Question

a resistance of the Potentiometer wire is 8 ohm and its length is 8 metre a resistance box and a 2 volt battery are connected in series with it what should be the resistance in the box if it is desire to have the potential difference of a 1 volt upon mm

Answer 1

Answer:

In the picture it's given 1uV/mm. However, you've asked for 1 V/mm. I'll do both.

E = IRwire/lwire

=> E = I (8/8) = I = 2/(8+R)

If E = 1uV/mm, then

10^-6/10^-3 = 2/(8+R)

=> R = 1992 ohm = 1.992 mohm

If E = 1 V/mm, then

1/10^-3 = 2/(8+R)

=> R = -7.998 ohm which is impossible. So, E can't be 1 V/mm. If you observe carefully, for a feasible R, E needs to be lesser 2.5 mV/cm i.e 0.25 V/m i.e 0.00025 V/mm.

I hope the matter is clear.

Answer 2

Answer:

     E = IRwire/lwire

=> E = I (8/8) = I = 2/(8+R)

If E = 1μV/mm, then

    10^-6/10^-3 = 2/(8+R)

=>                  R = 1992 Ω

                        = 1.992 mΩ

Explanation:

Answer is R = 1992 Ω