A potentiometer wire of length 100 cm has a resistance of 30 ohms. It is connected in series with a resistance of 20 ohms and an accumulator of emf 10 V having negligible internal resistance. A source of 2.4 V is balanced against a length L of the potentiometer wire. What is the value of L?

Concept:

• A potentiometer is an electronic device that measures the EMF (electromotive force) of a cell as well as the cell’s internal resistance
• The null point of the potentiometer is the point on the slide wire when the galvanometer shows zero deflection.

Working principle of potentiometer

• The potential difference across the length of the potentiometer wire is directly proportional to its length
•  when a steady current is passed through a uniform wire, the potential drop per unit length or potential gradient is constant.
  • ϵ ∝ L
  • ε = ϕL --- (1)
  • ϕ = potential gradient, ϵ = emf of cell / voltage across the wire.
• \(I = \frac VR\)
• For equivalent resistance in series combinations
  • R_eq = R₁ + R₂ + - - -
• Where, I = current, V = voltage, R = Resistance

Calculation:

Given, V = 10 V, resistance in potentiometer R₁ = 30 Ω, R₂ = 20 Ω (connected in series)

Resistance connected in series 

So, R = R₁ + R₂ = 30 + 20 = 50 Ω 

The current passing through the potentiometer wire 

\(I = \frac VR = \frac {10}{50} \) = 0.2 A

The potential difference across the potentiometer wire (R = 30Ω) 

V = IR = 0.2 × 30 = 6 V

Length of wire, l = 100 cm

From equation 1

\(ϕ = \frac Vl = \frac 6{100}\) = 0.06

The emf 2.4 V is balanced against the length L of the wire

Then, 2.4 = ϕL

\(⇒ L = \frac {2.4}{\phi } = \frac{2.4}{0.06}\) = 40 cm