Question 482: Current Electricity

Question

A 2.5 V battery is connected to a potentiometer wire. A cell of e.m.f. 1.08 V is balanced by the voltage drop across 2.16 m of wire. The length of the potentiometer wire is

Accepted Answer

Given:
Battery across potentiometer wire: $ V = 2.5 \, 	ext{V} $
A cell of emf $ E = 1.08 \, 	ext{V} $ balances at length $ l = 2.16 \, 	ext{m} $
Need: total length $L$ of potentiometer wire.
Step 1: Potential gradient
The potential drop per unit length is constant:
$ k = \frac{V}{L} $
Step 2: Balance condition
For balance,
$ E = k \cdot l = \frac{V}{L} \cdot l $
$ E = \frac{V \cdot l}{L} $
Step 3: Solve for $L$
$ L = \frac{V \cdot l}{E} $
$ L = \frac{2.5 \cdot 2.16}{1.08} $
$ L = \frac{5.4}{1.08} = 5.0 \, 	ext{m} $
Ã¢ÂÂ Final Answer:
The length of the potentiometer wire is:
Option C: 5 m

Answer

2.5 m

Answer

3 m

Answer

5 m

Answer

6 m

Current Electricity

A 2.5 V battery is connected to a potentiometer wire. A cell of e.m.f. 1.08 V is balanced by the voltage drop across 2.16 m of wire. The length of the potentiometer wire is

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