CBSE Class 12-science Answered

Potentiometer is a mainly used to measure emf of a given cell, internal resistance of a cell and also to compare emf's of cells.

Potentiometer consists of a long resistive wire l made up of mangnine or constantan and a battery of known voltage V called driver cell. Connection of these two forms the primary circuit.

One terminal of another cell (whose emf E is to be measured) is connected at one end of the main circuit and the other terminal at any point on the resistive wire through a galvanometer G. This forms the secondary circuit.

When there is no potential difference between two points there no electric current will flow, this is the basic potentiometer principle.

Now the potentiometer wire AB is actually is a wire with high resistivity (?) with uniform cross section area. Thus this wire has a uniform resistance throughout.

Now this potentiometer end terminals are connected to a cell of high emf V (neglecting its internal resistance) called the driver or the auxiliary cell. Let a current I flow through the potentiometer wire. Let R = total resistance of the potentiometer wire.

Then by Ohm’s law V = IR

We know

R = ? l /A

Thus,

V = I ? l /A

As ?, A are constant and I is kept constant by a rheostat (not shown in the figure) so I?/A = k ( constant )

V = k l 

Now suppose a cell of lower emf than the auxiliary is put in the circuit as shown above. Say it has a emf E. Now in the potentiometer wire say at a length = x the potential difference is E.

Thus

E = I ?x/A = kx

When this cell be put in the ckt as shown in the fig with a jockey connected to the corresponding length = x , there will be no flow of current through the galvanometer because when pot. difference is equal no current flows. So galvanometer G would show no deflection. Then the length x is called the length of the null point.

Now by knowing the constant k and noting down the x, we may find the emf of the unknown cell.

Secondly emf of two cells may also be compared, Let the first cell of emf E₁ give a

null point at a length = l₁ 

And the second cell show a null point at length = l₂ 

Then,

E₁ /E₂ = l₁ /l₂ 

• Measure s internal resistance of a cell

• The cell of emf E (internal resistance r) is connected across a resistance box (R) through key K ₂.

• K ₂ ? open, balance length is obtained at length AN₁ = l ₁

E= ? l ₁ (3)

• K ₂ ? closed

• Let V be the terminal potential difference of cell and the balance is obtained at AN₂ = l ₂

? V = ? l ₂ (4)

From equations (3) and (4),

From (5) and (6),