Class 12 Physics
CBSE and UP Board
Current Electricity
Additional Exercise
Chapter-3 Exercise 3.23
NCERT Solutions Class 12 Physics Chapter 3 Question 23
Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?
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Resistance of the standard resistor, R = 10.0 Ω
Balance point for this resistance, l₁ = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E₁ = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l₂ = 68.5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
E₁/E₂ =l₁/l₂
=> iR/iX =l₁/l₂
X= (l₁/l₂ ) x R
= 68.5/58.3 x 10 = 11.749 Ω
Therefore, the value of the unknown resistance, X, is 11.75Ω.
If we fail to find a balance point with the given cell of emf, ε , then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.