📌 1. Question Description
Potentiometer
A potentiometer is a device used to measure the potential difference or
EMF of a cell accurately by using a null method.
Its working principle is based on the fact that the potential drop across
a uniform resistance wire is directly proportional to its length when a
constant current flows through the wire.
V ∝ l
The potentiometer is more accurate than a voltmeter because at the balance
point it draws no current from the cell being measured.
It is mainly used for:
- Comparison of EMFs of two cells.
- Determination of internal resistance of a cell.
- Measurement of potential difference.
⚡ 2. Principle of Potentiometer
Consider a uniform wire of length L and resistance R carrying a constant
current I.
Using Ohm's law:
V = IR
Resistance of a portion of wire of length l is proportional to l.
Therefore potential drop across length l is:
V ∝ l
Hence:
V/l = constant = k
Here
k is called the
potential gradient.
k = V/L
The balance condition is obtained when the potential drop along the
potentiometer wire becomes equal to the EMF being measured.
📐 3. Potential Gradient
Potential gradient is the potential drop per unit length of the
potentiometer wire.
k = V/L
where:
- k = potential gradient
- V = potential difference across the potentiometer wire
- L = total length of potentiometer wire
Important: Smaller potential gradient gives greater sensitivity
because a small potential difference produces a larger balance length.
🔬 4. Animated Potentiometer Practical
Cell EMF:
1.50 V
Potential Gradient:
0.015 V/m
Balance Length:
100.00 m
Condition:
BALANCE POINT ✔
🔋 5. Comparison of EMFs of Two Cells
Suppose two cells have EMFs ε₁ and ε₂. Let their balance lengths be
l₁ and l₂ respectively.
At balance:
ε₁ = kl₁
and
ε₂ = kl₂
Dividing:
ε₁/ε₂ = l₁/l₂
Therefore:
ε₁ : ε₂ = l₁ : l₂
Thus, the EMFs of two cells can be compared by comparing their
corresponding balance lengths.
🧪 6. Interactive EMF Comparison
ε₁ / ε₂ =
1.500
l₁ / l₂ =
1.500
Calculated l₂ =
50.00 cm
RELATION SATISFIED ✔
🔌 7. Determination of Internal Resistance of a Cell
Let ε be the EMF of the cell and r its internal resistance.
When the cell is connected to an external resistance R, the terminal
potential difference is V.
The current is:
I = V/R
Also:
ε = V + Ir
Therefore:
ε = V + (V/R)r
Hence:
r = R(ε - V)/V
Therefore the internal resistance is:
r = R(ε/V - 1)
📏 8. Potentiometer Method for Internal Resistance
Let l₁ be the balance length when the cell is not supplying current.
Then:
ε = kl₁
When an external resistance R is connected, let l₂ be the new balance
length.
V = kl₂
Therefore:
ε/V = l₁/l₂
Using:
r = R(ε/V - 1)
we get:
r = R(l₁/l₂ - 1)
or:
r = R(l₁-l₂)/l₂
⭐ Important Formula:
r = R(l₁ - l₂)/l₂
🔬 9. Interactive Internal Resistance Practical
External Resistance R:
5.00 Ω
l₁:
80.00 cm
l₂:
60.00 cm
Internal Resistance r:
1.667 Ω
r = R(l₁-l₂)/l₂
⭐ 10. Advantages of Potentiometer
✔ High Accuracy
It uses a null method and can measure EMF accurately.
✔ No Current Draw
At balance, no current is drawn from the test cell.
✔ EMF Comparison
It can compare the EMFs of two cells directly.
✔ Internal Resistance
It can determine the internal resistance of a cell.
⚠️ 11. Important Precautions
- The potentiometer wire should be uniform.
- The current through the potentiometer wire should remain constant.
- The jockey should not be dragged continuously over the wire.
- The jockey should be pressed gently.
- The balance point should preferably lie near the middle of the wire.
- Connections should be tight and clean.
- The driver cell should have sufficient EMF.
- The positive terminal of the test cell must be connected to the appropriate end of the potentiometer wire.
📝 12. MCQ Practice
1. The principle of potentiometer is based on:
A. V ∝ 1/l
B. V ∝ l
C. V ∝ l²
D. V = constant
✔ Answer: B
2. At the balance point of a potentiometer:
A. Maximum current flows through galvanometer
B. No current flows through galvanometer
C. Infinite current flows
D. Current is maximum in test cell
✔ Answer: B
3. Potential gradient is given by:
A. k = VL
B. k = V/L
C. k = L/V
D. k = IR
✔ Answer: B
4. If two cells have balance lengths l₁ and l₂, their EMF ratio is:
A. ε₁/ε₂ = l₂/l₁
B. ε₁/ε₂ = l₁/l₂
C. ε₁ε₂ = l₁l₂
D. ε₁ = ε₂
✔ Answer: B
5. The potentiometer is more accurate than a voltmeter because:
A. It has high resistance only
B. It uses a null method
C. It has no wire
D. It produces current
✔ Answer: B
6. Internal resistance of a cell using potentiometer is:
A. R(l₂-l₁)/l₁
B. R(l₁-l₂)/l₂
C. Rl₂/l₁
D. Rl₁/l₂
✔ Answer: B
7. When external resistance is connected to a cell, terminal voltage:
A. Increases
B. Decreases
C. Becomes infinite
D. Always becomes zero
✔ Answer: B
8. For greater sensitivity, potential gradient should be:
A. Large
B. Small
C. Infinite
D. Zero current only
✔ Answer: B
9. The SI unit of potential gradient is:
A. V m⁻¹
B. Ω m
C. A m⁻¹
D. C m⁻¹
✔ Answer: A
10. If l₁ = 80 cm, l₂ = 60 cm and R = 5Ω, internal resistance is:
A. 1.0Ω
B. 1.67Ω
C. 2.5Ω
D. 3.0Ω
✔ Answer: B
🟣 13. Assertion–Reason Practice — 5 Questions
Choose the correct option:
A. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
B. Both Assertion and Reason are true, but Reason is not the correct explanation.
C. Assertion is true, but Reason is false.
D. Assertion is false, but Reason is true.
Assertion (A):
A potentiometer is more accurate than a voltmeter for measuring EMF.
Reason (R):
At the balance point, the potentiometer draws no current from the test cell.
Answer: A
Explanation: Since no current is drawn from the test cell at balance,
there is no voltage drop due to its internal resistance.
Assertion (A):
The potential drop across a uniform potentiometer wire is proportional
to its length.
Reason (R):
The resistance of a uniform wire is proportional to its length.
Answer: A
Explanation: For constant current, V = IR and R ∝ l. Therefore,
V ∝ l.
Assertion (A):
A smaller potential gradient makes a potentiometer more sensitive.
Reason (R):
For a given small potential difference, a smaller gradient produces a
larger balance length.
Answer: A
Explanation: Since ε = kl, smaller k gives larger l for the same ε.
Assertion (A):
The terminal potential difference of a cell is smaller than its EMF when
the cell supplies current.
Reason (R):
Some potential is lost across the internal resistance of the cell.
Answer: A
Explanation: When current flows, the internal resistance causes
a voltage drop Ir, so V = ε - Ir.
Assertion (A):
When an external resistance is connected to a cell, the balance length
for its terminal voltage becomes larger.
Reason (R):
Terminal voltage is always greater than the EMF of a cell.
Answer: C
Explanation: The terminal voltage becomes smaller than EMF when
the cell supplies current. Therefore the balance length becomes smaller.
🟢 14. 2 Marks — 6 Questions
Q1
State the principle of a potentiometer.
Q2
Define potential gradient.
Q3
Why is a potentiometer more accurate than a voltmeter?
Q4
Write the relation used for comparing the EMFs of two cells.
Q5
What is meant by internal resistance of a cell?
Q6
Write the formula for determining internal resistance using a potentiometer.
🟡 15. 3 Marks — 6 Questions
Q1
Explain the principle of a potentiometer.
Q2
Explain the meaning of potential gradient and write its expression.
Q3
Explain how the EMFs of two cells are compared using a potentiometer.
Q4
Why does a potentiometer use a null method?
Q5
Derive the expression ε₁/ε₂ = l₁/l₂.
Q6
Explain why the terminal potential difference of a cell is less than
its EMF when supplying current.
🟠 16. 4 Marks — 6 Questions
Q1
Explain the construction and working of a potentiometer.
Q2
Derive the expression for potential gradient.
Q3
Explain the experimental method for comparing the EMFs of two cells.
Q4
Derive the expression r = R(l₁-l₂)/l₂ for internal resistance.
Q5
A cell gives a balance length of 80 cm in open circuit and 60 cm when
connected across a 5Ω resistance. Calculate its internal resistance.
Q6
Write four precautions for a potentiometer experiment.
🔴 17. 5 Marks — 6 Questions
Q1
Explain the principle, construction and working of a potentiometer.
Q2
Describe the experimental procedure for comparing the EMFs of two cells
using a potentiometer and derive the required formula.
Q3
Explain how the internal resistance of a cell is determined using a
potentiometer.
Q4
Explain why the potentiometer is preferred over a voltmeter for measuring
the EMF of a cell.
Q5
Two cells have balance lengths 60 cm and 90 cm respectively. If the EMF
of the first cell is 1.2 V, calculate the EMF of the second cell.
Q6
Explain the factors affecting the sensitivity of a potentiometer.
🔵 18. 6 Marks — 6 Questions
Q1
Explain the construction, principle and working of a potentiometer in
detail. Derive the relation between potential difference and balance length.
Q2
Describe the complete experiment for comparison of EMFs of two cells
using a potentiometer and derive ε₁/ε₂ = l₁/l₂.
Q3
Describe the complete experiment for determination of internal resistance
of a cell using a potentiometer. Derive the required expression.
Q4
Explain the advantages of a potentiometer over a voltmeter and discuss
the factors affecting its sensitivity.
Q5
A cell has an open-circuit balance length of 80 cm. When connected to an
external resistance of 5Ω, the balance length becomes 60 cm. Calculate its
internal resistance and explain the complete calculation.
Q6
Explain the complete principle, mathematical derivations, applications,
advantages, limitations and precautions of a potentiometer.
🧮 19. Numerical Practice
Example 1 — EMF Comparison
Two cells have balance lengths 60 cm and 90 cm respectively.
The EMF of the first cell is 1.2 V. Find the EMF of the second cell.
ε₁/ε₂ = l₁/l₂
1.2/ε₂ = 60/90
ε₂ = 1.8 V
✔ Answer: ε₂ = 1.8 V
Example 2 — Internal Resistance
A cell has an open-circuit balance length of 80 cm and a loaded balance
length of 60 cm. An external resistance of 5Ω is connected across the cell.
Find its internal resistance.
r = R(l₁-l₂)/l₂
r = 5(80-60)/60
r = 100/60
r = 1.67Ω
✔ Answer: r ≈ 1.67Ω
🚀 20. Quick Revision
Principle:
Potential drop across a uniform wire is directly proportional to its
length.
Potential Gradient:
k = V/L
Balance Condition:
ε = kl
Comparison of EMFs:
ε₁/ε₂ = l₁/l₂
Internal Resistance:
r = R(l₁-l₂)/l₂
Higher Sensitivity:
Smaller potential gradient.
Major Advantage:
No current is drawn from the test cell at balance.
