Physics 9702 · Paper 2

Potentiometer Calculations & Null Methods

At balance the p.d. along the wire equals the test cell's e.m.f., so every potentiometer calculation is one length ratio.

The balance calculation, and the driver trap

At balance: $E = \dfrac{L_{\text{balance}}}{L_{\text{whole wire}}} \times V_{\text{wire}}$ , the fraction of the wire used times the p.d. across the whole wire.

Worked example

Warm-up: a driver with no internal resistance

A driver cell of e.m.f. 1.2 V and negligible internal resistance is connected across a 150 cm nichrome wire. The galvanometer shows zero when the contact is 64 cm from the left-hand end. Find the e.m.f. of the test cell X.

The whole wire carries the full 1.2 V, so: $E = \dfrac{64}{150} \times 1.2 = 0.51$ V.

Answer

E = 0.51 \text{ V}

(9702/23/M/J/25 Q7)

The trap: $V_{\text{wire}}$ is not automatically the driver's e.m.f. If the driver has internal resistance, some of its e.m.f. is lost inside it before the wire gets any.

Worked example

A real exam potentiometer, trap included

A driver cell of e.m.f. 1.5 V and internal resistance 0.50 Ω drives a uniform wire XY of length 0.96 m and resistance 0.50 Ω. The galvanometer reads zero when the contact Z is 0.64 m from X. Find the e.m.f. E of the test cell.

Driver loop current: $I = \dfrac{1.5}{0.50 + 0.50} = 1.5$ A.
P.d. across the whole wire: $V_{\text{wire}} = 1.5 \times 0.50 = 0.75$ V. Only half the driver's e.m.f. reaches the wire; the rest is lost volts inside the driver.
Balance: $E = \dfrac{0.64}{0.96} \times 0.75 = 0.50$ V.

Answer

E = 0.50 \text{ V}

(9702/11/O/N/24 Q37)

Common mistake

Using the driver's 1.5 V directly:

1.5 \times 0.64/0.96 = 1.0

V, and 1.0 V is one of the answer options. Always ask first: how much p.d. actually reaches the wire?

Same idea in explain form: if the driver cell develops internal resistance, the wire's p.d. drops. Each centimetre of wire now matches fewer volts, so the balance point moves further along the wire to match the same test e.m.f. The same 2025 paper asks exactly this. (9702/23/M/J/25 Q7)

Comparing two e.m.f.s cancels the unknowns

Balance cell X, note its balance length. Replace it with cell Y and balance again. The wire's p.d. is the same both times, so it cancels:

\dfrac{E_X}{E_Y} = \dfrac{L_X}{L_Y}

Symbols

$E_X, E_Y$ = the two e.m.f.s being compared (V)
$L_X, L_Y$ = their balance lengths on the same wire (m)

Balance cell X and note the balance point. Then put cell Y in its place and balance again. The driver and the wire did not change, so the two e.m.f.s are in the same ratio as the two lengths.

If one cell's e.m.f. is accurately known (a standard cell), the other follows with the same accuracy. No need to know the driver's e.m.f. or the wire's resistance at all.
The same ratio works for comparing two p.d.s: connect each in turn, take the ratio of the balance lengths.

Worked example

Test cell against standard cell

On one potentiometer, a test cell balances at 22.5 cm and a standard cell of e.m.f. 1.434 V balances at 34.6 cm. Find the test cell's e.m.f.

$E = 1.434 \times \dfrac{22.5}{34.6} = 0.93$ V.

Answer

E \approx 0.93 \text{ V}

Two balance lengths and one known cell: the answer is just the ratio of the lengths. All the unknown values cancel.

Null methods with two dividers

The wire is not the only null method. Another version uses two potential dividers side by side, and the exam tests this one too.

Two dividers, one supply. The galvanometer compares the two middle points. When it reads zero, the two chains split the supply in the same ratio.

The galvanometer connects the two middle points. It reads zero only when the two points are at the same voltage. That happens when the two chains split the supply in the same ratio: $\dfrac{R_1}{R_2} = \dfrac{X}{Y}$ .
Three of the four resistances known, galvanometer at zero: the fourth follows from the ratio. A 2025 paper asks you to explain exactly this. (9702/22/M/J/25 Q6)

Every null method is the same idea: arrange the circuit so the meter reads zero. At zero current the meter takes nothing, no lost volts appear anywhere in the measuring branch, and the ratios you use are exact.

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Potentiometer Calculations & Null Methods

At balance the p.d. along the wire equals the test cell's e.m.f., so every potentiometer calculation is one length ratio.

The balance calculation, and the driver trap

At balance: $E = \dfrac{L_{\text{balance}}}{L_{\text{whole wire}}} \times V_{\text{wire}}$ , the fraction of the wire used times the p.d. across the whole wire.

Worked example

Warm-up: a driver with no internal resistance

The whole wire carries the full 1.2 V, so: $E = \dfrac{64}{150} \times 1.2 = 0.51$ V.

Answer

E = 0.51 \text{ V}

(9702/23/M/J/25 Q7)

The trap: $V_{\text{wire}}$ is not automatically the driver's e.m.f. If the driver has internal resistance, some of its e.m.f. is lost inside it before the wire gets any.

Worked example

A real exam potentiometer, trap included

Driver loop current: $I = \dfrac{1.5}{0.50 + 0.50} = 1.5$ A.
P.d. across the whole wire: $V_{\text{wire}} = 1.5 \times 0.50 = 0.75$ V. Only half the driver's e.m.f. reaches the wire; the rest is lost volts inside the driver.
Balance: $E = \dfrac{0.64}{0.96} \times 0.75 = 0.50$ V.

Answer

E = 0.50 \text{ V}

(9702/11/O/N/24 Q37)

Common mistake

Using the driver's 1.5 V directly:

1.5 \times 0.64/0.96 = 1.0

V, and 1.0 V is one of the answer options. Always ask first: how much p.d. actually reaches the wire?

Same idea in explain form: if the driver cell develops internal resistance, the wire's p.d. drops. Each centimetre of wire now matches fewer volts, so the balance point moves further along the wire to match the same test e.m.f. The same 2025 paper asks exactly this. (9702/23/M/J/25 Q7)

Comparing two e.m.f.s cancels the unknowns

Balance cell X, note its balance length. Replace it with cell Y and balance again. The wire's p.d. is the same both times, so it cancels:

\dfrac{E_X}{E_Y} = \dfrac{L_X}{L_Y}

Symbols

$E_X, E_Y$ = the two e.m.f.s being compared (V)
$L_X, L_Y$ = their balance lengths on the same wire (m)

Balance cell X and note the balance point. Then put cell Y in its place and balance again. The driver and the wire did not change, so the two e.m.f.s are in the same ratio as the two lengths.

If one cell's e.m.f. is accurately known (a standard cell), the other follows with the same accuracy. No need to know the driver's e.m.f. or the wire's resistance at all.
The same ratio works for comparing two p.d.s: connect each in turn, take the ratio of the balance lengths.

Worked example

Test cell against standard cell

On one potentiometer, a test cell balances at 22.5 cm and a standard cell of e.m.f. 1.434 V balances at 34.6 cm. Find the test cell's e.m.f.

$E = 1.434 \times \dfrac{22.5}{34.6} = 0.93$ V.

Answer

E \approx 0.93 \text{ V}

Two balance lengths and one known cell: the answer is just the ratio of the lengths. All the unknown values cancel.

Null methods with two dividers

The wire is not the only null method. Another version uses two potential dividers side by side, and the exam tests this one too.

Two dividers, one supply. The galvanometer compares the two middle points. When it reads zero, the two chains split the supply in the same ratio.

The galvanometer connects the two middle points. It reads zero only when the two points are at the same voltage. That happens when the two chains split the supply in the same ratio: $\dfrac{R_1}{R_2} = \dfrac{X}{Y}$ .
Three of the four resistances known, galvanometer at zero: the fourth follows from the ratio. A 2025 paper asks you to explain exactly this. (9702/22/M/J/25 Q6)