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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:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a common emitter (CE) amplifier having a voltage gain G, the transistor
- The temperature coefficient of resistance of an alloy used for making resistor is
- Four identical cells of emf Ԑ and internal resistance r are to be connected in series.
- Time period of oscillation of a magnet is 2 sec. When it is remagnetised
- Find the maximum velocity with which a train can be moved on a circular track
Topics: Current Electricity (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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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