(i) In the circuit diagram given below, AB is a uniform wire of resistance 15 Ω and length 1 m. It is connected to a cell E1 of emf 2V and negligible internal resistance and a resistance R. The balance point with another cell E2 of emf 75 mV is found at 30 cm from end A. Calculate the value of R.

(ii) Why is potentiometer preferred over a voltmeter for comparison of emf. of cells?
(iii) Draw a circuit diagram to determine internal resistance of a cell in the laboratory.
OR
(i) Plot a graph showing variation of voltage vs the current drawn from the cell.
How can one get information from this plot about the emf of the cell and its internal resistance?
(ii) Two cells of emf’s E1 and E2 and internal resistance r1 and r2 are connected in parallel. Obtain the expression for the emf and internal resistance of a single equivalent cell that can replace this combination?
Given:
Resistance of the wire, r=15 Ω
Length of potentiometer wire, l = 1 m
EMF of cell 1, E1 = 2 V
EMF of cell 2, E2 = 75 mV
Balancing point, L=30 cm=0.3 m
Calculation EMF per unit length is given by,

Current is given by,

Potential is given by,

EMF per unit length of the wire is given by,

(ii) The reasons for selecting potentiometer over voltmeter for comparison are as follows:
• The potentiometer is more sensitive in comparison with the voltmeter.
• The voltmeter measures the terminal voltage whereas the potentiometer measures the actual emf.
• The potentiometer is used to measures the internal resistance of the cell but the voltmeter cannot measure the internal resistance.
(iii)

OR
(i) The graph showing variation of voltage vs the current drawn from the cell is given by.

The potential of the cell is given by,
![]()
Where,
E=EMF of the cell
I = Current
r = internal resistance
(ii) When two cells are connected in parallel combination, Applying Kirchhoff’s law
I =I1 +I2
The equivalent emf of two cells connected in parallel is given by,
![]()
The equivalent internal resistance,
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