drops below the value the formula gives. Simple rule: the formula stays accurate only if the thing you connect has a resistance much larger than R2. The numbers on the effect-of-change page show exactly this.

The output can only reach the full supply if the other part can shrink to zero. That is why the sliding form is used when the full range matters.

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