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  • A potentiometer wire of length 100 cm has a total resistance of 10. It is connected in series with a resistance R and a cell of emf 2 volts and of n

A potentiometer wire of length 100 cm has a total resistance of 10\Omega. It is connected in series with a resistance R and a cell of emf 2 volts and of negligible internal resistance. A cell of emf 10 mV is balanced against a length of 40cm of potentiometer wire. The value of the external resistance R is:  

Option: 1

790 \Omega.


Option: 2

800 \Omega.


Option: 3

900 \Omega


Option: 4

890 \Omega


Answers (1)

best_answer

As shown in the figure, if R is the unknown resistance, the current in the circuit I=\frac{V}{(r+R)}=\frac{2}{(10+R)}
Now as the 100 cm wire has a resistance of 10 \Omega, the resistance of 40 cm of wire will be
40 \times(10 / 100)=4 ohm. Potential drop across 40 cm wire will be V = I \times 4
but here V = 10 mv (given)
Hence, 10 \times 10^{-3}=\frac{2}{(10+R)} \times 4$ i.e. $R=790 \Omega

Posted by

Suraj Bhandari

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