Video Lecture
Theory For Making Notes
Potentiometer
A potentiometer is an instrument which allows the measurement of potential difference without drawing any current from the circuit being measured. Hence, it acts as an infinite-resistance voltmeter.
The resistance between A and B is a uniform wire of length L, with a sliding contact C at a distance x and B. The potential difference V is measured by sliding the contact until the galvanometer G reads zero. The no-deflection condition of the galvanometer ensures that there is no current through the branch containing G and the potential difference to be measured. The length x for no-deflection is called as the balancing length.
If l is the resistance per unit length of the wire AB, under balanced condition (i.e., when no current is indicated by galvanometer G), the pd to be measured is given as
\displaystyle V={{V}_{{CB}}}={{V}_{{AB}}}\frac{{{{R}_{{CB}}}}}{{{{R}_{{AB}}}}}={{V}_{{AB}}}\frac{{\lambda \left( x \right)}}{{\lambda L}}
= \left( {\frac{{{{V}_{{\text{AB}}}}}}{L}} \right)\times x
The ratio VAB/L is called the potential gradient in the wire AB.
Comparing Two Cells using Potentiometer
A potentiometer is commonly used for comparison of emfs of cells. A battery X is connected across a long uniform wire AB. The cells E1 and E2 (whose emfs are to be compared) are connected as shown along with a galvanometer.
First the key S1 is pressed which brings E1 in the circuit. The sliding contact C is moved till the galvanometer shows no deflection. Let the length AC = L1. Next S1 is opened and S2 is closed. This brings E2 in the circuit. Let C’ be the new null point and let AC’ = L2. Then, clearly,
\displaystyle \frac{{{{E}_{1}}}}{{{{E}_{2}}}}=\frac{{{{L}_{1}}}}{{{{L}_{2}}}}
Measurement of Internal Resistance of a Cell
A potentiometer can be used to find the internal resistance of a cell. First the emfE of the cell is balanced against a length AC = L. A known resistance R is then connected across the cell as shown. The terminal voltage V is now balanced against a smaller length AC’ = L’.
Then, \displaystyle \frac{E}{V}=\frac{L}{{{L}’}}
But we know that \displaystyle \frac{E}{V}=\frac{{R+r}}{R}
\displaystyle \frac{{R+r}}{R}=\frac{L}{{{L}’}}
\displaystyle r=\left( {\frac{L}{{{L}’}}-1} \right)R
Illustration
A battery of emf 4 V is connected across a 10 m long potentiometer wire having a resistance per unit length 1.6 W m–1. A cell of emf 2.4 V is connected so that its negative terminal is connected to the low potential end of the potentiometer wire and the other end is connected through a galvanometer to a sliding contact along the wire. It is found that the no-deflection point occurs against the balancing length of 8 m. Calculate the internal resistance of the 4 V battery.
Solution
emf of cell = (potential gradient) X (balancing length)
\displaystyle {{E}_{1}}=\frac{{{{V}_{{AB}}}}}{L}\times x
or \displaystyle 2.4=\frac{{{{V}_{{AB}}}}}{{10}}\times 8
⇒ VAB = 3 V
Consider the loop containing E. Applying potential divider concept,
\displaystyle {{V}_{{AB}}}=E\frac{{{{R}_{{AB}}}}}{{{{R}_{{AB}}}+r}}
⇒ \displaystyle 3=4\frac{{1.6\times 10}}{{1.6\times 10+r}}
⇒ r = 16/3 Ω
Note that as there is no current through the cell and galvanometer, the battery E, the internal resistance r and the potentiometer wire AB are in series.
Practice Questions (Level-1)
Q.1
In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of 2 m, when the cell is shunted by a 5W resistance and at a length of 3 m, when the cell is shunted by a 10 W resistance. The internal resistance of the cell is
(a) 1.5 W (b) 10 W (c) 15 W (d) 1 W
Ans : (b)
Q.2
A resistance of 4W and a wire of length 5 metres and resistance 5W are joined in series and connected to a cell of e.m.f. 10V and internal resistance 1W. A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. of each cell is
(a) 1.5 V
(b) 3.0 V
(c) 0.67 V
(d) 1.33 V
Ans. (b)
Practice Questions (Level-2)
Q.1
If the length of the potentiometer wire is decreased, then its sensitivity :
(a) Increases
(b) Decreases
(c) Remains unaffected
(d) Either (b) or (c)
Ans: (b)
Q.2
In an experiment to measure the internal resistance of a cell by a potentiometer it is found that all the balance points at a length of 2 mtr. When the cell is shunted by a 5 ohm resistance, and is at a length of 3 m when the cell is shunted by a 10 ohm resistance, the internal resistance of the cell is then :
(a) 1.5 \displaystyle \Omega
(b) 10 \displaystyle \Omega
(c) 15 \displaystyle \Omega
(d) 1 \displaystyle \Omega
Ans: (b)
Q.3
A 10 m long wire of resistance 20 \displaystyle \Omega is connected in series with a battery of e.m.f. 3 V (negligible internal resistance) and a resistance of 10 \displaystyle \Omega . The potential gradient along the wire in volt per metre is :
(a) 0.02 (b) 0.1 (c) 0.2 (d) 1.2
Ans: (c)
Q.4
In a potentiometer experiment the balancing point with a cell is at length 240 cm. On shunting the cell with a resistance of 2 \displaystyle \Omega the balancing length becomes 120 cm. The internal resistance of the cell is :
(a) 1 \displaystyle \Omega
(b) 0.5 \displaystyle \Omega
(c) 4 \displaystyle \Omega
(d) 2 \displaystyle \Omega
Ans: (d)
Q.5
In the above question if the jockey touches the wire at a distance of 560 cm from A, what will be the current in the galvanometer ?
(a) \displaystyle \frac{{3E}}{{22r}}
(b) \displaystyle \frac{{13E}}{{22r}}
(c) \displaystyle \frac{{5E}}{{22r}}
(d) \displaystyle \frac{{7E}}{{24r}}
Ans : (a)
Q. 6
Consider the potentiometer circuit arranged as in fig. The potentiometer wire is 600 cm long, and its resistance is 15 r.
At what distance from point A should the jockey touch the wire to get zero deflection in the galvanometer ?
(a) 400cm
(b) 240 cm
(c) 320 cm
(d) 560cm
Ans. (c)