Question

A potentiometer wire of length 100cm has a resistance of 10 ohm. It is connected in series with a resistance & a cell of emf 2v & of negligible internal resistance. A source of emf 10mv is balanced against a length of 40 cm of the Potentiometer wire . What is the value of external resistance ?

Answer 1

Answer:

As the source of e.m.f. E

′

=10mV=10×10

−3

V is balanced by a length of 40cm of the potentiometer wire, it follows that 10×10

−3

=J resistance of 40cm of the potentiometer wire.

If I is current through the potentiometer wire then

J=

R+10

E

=

R+10

2

Now resistance of 40cm of the potentiometer wire =

100

10

×40=4Ω

10×10

−3

=

R+10

2

×4

⇒R=790Ω

Hence, resistance is 790Ω

solution