For a potenitometer experiment, the emf of a battery in the primary circuit is 20 V and its internal resistance is `5 Omega`. There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from `120 Omega " to " 170 Omega`. Resistance of the potentiometer wire is `75 Omega`. The following potential differences can't be measured using this potentiometer

Options

A. 5 V
B. 6 V
C. 7 V
D. 8 V

Answer

D

Explanation

Maximum external resistance put across the potentiometer battery is `(170+75) Omega`.
As external resistance `=r=5 Omega`
So, current in potentiometer wire
`=I=(E )/((R+r))=(20)/(250)=0.08 A`
Potential drop occurs across wire of potentiometer =IR
`=0.08xx75=6V`
Also, minimum resistance `=120+75 Omega`
` therefore " " l=(20)/(200)=0.1 A`
Potential drop `=lR=0.1xx75=7.5 V`
So, upto 7 V it an be measured with the help of potentiometer.