The potential gradient of a potentiometer is 3 mV/cm. It is used to measure the potential difference across a resistance of 20 Ω. If a length of 60 cm of the potentiometer wire is required to get the null point, then the current passing through the 20 Ω resistor is:

CONCEPT:

Potentiometer:

• A potentiometer is a device that is mainly used to measure the emf of a given cell and to compare the EMFs of cells.
  • It is also used to measure the internal resistance of a given cell.
• Potentiometer consists of a long resistive wire AB of length L (about 6m to 10 m long) made up of manganin or constantan and a battery of known voltage e and internal resistance r called supplier battery or driver cell. Connection of these two forms a 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.

Potential gradient (x):

• The potential difference (or fall in potential) per unit length of wire is called a potential gradient.
• It is given as,

\(⇒ x = \frac{e}{R+R_h+r}.\frac{R}{L}\)

Where e = emf of the primary cell, R = resistance of potentiometer wire, R_h = resistance of rheostat, and r = internal resistance of the primary cell

CALCULATION:

Given x = 3 mV/cm = 0.3 V/m, R = 20 Ω, and l = 60 cm = 0.6 m

Let the current in the 20 Ω resistance be I.

• So potential difference across 20 Ω resistance is given as,

⇒ V = IR     -----(1)

• If the potential gradient is x and the null point comes at length l, then,

⇒ V = xl     -----(2)

By equation 1 and equation 2,

\(⇒ I=\frac{xl}{R}\)

\(⇒ I=\frac{0.3×0.6}{20}\)

⇒ I = 9 × 10^-3 A

⇒ I = 9 mA

• Hence, option 2 is correct.