Question

A potonometer wire has length 4 m and resistance 8Ω the resistance that must be connected in series with the wire and an accumulator of EMF 2v so as to get a potential gradient m v per CM on the wire is?​

Answer 1

=> R = 32 Ω

Explanation:

Given, l = 4 m

R = potentiometer wire resistance = 8 Ω

potential gradient = dV/ dr = 1 mV/cm

so, for 400cm,

∆V = 400 × 1 × 10^-3 = 0.4 V

letter resistor Rs connected in series,

so as ∆V = ( V/ R + Rs ) × R

=> 0.4 = [ 2/ ( 8+R ) ] × 8

=> 8+R = 16/0.4 = 40

=> R = 32 Ω

∴To get a potential gradient 1 mV per cm on the wire is 32 Ω !

Answer 2

Answer:

♣️To get a potential gradient 1mV per cm on the wire is 32Ω♣️

Hope it will be helpful ☺️ thank you!!!