Non-Uniform Potentiometer Calibration: Solving potentiometer problems where the wire's resistance per unit length is non-uniform and varies continuously as a function of distance along the wire.

1. What is a Non-Uniform Potentiometer?

In an ordinary potentiometer, the wire has uniform resistance per unit length. Therefore, potential drop is directly proportional to length.

V ∝ l

But in a non-uniform potentiometer, resistance per unit length changes continuously along the wire.

Let the resistance per unit length at distance x be:

r(x) = resistance per unit length

Important: For a non-uniform wire, we cannot directly use V₁/V₂ = l₁/l₂. We must first calculate the resistance of the required portion.

2. Basic Principle

Consider a potentiometer wire of length L. Its resistance per unit length varies as:

r(x)

A small element of length dx has resistance:

dR = r(x)dx

Therefore, resistance from x = 0 to x = l is:

R(l) = ∫₀ˡ r(x)dx

If a constant current I flows through the wire, the potential drop across this portion is:

V(l) = I ∫₀ˡ r(x)dx

This is the most important equation for a non-uniform potentiometer.

3. Uniform vs Non-Uniform Potentiometer

Uniform Wire	Non-Uniform Wire
r = constant	r = r(x)
R = rl	R = ∫r(x)dx
V ∝ l	V is generally not proportional to l
Simple length ratio works	Integral/resistance ratio is required

4. General Calibration Formula

Suppose the total length of the potentiometer is L and:

r(x) = f(x)

The total resistance is:

R_total = ∫₀ᴸ f(x)dx

For a point at distance l:

R(l) = ∫₀ˡ f(x)dx

If the total potential difference across the wire is V_AB, then:

V(l) = V_AB × R(l)/R_total

Therefore:

V(l) = V_AB × [∫₀ˡ f(x)dx] / [∫₀ᴸ f(x)dx]

5. Important Case: r(x) = r₀(1 + αx)

Suppose the resistance per unit length varies linearly:

r(x) = r₀(1+αx)

Resistance from 0 to l is:

R(l) = ∫₀ˡ r₀(1+αx)dx

Therefore:

R(l) = r₀[l + αl²/2]

The total resistance is:

R(L) = r₀[L + αL²/2]

Hence potential at x = l:

V(l) = V × [l+αl²/2] / [L+αL²/2]

Notice that the r₀ term cancels from the final voltage ratio.

6. Case: r(x) = kx

Suppose resistance per unit length increases linearly with distance:

r(x)=kx

Resistance from 0 to l:

R(l)=∫₀ˡ kx dx

Therefore:

R(l)=kl²/2

Total resistance:

R(L)=kL²/2

Therefore:

V(l)/V = l²/L²

For r(x)=kx, the potential varies as x², not x.

7. Case: r(x) = k/x

If:

r(x)=k/x

Then:

R(l)=∫₀ˡ k/x dx

This integral diverges at x = 0.

Therefore, the ideal mathematical form r = k/x cannot be used starting exactly from x = 0. A finite lower limit such as x = a > 0 must be used.

For a ≤ x ≤ l:

R(a→l)=k ln(l/a)

8. Non-Uniform Potentiometer Calibration

Calibration means determining the unknown potential difference or emf by using a known reference and the actual resistance distribution of the wire.

For a uniform potentiometer:

E ∝ l

For a non-uniform potentiometer:

E ∝ R(l)

Therefore, if a standard cell of emf E_s balances at x = l_s and an unknown cell balances at x = l_x:

E_x/E_s = R(l_x)/R(l_s)

Hence:

E_x = E_s × R(l_x)/R(l_s)

9. Solved Example 1

A potentiometer has resistance per unit length:

r(x)=2x Ω/m

Find the ratio of potential drop across the first 2 m to that across the first 4 m.

Solution

For 0 to l:

R(l)=∫₀ˡ 2x dx=l²

Therefore:

R(2)=4 Ω

R(4)=16 Ω

Since potential drop is proportional to resistance:

V₂/V₄=4/16

V₂/V₄ = 1/4

10. Solved Example 2

A non-uniform potentiometer has:

r(x)=r₀(1+x/L)

If the total potential drop is V, find the potential at x=L/2.

Solution

Resistance from 0 to x:

R(x)=r₀[x+x²/(2L)]

For x=L:

R(L)=r₀[L+L/2]

R(L)=3r₀L/2

For x=L/2:

R(L/2) = r₀[L/2+L/8]

R(L/2)=5r₀L/8

Therefore:

V(L/2)/V = (5/8)/(3/2)

V(L/2)=5V/12

11. Comparison with a Standard Cell

Suppose a standard cell has emf E_s and balances at x=l₁. An unknown cell has emf E_x and balances at x=l₂.

For a non-uniform wire:

E_s ∝ R(l₁)

E_x ∝ R(l₂)

Therefore:

E_x = E_s × R(l₂)/R(l₁)

Never replace the resistance ratio by l₂/l₁ unless the wire is uniform.

12. Potential Distribution

For a uniform potentiometer, the potential-distance graph is a straight line.

V(x) ∝ x

For a non-uniform wire, the graph depends on r(x).

Since:

dV/dx = I r(x)

Very Important: The slope of the potential-distance graph at any point is proportional to the local resistance per unit length.

Therefore:

dV/dx = I r(x)

13. Differential Calibration Equation

For a small length dx:

dR=r(x)dx

The voltage drop is:

dV=I dR

Therefore:

dV=I r(x)dx

Hence:

dV/dx=I r(x)

This equation directly connects the non-uniform resistance distribution with the potential gradient.

14. Quick Method for Numerical Problems

Step 1: Write r(x).

Step 2: Find resistance up to the balance point:

R(x)=∫r(x)dx

Step 3: Find total or reference resistance.

Step 4: Use:

V(x)/V_total=R(x)/R_total

Step 5: For standard-cell comparison:

E_x/E_s = R(x_x)/R(x_s)

15. Important Formula Sheet

Quantity	Formula
Small resistance	dR = r(x)dx
Resistance up to x	R(x)=∫₀ˣr(x)dx
Total resistance	R(L)=∫₀ᴸr(x)dx
Voltage up to x	V(x)=I∫₀ˣr(x)dx
Potential gradient	dV/dx=Ir(x)
Calibration	Eₓ/Eₛ=R(xₓ)/R(xₛ)
r(x)=kx	V(x)/V=L? → x²/L²

Section A — 1 Mark
10 MCQs

Q1. For a non-uniform potentiometer, resistance per unit length is: 1 Mark

A) Always constant

B) A function of position

C) Always zero

D) Infinite

Correct Answer: B

Q2. A small element dx of a non-uniform potentiometer has resistance: 1 Mark

A) r/x

B) r+dx

C) r(x)dx

D) x/r

Correct Answer: C

Q3. The resistance from 0 to x is: 1 Mark

A) r(x)x

B) ∫₀ˣr(x)dx

C) r/x

D) x/r

Correct Answer: B

Q4. For a uniform potentiometer, potential drop is proportional to: 1 Mark

A) Length

B) Length²

C) 1/length

D) Length³

Correct Answer: A

Q5. For a non-uniform potentiometer, generally: 1 Mark

A) V ∝ x always

B) V is independent of x

C) V is not necessarily proportional to x

D) V=0

Correct Answer: C

Q6. The potential gradient is: 1 Mark

A) Ir(x)

B) I/r(x)

C) r(x)/I

D) I+r(x)

Correct Answer: A

Q7. If r(x)=kx, then resistance from 0 to x is proportional to: 1 Mark

A) x

B) x²

C) 1/x

D) √x

Correct Answer: B

Q8. In potentiometer calibration, emf is proportional to: 1 Mark

A) Balance-point resistance

B) Temperature only

C) Length only in every case

D) Current²

Correct Answer: A

Q9. For a non-uniform wire, the correct relation is: 1 Mark

A) dR=r(x)dx

B) dR=dx/r(x)

C) dR=r(x)+dx

D) dR=xr(x)

Correct Answer: A

Q10. The slope of V-x graph at a point is proportional to: 1 Mark

A) Local resistance per unit length

B) Total length only

C) Current²

D) Voltage²

Correct Answer: A

Section B — 2 Marks
10 Conceptual + Numerical Questions

Q11. Write the expression for resistance of a non-uniform wire from x=0 to x=l. 2 Marks

R(l)=∫₀ˡr(x)dx

Q12. Why can length ratio not generally be used in a non-uniform potentiometer? 2 Marks

Because resistance per unit length changes with position. Hence potential drop is proportional to the integrated resistance, not simply to length.

Q13. A potentiometer has r(x)=2x Ω/m. Find resistance from 0 to 3 m. 2 Marks

R=∫₀³2x dx

R=[x²]₀³=9 Ω

R = 9 Ω

Q14. For r(x)=kx, what is the potential ratio V(x)/V(L)? 2 Marks

R(x)=kx²/2

R(L)=kL²/2

Therefore:

V(x)/V(L)=x²/L²

Q15. A uniform potentiometer has total voltage 10 V across 100 cm. Find the voltage drop per cm. 2 Marks

k=10/100

0.1 V/cm

Q16. State the differential equation for potential along a non-uniform potentiometer. 2 Marks

dV/dx=Ir(x)

Q17. If r(x) is doubled everywhere, what happens to the potential fraction at a fixed position? 2 Marks

The potential fraction remains unchanged because both numerator and denominator of the resistance ratio are multiplied by the same factor.

Q18. A standard cell of 1.0 V balances at resistance 5 Ω. An unknown cell balances at resistance 8 Ω. Find its emf. 2 Marks

Eₓ/Eₛ=Rₓ/Rₛ

Eₓ=1×8/5

Eₓ = 1.6 V

Q19. What does a larger slope of the V-x graph indicate? 2 Marks

A larger slope means a larger local potential gradient and therefore a larger local resistance per unit length, assuming constant current.

Q20. For r(x)=r₀ constant, show that the non-uniform formula reduces to the ordinary potentiometer formula. 2 Marks

R(x)=∫₀ˣr₀dx=r₀x

Therefore:

V(x)=V×x/L

Hence:

V ∝ x

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the potential at distance x for a potentiometer whose resistance per unit length is r(x). 3 Marks

For a small element:

dR=r(x)dx

Therefore:

dV=I dR=Ir(x)dx

Integrating from 0 to x:

V(x)=I∫₀ˣr(x)dx

If total voltage is V:

V=I∫₀ᴸr(x)dx

Hence:

V(x)/V = [∫₀ˣr(x)dx]/[∫₀ᴸr(x)dx]

Q22. A potentiometer has r(x)=3x Ω/m. Its length is 4 m and total potential drop is 12 V. Find the potential drop at x=2 m. 3 Marks

Resistance from 0 to x:

R(x)=∫₀ˣ3x dx=3x²/2

At x=2:

R(2)=6 Ω

At x=4:

R(4)=24 Ω

Therefore:

V(2)=12×6/24

V(2)=3 V

Q23. The resistance per unit length of a wire varies as r(x)=r₀(1+x/L). Find the fraction of total voltage at x=L/2. 3 Marks

For x:

R(x)=r₀[x+x²/(2L)]

At L:

R(L)=3r₀L/2

At L/2:

R(L/2)=5r₀L/8

Therefore:

V(L/2)/V = (5/8)/(3/2)

V(L/2)/V=5/12

Q24. A standard cell of emf 1.5 V balances at x=20 cm. An unknown cell balances at x=30 cm. The potentiometer has r(x)=kx. Find the unknown emf. 3 Marks

For:

r(x)=kx

Resistance up to x:

R(x)=kx²/2

Therefore:

Eₓ/Eₛ=xₓ²/xₛ²

Hence:

Eₓ=1.5×(30²/20²)

Eₓ=1.5×9/4

Eₓ=3.375 V

Q25. Show mathematically that for a uniform potentiometer the potential gradient is constant. 3 Marks

For a uniform wire:

r(x)=r₀

Therefore:

dV/dx=Ir₀

Since I and r₀ are constant:

dV/dx=constant

Thus the V-x graph is a straight line.

Q26. A non-uniform potentiometer has r(x)=2+0.5x Ω/m. Find the resistance from x=0 to x=4 m. 3 Marks

R=∫₀⁴(2+0.5x)dx

R=[2x+0.25x²]₀⁴

R=8+4

R=12 Ω

Q27. For the wire in Q26, find the potential fraction at x=2 m relative to the total potential across 4 m. 3 Marks

At x=2:

R(2)=2(2)+0.25(2²)=5 Ω

Total:

R(4)=12 Ω

Therefore:

V(2)/V=5/12

V(2)=5V/12

Q28. A potentiometer is calibrated using a standard cell of emf 1.2 V. The resistance up to its balance point is 4 Ω. An unknown cell balances at a point where resistance up to the balance point is 7 Ω. Find its emf. 3 Marks

For the same current:

Eₓ/Eₛ=Rₓ/Rₛ

Therefore:

Eₓ=1.2×7/4

Eₓ=2.1 V

Q29. Explain why the balance-point length cannot directly be used for calibration when the potentiometer wire is non-uniform. 3 Marks

In a uniform potentiometer:

R∝x

Therefore:

E∝x

But in a non-uniform potentiometer:

R(x)=∫₀ˣr(x)dx

Hence:

E∝∫₀ˣr(x)dx

Therefore two equal lengths do not necessarily have equal resistance.

Calibration must be done using the resistance integral, not merely the balance-point length.

Q30. Describe the complete method for solving a non-uniform potentiometer calibration problem. 3 Marks

Step 1: Write the resistance per unit length r(x).

Step 2: For the required position x, calculate:

R(x)=∫₀ˣr(x)dx

Step 3: Calculate the reference or total resistance.

Step 4: Use:

V(x)/V_total=R(x)/R_total

Step 5: For standard-cell calibration:

Eₓ/Eₛ=Rₓ/Rₛ

Step 6: Substitute numerical values and obtain the unknown emf.

Main idea: In a non-uniform potentiometer, replace "length ratio" by "resistance ratio".

Quick Revision

dR=r(x)dx

R(x)=∫₀ˣr(x)dx

dV=Ir(x)dx

dV/dx=Ir(x)

V(x)/V_total = R(x)/R_total

Eₓ/Eₛ=Rₓ/Rₛ

Most Important Exam Point:

Uniform potentiometer:

E∝x

Non-uniform potentiometer:

E∝∫r(x)dx

Therefore, always calculate the resistance corresponding to the balance position before doing calibration.

Non-Uniform Potentiometer Calibration
CBSE + Competitive Physics Practice | 30 Questions with Answers & Solutions