A potentiometer wire has length `4 m` and resistance `8 Omega`. The resistance that must be connected in series with the wire and an accumulator of e.m.f. `2 V`, so as the get a potential gradient `1 mV` per cm` on the wire is

Options

A. `40 Omega`
B. `44 Omega`
C. `48 Omega`
D. `32 Omega`

Answer

D

Explanation

We need a potential gradient of `1 xx 10^(-3)V//cm`, so the total potential difference across the length of potentiometer wire of 400cm will be 0.4V. Current needed to generate a potential difference of 0.4V across the potentiometer wire of resistance `8 Omega` is I= 0.4/8= 0.05A. Hence, the potentiometer accumulator needs to drive a current of 0.05A in the circuit. Let R be the resistance needed to be connected in series then current can be written as:
`i= (2)/(8+R) = 0.05 rArr R + 8 = 40 rArr R = 32 Omega` Thus, option (d) is correct