# Answer to Question #92218 in Electrical Engineering for Anurag

Question #92218
A potentiometer wire of length 100cm has a resistance of 10Ω. It is connected in series to to a resistance R and a cell of emf 2V and negligible internal resistance. A source of emf of 10mV is balanced by a length of 40cm of the potentiometer wire. What is the value of the resistance R? A.526.67 ohm B.70 ohm C.1580 ohm D.zero ohm

Length of potentiometer wire=100cm

Total resistance of wire(r)=10@$\Omega@$

Resistance of 40 cm wire(r1)=4Ω

EMF of driver cell(E)=@$2V@$

External resistance of circuit(R)=R

Current(I) in the circuit is given by @$\frac{E}{r+R}@$ =@$\frac{2}{10+R}@$ A

Voltage drop across 40 cm potentiometer wire=@$Ir_1@$ =

@$\frac{2}{10+R}\times4@$ =@$\frac{8}{R+10}V@$

This voltage drop is equal to the balancing EMF of @$10mv@$ or @$10\times10^{-3}@$ or @$10^{-2}V@$

so,

@$10^{-2}=\frac{8}{10+R} \implies 10+R=800 \implies R=790\Omega@$

It is the right answer.

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