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Question : 29
Total: 37
(a) Explain the principle of working of a potentiometer.
(b) In a potentiometer, a standard source of emf 5V and negligible internal resistance maintain a steady current through the potentiometer wire of length 10 m . Two primary cells of emf E 1 and E 2 are joined together in a series with (i) same polarity and (ii) opposite polarity. The combination is connected to the potentiometer circuit in each case. The balancing length of the wire in the two cases are found to be 700 cm and 100 cm , respectively.
Find the values of emf of the two cells.
(b) In a potentiometer, a standard source of emf 5
Find the values of emf of the two cells.
Solution:
(a) Working principle of potentiometer: The Potentiometer is an electric instrument that used to measure the potential difference of a given cell and the internal resistance of a cell. It is also used to compare EMFs of different cells.
The potentiometer consists of a long resistive wireL having uniform cross-section and a battery of known emf V . This voltage is called as driver cell voltage. Two ends of the resistive wire L is connected to the battery terminals. This is a primary circuit arrangement. One terminal of another cell (whose emf E is to be measured) is at one end of the primary circuit and another end of the cell terminal is connected to any point on the resistive wire using a jockey J and a galvanometer G . This arrangement is a secondary circuit.
The basic working principle of this circuit is based on the fact that the potential drop across any portion of the wire is directly proportional to the length of the wire.
If the current through the potentiometer is I andR is the total resistance of the potentiometer, then V = IR .
The resistivityρ and area of cross-section A are constant. Also, the Current I is kept constant using the rheostat.
R ∝
V = ρ
= K L (where K = is a constant)
Now, the jockey is so adjusted that for lengthX of the wire there is no current flow through the galvanometer.
So,E = K X
K is known, X is measurable by a scale. So, E can be evaluated from the equation.
When EMF of two cells are to be considered, then in the above similar way
E 1 = KX 1
E 2 = KX 2
∴
=
(b) Potential gradient= k =
= 0.5 V ∕ m
WhenE 1 and E 2 are joined with same polarity, then
E 1 − E 2 = k ×
= 3.5 V .......(1)
WhenE 1 and E 2 are joined with opposite polarity, then
E 1 + E 2 = k ×
= 7 V ......(2)
Adding eqns (1) and (2)
2 E 1 = 10.5 V
∴ E 1 = 5.25 V
Subtracting eqn (2) from eqn (1)
2 E 2 = 3.5 V
∴ E 2 = 1.75 V
The potentiometer consists of a long resistive wire
The basic working principle of this circuit is based on the fact that the potential drop across any portion of the wire is directly proportional to the length of the wire.
If the current through the potentiometer is I and
The resistivity
Now, the jockey is so adjusted that for length
So,
When EMF of two cells are to be considered, then in the above similar way
(b) Potential gradient
When
When
Adding eqns (1) and (2)
Subtracting eqn (2) from eqn (1)
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