Q) In the circuit diagram given below, AB is a uniform wire of resistance 15   Ω 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,
The answer is as follows:

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


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

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