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MHT-CET Physics · Current Electricity

The Potentiometer

A steady current through a long uniform wire gives a constant potential drop per unit length, the potential gradient k; a cell connected against the wire balances at the length where kl equals its e.m.f., drawing no current, so the potentiometer measures e.m.f. directly and compares two cells by their balancing lengths.

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Why this matters

18 PYQs, 5 of them HARD. Ten are the potential gradient — finding it when a series resistance shares the driving cell's voltage, the e.m.f. a length balances, and what happens to the null point when the wire is made longer. Eight compare cells: the ratio of two e.m.f.s from their sum and difference, and a cell's internal resistance from the balancing lengths with two different shunts. Two cards.

Concept 1 of 2: Potential Gradient and the Balancing Length

The driving cell sends a current I through the wire, whose resistance is R_w over its length L. The drop across the wire is IR_w, and it is spread evenly, so k = IR_w/L volts per metre. When a resistance sits in series with the wire, it takes its share of the driving voltage first: I = E/(R_w + R + r). A test cell of e.m.f. E′ balances at l = E′/k. Make the wire longer at the same voltage across it and k falls, so every balancing length grows in proportion. At balance the test cell carries no current, which is why the potentiometer reads e.m.f., not terminal voltage.

Definition

  • k=VwireL=IRwLk = \dfrac{V_{\text{wire}}}{L} = \dfrac{I R_w}{L}, with I=ERw+R+rI = \dfrac{E}{R_w + R + r}.
  • A cell balances at l=E′kl = \dfrac{E'}{k}.
  • Wire lengthened, same voltage across it: k falls, balancing length grows in proportion (L/5 on L ⇒ 3L/10 on 3L/2).
  • Resistance per unit length given: I=E′l⋅(R/L)I = \dfrac{E'}{l \cdot (R/L)}.
  • At balance the test cell draws no current: the reading is its e.m.f.

Potential gradient

k=ERw+R+r⋅RwL,E′=klk = \frac{E}{R_w + R + r} \cdot \frac{R_w}{L}, \qquad E' = kl

Worked example

A 4 m wire of 8 Ω is driven by a 2 V cell through a 2 Ω resistor (no internal resistance). Gradient, and the e.m.f. balanced at 2.5 m?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q40Easy

Example 1 · Current Electricity · Potentiometer

The potentiometer wire is 5 m long and potential difference of 4 V is maintained between the ends. The e.m.f. of the cell which balances against a length of 200 cm of the potentiometer wire is

Forgetting the series resistance in the gradient

When a resistance is in series with the wire, only the wire's share of the driving voltage sets k. Find I from the whole circuit first, then k = IR_w/L.

Thinking a longer wire moves the null point closer

At the same voltage a longer wire has a smaller gradient, so a cell needs MORE length to balance. The null point moves further along.
Drill 9 more on potential gradient and the balancing length

Concept 2 of 2: Comparing E.m.f.s and Finding Internal Resistance

Balancing lengths are proportional to the e.m.f. balanced. Two cells assisting each other balance at l₁ ∝ E₁ + E₂; opposing, at l₂ ∝ E₁ − E₂. Divide: E₁/E₂ = (l₁ + l₂)/(l₁ − l₂). For internal resistance, the cell alone balances at l₀ ∝ E; shunted by R it drives a current and its terminal voltage ER/(R + r) balances at l, so r = R(l₀ − l)/l. If l₀ is not given, two shunts R₁ and R₂ give two equations: l ∝ ER/(R + r), and the ratio of the two eliminates E.

Definition

  • Sum and difference: E1E2=l1+l2l1−l2\dfrac{E_1}{E_2} = \dfrac{l_1 + l_2}{l_1 - l_2} (64 cm and 32 cm ⇒ 3:1).
  • One cell, then opposed: E1E2=l1l1−l2\dfrac{E_1}{E_2} = \dfrac{l_1}{l_1 - l_2}.
  • Internal resistance: r=R l0−llr = R\,\dfrac{l_0 - l}{l}.
  • Two shunts: l1l2=R1(R2+r)R2(R1+r)\dfrac{l_1}{l_2} = \dfrac{R_1(R_2 + r)}{R_2(R_1 + r)}; solve for r (5 Ω at 200 cm, 15 Ω at 300 cm ⇒ r = 5 Ω).

Internal resistance

r=R l0−llr = R\,\frac{l_0 - l}{l}

Worked example

Two cells balance at 75 cm assisting and 25 cm opposing. E₁/E₂?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q38Moderate

Example 2 · Current Electricity · Potentiometer

In potentiometer experiments, two cells of e.m.f. E1E_1 and E2E_2 are connected in series E1>E2E_1 > E_2, the balancing length is 64 cm of the wire. If the polarity of E2E_2 is reversed, the balancing length becomes 32 cm. The ratio E1/E2E_1/E_2 is

Using the sum formula for 'one cell, then opposed'

If the first length is E₁ ALONE, the ratio is l₁/(l₁ − l₂), not (l₁ + l₂)/(l₁ − l₂). Read which combination each length balances.

Taking the shunted length for the e.m.f.

With a shunt the cell delivers current, so the balance reads its terminal voltage, smaller than E. Only the unshunted length l₀ measures E.
Drill 7 more on comparing e.m.f.s and finding internal resistance

Summary — formulas & gotchas at a glance

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Formulas (2)

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