A potentiometer wire of length 100cm has a resistance of 10Ω. It is connected in series with a resistance and a cell of emf 2V having negligible internal reistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire.The value of external resistance is
Options
(a) 760Ω
(b) 640Ω
(c) 790Ω
(d) 840Ω
Correct Answer:
790Ω
Explanation:
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Let the resistance be r and current be I .According to loop rule :
E=V+Ir
2=I(10+r)
And also
I*(40/100)=0.01V
Here you get I=2.5*10^-3ampere
Put this Vue in eqn 2=I(10+r) and get solution