Current Electricity Charging Discharging Of Capacitors Question 300
Question: The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the e.m.f of a battery whose internal resistance is 0.5 W. If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is
[AIEEE 2003]
Options:
A) $ \frac{30E}{100} $
B) $ \frac{30E}{100.5} $
C) $ \frac{30E}{(100-0.5)} $
D) $ \frac{30(E-0.5i)}{100} $ , where i is the current in the potentiometer
Show Answer
Answer:
Correct Answer: A
Solution:
From the principle of potentiometer $ V\propto l $
Therefore $ \frac{V}{E}=\frac{l}{L} $ ; where V = emf of battery, E = emf of standard cell, L = Length of potentiometer wire $ V=\frac{El}{L}=\frac{30E}{100} $ .
BETA
