Question

a potentiometer wire of length 100 cm having resistance of 10 ohm is connected in series witha. resistance r and cell of emf 2v of negligible internal resistance a source emf of 10mv is balanced against a length of 40cm of potentiuomeneter what is value fo external resistanc e

Answer 1

given, resistance of potentiometer wire, R = 10 ohm

length of potentiometer wire , l = 100cm

emf of cell, connected in series with potentiometer wire, E = 2V.

balancing length is obtained at 40cm.

so, I = E/(R + r)

where I is the current through potentiometer .

I = 2/(10 + r) ....(1)

now resistance of 40cm wire = 40/100 × 10 = 4 ohms.

given, emf of the cell is balanced by a length of wire, E' = 10mV = 10^-2 volts

using Ohm's law,

V = IR

10^-2 = I × 4 ohms

or, I = 2.5 × 10^-3 A

now putting the value of I in equation (1),

2.5 × 10^-3 = 2/(10 + r)

10 + r = 2/(2.5 × 10^-3) = 2000/2.5 = 800

or, r = 790 ohms

hence, answer is 790 ohms.

Answer 2

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