Question

The potentiometer wire of length 100 cm has a resistance of 10 ohm it is connected in series with the resistance 5 ohm and an accumulator of emf 3 volt having negligible resistance of source of 1.2 world is balanced and intel and their of the potentiometer wire find the value of

Answer 1

Here, let current through potentiometer wire = J

Applying Ohm's Law

=>  J = E/(R + r)

where, R = resistance in series with cell =  5 ohm

            r = resistance of potentiometer wire = 10 ohm

            E  =  emf of accumulator  =  3 V

=>  J =  3/(10 + 5)

       =  0.2  A

As, the source of emf of 1.2 V is balanced by the length of potentiometer wire of length L .

Hence, 1.2 V = J * resistance of L cm of potentiometer wire

=> 1.2 V = 0.2 * resistance of L cm of potentiometer wire

=>  resistance of L cm of potentiometer wire = 1.2/0.2

                                                                    =  6 ohm

As,  10 ohm resistance potentiometer wire has length = 100 cm

=> 6 ohm resistance potentiometer wire has length,L = 100 * 6/10

                                                                         =  60  cm

Thus, value of length L of potentiometer wire = 60 cm

Answer 2

Answer:790

Explanation:

Resistance of 40cm length of potentiometer wire

=10/100×40=4ohm

Current

I=10mV/4=10×10^-3/4=2.5×10^-3

R=(R+10)ohm

2.5×10^-3A =2V/R+10

R+10=2/2.5×10^-3=800

R+10=800,R=800-10=790ohm