Topic 10 of 10 • Chapter: Current Electricity • Physics
Potentiometer – Complete Notes, Revision, Important Questions & Downloads
The potentiometer is an analytical tool primarily tested in NEET through circuit balancing problems. You will study its core principle through the subtopic Potential Gradient, understanding how voltage drops evenly across the wire ($E = xl$). Then, in Application of Potentiometer, you will apply this to compare cell emfs, calculate internal resistance, and calibrate meters.
PhysicsClass 12Current Electricity
Expected QuestionsQ
1-2
Typical questions in recent years
Time Required⏱
2.5 Hours
For full mastery
Difficulty⚡
Hard
Requires circuit analysis
NRI USA Curriculum GapUS
High Gap
US curricula favor digital multimeters
2Subtopics
25+Practice Questions
4Free Downloads
2.5 HoursPrep Time
NEET Weightage & Trends
Chapter 19: Current Electricity| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2023 | 1 | 4 | |
| 2022 | 2 | 8 | |
| 2021 | 1 | 4 | |
| Expected Total | 1 | 4 |
Questions often combine internal resistance calculation with a change in the primary circuit.
Sensitivity manipulation (changing wire length or primary resistance) is a frequent theoretical question.
Balancing length ratio problems are the most common calculation type.
Sensitivity manipulation (changing wire length or primary resistance) is a frequent theoretical question.
Balancing length ratio problems are the most common calculation type.
1.3
Avg Questions / Year
5.2
Total Marks (6 yrs)
Mixed
Pattern
Hard
Difficulty
Preparation Strategy
1
Master the Potential Gradient Formula Memorize $x = i\rho$ and $x = V/L$. Ensure you check if the primary circuit current changes when the secondary circuit is modified. The trap is assuming $x$ remains constant if the primary driver cell's series resistance is altered. Recognize this when a rheostat is adjusted.
2
Understand the Null Point Condition Memorize that at the null point, no current flows through the galvanometer, meaning the secondary cell is in an open circuit state. The trap is using $V = E - ir$ with a non-zero current. Look for 'galvanometer shows zero deflection' to apply $E = xl$.
3
Internal Resistance Calculation Memorize $r = R(l_1/l_2 - 1)$. The condition to check is whether $l_1$ is with the key open and $l_2$ is with the key closed. The trap is swapping $l_1$ and $l_2$ or forgetting to multiply by the external shunt $R$. Recognize this when a resistance box is connected across the unknown cell.
Study Materials
PDF · Cheat Sheet · MCQ Set · PYQTopic Breakdown
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision
Concept → Trap → Example1) Potential Gradient
Theory & FormulaPotential gradient x = V/L = iR/L = (e / (R + R_h + r)) * (R/L)
- It is the fall of potential per unit length of the potentiometer wire.
- Depends directly on the primary circuit current and the wire's resistance per unit length.
- Trap: Assuming $x$ changes when a cell in the secondary circuit is replaced. $x$ only depends on the primary circuit.
Example (NEET-style)If a 10m wire has 20 ohms resistance and the driver cell provides 2V with no internal resistance, $x = 2/10 = 0.2$ V/m.
2) Application of Potentiometer
Circuit ApplicationInternal resistance r = ((l_1 - l_2) / l_2) * R'
- Used to compare emfs: $E_1/E_2 = l_1/l_2$.
- To find internal resistance, $l_1$ is the balance length on open circuit, $l_2$ on closed circuit with shunt $R'$.
- Trap: Using the wrong length in the denominator; $l_2$ (the shorter length) is always in the denominator for internal resistance.
Example (NEET-style)If $l_1 = 125$ cm, $l_2 = 100$ cm, and $R' = 20\Omega$, then $r = ((125-100)/100)\times 20 = 5\Omega$.
Curriculum Context
Note for International StudentsNull Deflection Method
US AP Physics 2 curriculum typically focuses on digital multimeters for voltage measurement, whereas NEET heavily emphasizes the analytical analog approach of the potentiometer.
- NEET requires calculating the exact balance length where galvanometer current is zero.
- US exams rarely test the internal circuitry of a potentiometer.
- The concept of drawing zero current to measure true emf is a staple of the NEET syllabus.
Meter Calibration
The explicit process of standardizing a potentiometer and using it to calibrate ammeters and voltmeters is a staple of Indian board and NEET syllabi, but rarely covered in standard US high school physics.
- Standardization using a known standard cell ($E_0$).
- Calculating errors in practical ammeters and voltmeters.
- Using a $1\Omega$ standard resistance coil for ammeter calibration.
Previous Year Questions
2 Questions1[NEET 2015] A potentiometer wire has length 4 m and resistance 8 $\Omega$. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1 mV per cm on the wire is:Potential Gradient
32 $\Omega$
40 $\Omega$
44 $\Omega$
48 $\Omega$
Target potential gradient $x = 1$ mV/cm = $10^{-3} / 10^{-2}$ V/m = 0.1 V/m. Total voltage drop across the 4m wire must be $V = x \times L = 0.1 \times 4 = 0.4$ V. The current in the primary circuit is $I = V / R_{wire} = 0.4 / 8 = 0.05$ A. The total resistance of the primary circuit is $R_{total} = E / I = 2 / 0.05 = 40 \Omega$. The series resistance $R$ is $R_{total} - R_{wire} = 40 - 8 = 32 \Omega$. The wrong options result from calculation errors or forgetting to subtract the wire's own resistance.
2[NEET 2016] A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emfs is:Application of Potentiometer
3:2
5:1
5:4
3:4
When cells assist each other: $E_1 + E_2 = x \times 50$. When cells oppose each other: $E_1 - E_2 = x \times 10$. Dividing the two equations gives $(E_1 + E_2) / (E_1 - E_2) = 50 / 10 = 5$. Solving this proportion: $E_1 + E_2 = 5E_1 - 5E_2 \implies 4E_1 = 6E_2 \implies E_1/E_2 = 6/4 = 3/2$. The incorrect options come from simply taking the ratio of the lengths (5:1) without setting up the sum and difference equations.
Practice Questions
Click "Reveal Answer" after attempting1In the primary circuit of a potentiometer, a battery of 2 V and a rheostat of 22 $\Omega$ are connected. If the length of the potentiometer wire is 10 m and its resistance is 18 $\Omega$, find the potential gradient in the wire.
0.09 V/m
0.18 V/m
0.20 V/m
0.05 V/m
👁 Reveal Answer
Option 1 is correct. The total resistance in the primary circuit is $R_{eq} = 18 + 22 = 40 \Omega$. The current in the circuit is $I = V/R_{eq} = 2/40 = 0.05$ A. The potential drop across the potentiometer wire is $V_{wire} = I \times R_{wire} = 0.05 \times 18 = 0.9$ V. The potential gradient is $x = V_{wire}/L = 0.9/10 = 0.09$ V/m.
2A cell gives a balance with 125 cm of a potentiometer wire. When the cell is shunted by a resistance of 20 $\Omega$, the balance length shifts to 100 cm. What is the internal resistance of the cell?
5 $\Omega$
10 $\Omega$
4 $\Omega$
2.5 $\Omega$
👁 Reveal Answer
Option 1 is correct. The internal resistance is given by the formula $r = R \left(\frac{l_1 - l_2}{l_2}\right)$. Here, $l_1 = 125$ cm (unshunted), $l_2 = 100$ cm (shunted), and $R = 20 \Omega$. Substituting the values: $r = 20 \times \left(\frac{125 - 100}{100}\right) = 20 \times \left(\frac{25}{100}\right) = 20 \times 0.25 = 5 \Omega$.
3To increase the sensitivity of a potentiometer, one should:
Increase the length of the potentiometer wire
Decrease the length of the potentiometer wire
Increase the current in the primary circuit
Replace the wire with one of higher specific resistance
👁 Reveal Answer
Option 1 is correct. The sensitivity of a potentiometer increases when its potential gradient decreases, meaning a smaller voltage corresponds to a larger, more measurable balancing length. Since potential gradient $x = V/L$, increasing the length $L$ decreases $x$, thereby increasing the sensitivity. Increasing current would increase $x$, reducing sensitivity.
4In a potentiometer experiment, the balancing length for a cell is 560 cm. When a 10 $\Omega$ resistor is connected in parallel to the cell, the balancing length changes by 60 cm. Find the internal resistance of the cell.
1.2 $\Omega$
1.0 $\Omega$
1.5 $\Omega$
2.0 $\Omega$
👁 Reveal Answer
Option 1 is correct. The initial balancing length is $l_1 = 560$ cm. The new balancing length $l_2$ is shorter because the cell is supplying current, so $l_2 = 560 - 60 = 500$ cm. Using the internal resistance formula $r = R \left(\frac{l_1 - l_2}{l_2}\right) = 10 \times \left(\frac{560 - 500}{500}\right) = 10 \times \frac{60}{500} = 1.2 \Omega$.
Physics Revision Checklist
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Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
Frequently Asked Questions
Notes · Downloads · Revision · Important QuestionsWhy is a potentiometer preferred over a voltmeter for measuring emf?
A potentiometer is based on the null deflection method. At the balance point, it draws no current from the cell whose emf is being measured. A practical voltmeter has finite resistance and draws a small current, which drops some voltage across the cell's internal resistance, measuring terminal voltage rather than true emf.
What happens if the driver cell's emf is less than the cell to be measured?
If the emf of the driver cell (primary circuit) is less than the emf of the unknown cell (secondary circuit), the potential drop across the entire potentiometer wire will be less than the unknown emf. Therefore, a balance point (null deflection) cannot be obtained anywhere on the wire.
How does the length of the potentiometer wire affect its sensitivity?
Increasing the length of the potentiometer wire (for a given potential difference) decreases the potential gradient (V/L). A smaller potential gradient means a given emf will balance at a longer length, allowing for more precise measurement and reducing percentage error, thus increasing sensitivity.
Does the cross-sectional area of the wire need to be uniform?
Yes, the cross-sectional area of the potentiometer wire must be uniform throughout its length. If it is not uniform, the resistance per unit length will vary, causing the potential gradient to be non-uniform. This would make the basic principle $E = xl$ invalid.
Can a potentiometer measure alternating current (AC)?
A standard DC potentiometer cannot measure AC directly because it relies on a steady DC driver cell and a galvanometer that detects unidirectional current. For AC, a specialized AC potentiometer must be used, which requires balancing both the magnitude and phase of the voltage.
What is the purpose of the high resistance in series with the galvanometer?
A high resistance is sometimes connected in series with the galvanometer to protect it from large currents when the jockey is far from the balance point. Once the balance point is approached, this resistance is shorted out to achieve maximum sensitivity for the final precise adjustment.
Why do we use manganin or constantan for the potentiometer wire?
Manganin and constantan are alloys that have a high specific resistance (resistivity) and a very low temperature coefficient of resistance. This ensures that the resistance of the wire, and thus the potential gradient, does not change significantly if the wire heats up due to the current.
How is thermo emf measured using a potentiometer?
Thermo emfs are very small (of the order of microvolts). To measure them, the potential gradient of the potentiometer must be made very small by connecting a high resistance in series with the primary circuit. The measurement then follows the standard balancing procedure.
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