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Q) In the circuit diagram given below, AB is a uniform wire of resistance $15\Omega $ and length 1 m. it is connected to a cell ${E}_{1}$ of emf 2V and negligible internal resistance and a resistance R. the balance point with another cell ${E}_{2}$ of emf 75 mV is found at 30 cm from end A. Calculate the value of R.

The answer is as follows:

First, lets note the important Points.

A. Potentiometer

Resistance of the Potentiometer wire,R

_{AB}= 15 Ohm

Length of the potentiometer wire = 1 m

B. Driver Cell

Emf of the cell = 2 V

C. The cell

Emf of this cell = 75 mV=75*10

^{-3}V

Balancing point, L

_{1 = }30 cm=30*10

^{-2}m

Now,

Step 1 -

E = K * L

_{1}

75*10

^{-3}= K* 30*10

^{-2}

K= 0.25 V/m

Now, Current = 2/ R + R

_{AB}

= 2/ R+15

Now, V

_{AB}= R

_{AB}* Current

= 15* 2/R+15

= 30/R+15

Now, k = V

_{AB}/L

_{AB }0.25 = (30/R+15)/1

Upon solving we get the R = 105 Ohm