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Question:

An ideal battery of 4V and resistance R are connected in series in the primary circuit of a potentiometer of length 1m and resistance 5Ω. The value of R, to given a potential difference of 5mV across 10cm of potentiometer wire is:

480Ω

495Ω

490Ω

395Ω

Solution:

Let current flowing in the wire is i.
∴i=(4/R+5)A
If resistance of 10m length of wire is x then x=0.5Ω=5×0.1Ω
∴ΔV=P.d on wire=i.x
5×10⁻³=(4/R+5)(0.5)
∴4/R+5=10⁻²
∴R=395Ω