Question
A potentiometer wire of length 1 m and resistance 10Ω is connected in series with a cell of e.m.f 2 volt with internal resistance 1Ω and a resistance box including a resistance R. If the potential difference between the ends of the wire is 1 mV, then the value of R will be
Answer: Option C
:
C
i=V2R1=1×10−310=10−4AE=i[R+r+RL]2=10−4[R+1+10]20000=R+11R=20000−11=19989Ω
Was this answer helpful ?
:
C
i=V2R1=1×10−310=10−4AE=i[R+r+RL]2=10−4[R+1+10]20000=R+11R=20000−11=19989Ω
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution