A potentiometer wire of length 1 m has a resistance of `100Omega`. It is connected to a 6V battery in series with a resistance of `5Omega`. Determine the emf of the primary cell which gives a balance point at 40cm.

Options

A. 1.2 V
B. 1.8 V
C. 1.6 V
D. 1.9 V

Answer

C

Explanation

Given, length of wire, l = 1m = 100 cm
Resistance, `R = 10 Omega`
The emf of a battery, `E_(1)=6V`
`R_(1)=5 Omega`
`X = 40 cm`
`therefore " "` Current, `I=(E_(1))/(R+R_(1))=(6)/(10+5)=(6)/(15)=(22)/(5)A`
`V_(AB)=IR=(2)/(5)xx10=4 V`
`therefore` The emf of the primary cell `= (V_(AB))/(l)xx x=(4)/(100)xx40 = 1.6 V`