In a potentiometer arrangement, a cell of emf 1.5 V gives a balance point at 45.0 cm length of the wire. If the cell is replaced by another cell of emf 2.25 V, where will the balance point shift to?
67.5 cm
A potentiometer is a versatile electrical instrument used to measure the electromotive force (emf) of a cell or potential difference across a component without drawing any current from the circuit being measured. Its operation is based on the principle that the potential drop across any portion of a wire of uniform cross-sectional area and uniform composition is directly proportional to the length of that portion, provided a constant current flows through the wire.
In a typical potentiometer setup, a primary circuit supplies a steady current through a long wire of uniform resistance per unit length. A secondary circuit contains the cell whose emf is to be measured, connected in series with a galvanometer and a jockey. The negative terminal of the cell is connected to the low-potential end of the potentiometer wire, and the positive terminal is connected to the galvanometer and then to the jockey.
The jockey is moved along the potentiometer wire until the galvanometer shows zero deflection. This point is called the balance point or null point. At the balance point, the potential difference across the length of the potentiometer wire from the high-potential end (connected to the positive terminal of the cell) to the balance point is exactly equal to the emf of the cell. Since no current flows through the secondary circuit at the balance point, the internal resistance of the cell does not affect the measurement.
According to the principle of the potentiometer, the potential drop (\(V\)) across a length (\(l\)) of the wire is proportional to the length, assuming uniform wire and constant current. If the potential gradient along the wire is \(k\) (potential drop per unit length), then \(V = k l\). When measuring the emf (\(E\)) of a cell using a potentiometer, the balance point is found at a length \(l\) such that the potential drop across this length is equal to the cell's emf. Therefore, at the balance point:
\(E = k l\)
Since the potential gradient \(k\) is constant for a given potentiometer setup and primary circuit current, the emf of a cell is directly proportional to its balance length:
\(E \propto l\)
We are given two different cells and their corresponding balance points (or asked to find one). Let \(E_1\) and \(l_1\) be the emf and balance length for the first cell, and \(E_2\) and \(l_2\) be the emf and balance length for the second cell. Using the proportionality \(E \propto l\), we can write:
\(\frac{E_1}{E_2} = \frac{l_1}{l_2}\)
We are given:
Substitute the given values into the equation:
\(\frac{1.5 \text{ V}}{2.25 \text{ V}} = \frac{45.0 \text{ cm}}{l_2}\)
Now, we solve for \(l_2\):
\(l_2 = \frac{2.25 \text{ V}}{1.5 \text{ V}} \times 45.0 \text{ cm}\)
\(l_2 = \left(\frac{2.25}{1.5}\right) \times 45.0 \text{ cm}\)
\(l_2 = (1.5) \times 45.0 \text{ cm}\)
\(l_2 = 67.5 \text{ cm}\)
Thus, if the cell is replaced by another cell of emf 2.25 V, the balance point will shift to 67.5 cm.
| Parameter | Cell 1 | Cell 2 |
|---|---|---|
| Emf (E) | 1.5 V | 2.25 V |
| Balance Length (l) | 45.0 cm | \(l_2\) |
Using the proportionality \(E_1/E_2 = l_1/l_2\):
\(\frac{1.5}{2.25} = \frac{45.0}{l_2}\)
\(l_2 = \frac{2.25 \times 45.0}{1.5}\)
\(l_2 = 1.5 \times 45.0\)
\(l_2 = 67.5\) cm
The new balance point is at 67.5 cm.
| Concept | Description | Formula/Principle |
|---|---|---|
| Potentiometer Principle | Potential drop across a uniform wire is proportional to its length, with constant current. | \(V \propto l\) |
| Balance Point | Point on the wire where potential drop equals the cell's emf, resulting in zero galvanometer deflection. | \(E = \text{potential drop across balance length}\) |
| Emf Measurement | At balance point, \(E = k l\), where \(k\) is potential gradient. | \(E \propto l\) (for constant \(k\)) |
| Comparing Emfs | Comparing emfs of two cells using balance lengths. | \(\frac{E_1}{E_2} = \frac{l_1}{l_2}\) |
It is useful to compare a potentiometer with a voltmeter for measuring emf:
Factors affecting the sensitivity of a potentiometer:
Figure shows drift speed Vd of conduction electrons in a copper wire versus position (X) for the three sections. Then,

A. Radius of III > Radius of II > Radius of I
B. Electric Field in III > Electric Field in II > Electric Field in I
C. Radius of wire is same in all sections
D. Conductivity is same in all sections
Choose the correct answer from the options given below:
Which of the following circuits cannot be used to measure the resistance of resistor R?
The temperature at which the resistance of a conductor becomes 30% more than that of its resistance at 47°C will be:
(Given the value of the temperature coefficient of resistance of the conductor is 2 × 10-4 K-1.)
Cell having an emf E and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by:
In the potentiometer circuit, the balance point is at X. The balance point will be shifted right towards B when:
A. Resistance R is increased keeping all other parameters constant
B. Resistance S is increased keeping all other parameters constant
C. Cell P is replaced by another cell whose emf is lower than Q
D. The polarity of Q is reversed
Choose the correct answer from the options given below: