A potentiometer measures the internal resistance of a cell, compares e.m.f. of cells, and potential difference across a resistor using a long wire of uniform cross-section. It operates on the principle that potential difference is directly proportional to wire length under constant current. The document discusses construction, working, calibration, comparisons of e.m.f., determination of internal resistance, and includes problem-solving scenarios involving a potentiometer setup.

2. Potentiometer is a device used to measure the internal
resistance of a cell, to compare the e.m.f. of two cells and
potential difference across a resistor. It consists of a long wire
of uniform cross sectional area and of 10 m in length. The
material of wire should have a high resistivity and low
temperature coefficient. The wires are stretched parallel to
each other on a wooden board. The wires are joined in series
by using thick copper strips. A metre scale is also attached on
the wooden board.
It works on the principle that when a constant current flows
through a wire of uniform cross sectional area, potential
difference between its two points is directly proportional to
the length of the wire between the two points.

3. Potentiometer

4. • A potentiometer is a null type resistance
network device used for measuring potential
differences and comparison of emfs of
potential difference sources.

6. Principle of potentiometer
• A potentiometer works on the principle of balancing
one voltage against another in parallel with it.
• Construction and working of potentiometer
• It consists of a long resistance wire AB, sometimes 6 m
in length, and of uniform diameter. This resistance wire
R, a working battery and a variable resistance Rx all
connected in series. Rx enables us to develop a
convenient output voltage V between A and B,
maintaining a constant current I in the circuit. Now,
according to Ohm’s law I = V/R

7. • A source of emf ε which is
to be measured is
connected along with a
Galvanometer through
the sliding contact C. See
figure. As C slides over
AB, a variable potential
difference is available
between A and C. If r
represents the resistance
between A and C, the
potential drop between A
and C is given by

8. Now if the PD between A and B is greater than ε , then there must be
some point C between A and B, at which the potential drop across AC
is equal to ε. This point is found by sliding C over AB until the current
in the galvanometer is zero. The current in galvanometer becomes
zero because under such condition, called null condition, both the
ends of the galvanometer are at the same potential. So under null
conditions

9. • Here r is the value of the wire resistance of length AC at null
condition.
• Let the wire is of uniform cross-section A and AB = L and AC =
l, then
We know that resistance is directly proportional to the
length and inversely proportional to the cross-sectional area
of the wire.

10. • Divide one equation by the other

11. • Where L is the total length of the wire and l is
the length AC. This equation gives the
unknown emf in terms of the ratio of two
known lengths and voltage applied. Ordinarily,
the potential drop along the wire is calibrated
and the balanced point gives directly the
unknown emf ε

12. Comparison of emfs
• Potentiometer can also be
used for the comparison of
two emf sources. The
balancing length of emf ε1 of a
cell is found and compared
with the balancing length of a
standard cell which is also
found separately. If ε and ε1
are the two emf sources
nulled at lengths l and l1
respectively, then,
• Dividing one equation by
other,

13. Dividing one equation by other,

14. Internal resistance of primary cell
• Internal Resistance is the
resistance which is
present within the
battery that resists the
current flow when
connected to a circuit.
Thus it causes a voltage
drop when current flows
through it. It is the
resistance provided by
the electrolyte and
electrodes which is
present in a cell.

15. Relation between e.m.f(E)., potential
difference(V) , and internal
resistance(r) of a cell.
If a cell of emf E and internal resistance r, connected to an external resistance R,
then the circuit has the total resistance (R+r). The current I in the circuit is given by,
Hence,
This means, V is less than E by an amount equal to the fall of potential inside
the cell due to its internal resistance.
From the above equation,

16. Or; The internal resistance of the cell,

17. Determination of internal resistance of
a primary by potentiometer
• Using a potentiometer, we can
adjust the rheostat to obtain
the balancing lengths l1 and l2
of the potentiometer for open
and closed circuits
respectively.
• Then, E= k l1 and V = k l2 ;
where k is the potential
gradient along the wire.
• Now we can modify the equation for
getting the internal resistance of the
given cell, by using the above relations
as;

18. Problems related to potentiometer

22. Home assignment
• Following Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40
Ω maintaining a potential drop across the resistor wire AB. A standard cell which
maintains a constant emf of 1.02 V(for very moderate currents upto a few mA) gives a
balance point at67.3 cm length of the wire. To ensure very low currents drawn from the
standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted
close to the balance point. The standard cell is then replaced by a cell of unknown emf ε
and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
• (a) What is the value ε?
• (b) What purpose does the high resistance of 600 kΩ have?
• (c) Is the balance point affected by this high resistance?
• (d) Is the balance point affected by the internal resistance of thedriver cell?
• (e) Would the method work in the above situation if the driver cellof the potentiometer
had an emf of 1.0V instead of 2.0V?
• (f ) Would the circuit work well for determining an extremely smallemf, say of the order
of a few mV (such as the typical emf of athermo-couple)? If not, how will you modify
the circuit?