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  • The driver cell of a potentiometer has an emf of 2V and negligible internal resistance. The potentiometer wire has a resistance of 5 and is 1m

The driver cell of a potentiometer has an emf of 2V and negligible internal resistance. The potentiometer wire has a resistance of 5\Omega and is 1m long. The resistance which must be connected in series with the wire so as to have a potential difference of 5mV across the whole wire is

Option: 1

1985 \Omega


Option: 2

1990 \Omega


Option: 3

1995 \Omega


Option: 4

2000 \Omega


Answers (1)

best_answer

In order to have a potential drop of 5 mV = 5 \times 10^{-3} \mathrm{~V} across a wire of resistance 5 \Omega,, the current flowing in the wire should be 
  \mathrm{I}=\frac{5 \times 10^{-3}}{5}=1 \times 10^{-3} \mathrm{~A}
If R is the resistance to be connected in series with the wire, then 
  \frac{2}{R+5}=1 \times 10^{-3}
which gives R = 1995 \Omega

\therefore (C)

Posted by

Ajit Kumar Dubey

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