# Q) In the circuit diagram given below, AB is a uniform wire of resistance 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.

Good Day,

First, lets note the important Points.
A. Potentiometer
Resistance of the Potentiometer wire,RAB = 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*10-3V
Balancing point, L1 = 30 cm=30*10-2m

Now,
Step 1 -
E = K * L1
75*10-3 = K* 30*10-2
K= 0.25 V/m

Now, Current =  2/ R + RAB
= 2/ R+15

Now, VAB = RAB * Current
= 15* 2/R+15
=  30/R+15

Now, k = VAB/LAB

0.25 = (30/R+15)/1

Upon solving we get the R = 105 Ohm

• 48
Thank you:-)
• -2
Yeah
• -2
Principle used is v=kl,and answer is 105 omega

• 2
What are you looking for?