Potentiometer - Principle And Applications

Potentiometer - Principle And Applications

Vishal kumarUpdated on 02 Jul 2025, 05:34 PM IST

A potentiometer is an adaptable tool used to find out and also compare electricity potentials- mostly in DC currents. Basically, it operates under the assumption that voltage drop on a constant length of wire varies uniformly with its length provided there is a constant current flowing through it, therefore, allowing accurate measurement and not requiring withdrawal of power from the test setup.

Potentiometer - Principle And Applications
Potentiometer

In this article, we will discuss the concept of Potentiometer - Principle And Applications. It is an essential gadget in the learning of current electricity which is important for Class 12, NEET and JEE Main exams. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of fifteen questions have been asked on this concept. And for NEET seven questions were asked from this concept.

What is a Potentiometer?

The potentiometer is a device which does not draw current from the given circuit and still measures the potential difference. It is a device used to measure the e.m.f of a given cell and to compare the e.m.f of cells. It is also used to measure the internal resistance of a given cell

The potentiometer consists of wires of length 5 to 10 meters arranged on a wooden block as parallel strips of wires with 1-meter length each and the ends of wires are joined by thick coppers. The wire has a uniform cross-section and is made up of the same material. A driver circuit that contains a rheostat, key, and a voltage source with internal resistance r. The driver circuit sends a constant current (I) through the wire.

The potential across the wire AB having length L is given as V=IR, Where R is the resistance of the wire AB

Since the driver circuit sends a constant current (I) through the wire So

VR

Using R=ρLA we can say that RL since area and resistivity are constant.

Therefore we get V is proportional to length. I.e VL

The secondary circuit contains cells/resistors whose potential is to be measured. One end is connected to a galvanometer and another end of the galvanometer is connected to a jockey which is moved along the wire to obtain a point where there is no current through the galvanometer. So the potential of the secondary circuit is proportional to the length at which there is no current through the galvanometer. This is how the potential of a circuit is measured using the potentiometer.

Calibration of Potentiometer

In the potentiometer, a battery of known emf E is connected to the secondary circuit. A constant current I is flowing through AB from the driver circuit (that is the circuit above AB). The jockey is slide on potentiometer wire AB to obtain null deflection in the galvanometer. Let l be the length at which the galvanometer shows null deflection.

Since the potential of wire AB (V) is proportional to the length AB(L).

Similarly El

So we get

VE=LlV=ELl

Thus we obtained the potential of wire AB when a constant current is passing through it. This is known as calibration.

Potential Gradient

The potential difference per unit length of wire i.e x=VL

or Using

V=iR=(eR+Rh+r)Rx=VL=e(R+Rh+r)RL


For the above figure

l1 is the balancing length obtained when a cell with emf E1 is included in the secondary circuit. That is key is at position 1.l2 is the balancing length obtained when cell with emf E2 is included in the secondary circuit. That is key is at position 2.

So since Eαl we get
E1E2=l1l2

With the help of the above ratio, we can compare the emf of these cells.

Determine the Internal Resistance of a Cell

Note- The cell in the secondary circuit has emf E and internal resistance r

Here l1 is the balancing length obtained when key K is open that is we include only the cell in the secondary circuit. So corresponding potentials of wire of balancing length l1 is E. And we know that El1 (1)

Similarly, l2 is the balancing length obtained when key K is closed that is both cell and R is connected in the secondary circuit. So corresponding potentials of wire of balancing length l2 is V.

And we know that

Vl2 (2)
or we can say that
IRl2Er+RRl2

So taking the ratio of equation (1) to equation (2), we get

EV=l1l2EERr+R=l1l2r+RR=l1l2

Then the internal resistance is given by
r=(l1l2l2)Rr=(EV1)R

Comparison of Resistances

The balance point is at a length l1 cm from A when jockey J is plugged in between Y and X, while the balance point is at a length l2 cm from A when jockey J is plugged in between Y and Z.

Then we get a ratio of resistances as

R2R1=l2l1l1

With the help of this ratio, we can compare these resistances.

Solved Examples Based on Potentiometer - Principle And Applications

Example 1: In the given circuit of the potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at J1 so that there is no deflection in the galvanometer. Now the first battery (E1) is replaced by the second battery (E2) for working by making K1 open and K2 closed. The galvanometer gives them a null deflection at J2. The value of E1E2 is abwhere a = ______

1) 1

2) 2

3) 3

4) 4

Solution:

Length of AB=10 m

For battery E1, balancing length is l1 and l1=380 cm [from end A ]
For battery E2, balancing length is l2 and l1=760 cm [from end A ]

Now, we know that

E1E2=l1l2E1E2=380760=12=aba=1&b=2a=1

Hence, the answer is option (1).

Example 2: In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells \frac{\varepsilon_1}{\varepsilon_2} is :


1) 53
2) 85
3) 43
4) 32

Solution:

When the galvanometer is connected to the point (1)
ε1=ϕl1(1)

When the galvanometer is connected to point (2)
ε1+ε2=ϕl2(2)

Where l1=250 cm&l2=400 cm
ε1ε1+ε2=250400=58ε1+ε2ε1=851+ε2ε1=85ε2ε1=35ε1ε2=53

Hence, the answer is option (1).

Example 3: A DC main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1Ω. The battery terminals are connected to an external resistance ' R '. The minimum value of ' R '(in ohm), so that a current passes through the battery to charge it is :

1) 11

2) 9

3) 7

4) 0

Solution:

Given E=200 V
r=1ΩV=220 Vr=(EVV)R=1=(20220)RR=11Ω

Hence, the answer is (1).

Example 4: In a potentiometer experiment the balancing with a cell is at a length 240 cm. On shunting the cell with a resistance of 2Ω the balancing length becomes 120 cm The internal resistance (in Ω ) of the cell is:

1) 7

2) 2

3) 1

4) 0.5

Solution:

Determine the internal resistance

r=(l1l2l2)Rr=(EV1)REV=l1l2

wherein

The internal resistance of a cell is given by

r=R(l1l21)=R(l1l2l2)r=2[240120120]=2Ω
Hence, the answer is option (2).


Example 5: The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10Ω is connected in parallel to the cell, the balancing length changes by 60 cm. If the internal
N10Ω
resistance of the cell is 10 , where N is an integer then the value of N is:

1) 21

2) 24

3) 12

4) 42

Solution:

E560E×1010+r50010+r10=5650r=1.2=n10n=12

Hence, the answer is option (2).

Summary

A potentiometer is a gadget customarily utilized for ascertaining the potential difference (voltage) over a component in a circuit without sourcing any current from that particular element. It works based on the fact the loss of potential over a length of wire is directly proportional to that length. By employing a sliding contact along the wire with the intention of comparing an unknown voltage to a given reference electrical potential point the unknown voltage could be found precisely.

Frequently Asked Questions (FAQs)

Q: What is the importance of the "standardization" process in potentiometer measurements?
A:
Standardization in potentiometer measurements refers to the process of calibrating the potentiometer using a standard cell of known emf. This process:
Q: How does the resistance of the potentiometer wire affect its performance?
A:
The resistance of the potentiometer wire affects its performance in several ways:
Q: What are the limitations of using a potentiometer for voltage measurements?
A:
Limitations of using a potentiometer for voltage measurements include:
Q: How does the potentiometer principle apply to modern electronic circuits?
A:
The potentiometer principle is applied in modern electronics through:
Q: What is the relationship between the balancing length and the emf being measured in a potentiometer?
A:
In a potentiometer, the relationship between the balancing length (L) and the emf being measured (E) is directly proportional, given by the equation: E = kL, where k is the potential gradient along the wire. This linear relationship forms the basis for potentiometer measurements, allowing unknown emfs to be determined by comparing balancing lengths.
Q: How does the potentiometer method compare to the Wheatstone bridge method for resistance measurement?
A:
While both methods use null detection, the potentiometer method:
Q: How can a potentiometer be used to verify Kirchhoff's voltage law?
A:
To verify Kirchhoff's voltage law using a potentiometer:
Q: What is the role of the standard cell in potentiometer measurements?
A:
The standard cell in potentiometer measurements serves as a reference voltage source. Its roles include:
Q: How does the concept of "back emf" relate to potentiometer measurements?
A:
Back emf, or counter-electromotive force, is relevant to potentiometer measurements in that:
Q: How can a potentiometer be used to study the characteristics of solar cells?
A:
A potentiometer can be used to study solar cell characteristics by: