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
1
Expert's answer
2019-08-05T06:36:19-0400

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}\n\\implies 10+R=800\n\\implies R=790\\Omega"


It is the right answer.


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