A potentiometer wire has a length 10m and a resistance `20Omega`. Its terminals are connected to a battery of e.m.f. 4V and internal resistance `5Omega`. What are the distances at which null points are obtained when two cells of e.m.f.s 1.5V and 1.3 V are connected so as to a ) assist and b) oppose each other ?

Options

A. 4.75 m, 0.625 m
B. 8.75 m, 6.25 m
C. 8.75 m, 0.625 m
D. 3.75 m, 2.625 m

Answer

C

Explanation

`L=10m , R=20 Omega, E=4 V, r=5 Omega`,
`E_(1)=1.5V, E_(2)=1.3V`,
a) `l_(1)=?` For sum
b) `l_(2)=?` For difference.
`Irho ((E)/(R+r))(R)/(L)=((4)/(20+5))(20)/(10)`
`=0.32 V//m`
a)`E_(1)+E_(2)=I rho l_(1)`
`:.l_(1)=(E_(1)+E_(2))/(I rho )=(1.5+1.3)/(0.32)=(2.8)/(0.32)`
`l_(1)=8.75m`.
b)`E_(1)-E_(2)=I rho l_(2)`
`:.l_(2)=(E_(1)-E_(2))/(I rho )=(1.5-1.3)/(0.32)=(0.2)/(0.32)`
`l_(2)=0.625`m.