Question

Length of potentiometer wire is L. In primary circuit battery of 2.5 V and 10 ft resistance is combined in series. Balance length is \(\frac{L}{2}\) for emf 1.0 Ω. If the resistance of primary circuit becomes double then what is the new balance length?

Answer 1

Length of wire = L Potential (V) = 2.5 (V)
Let a resistance of R₁ ohm is placed in series with battery, then current

When we double the resistance R₁, connected in series with potentiometer then total resistance (R’) = R + 2R₁
R’ = 10 + 5 = 15 Ω
In this condition if current in potentiometer is i, then