Current Electricity Charging Discharging Of Capacitors Question 607
Question: The electromotive force of a primary cell is 2 volts. When it is short-circuited it gives a current of 4 amperes. Its internal resistance in ohms is
[MP PET 1995]
Options:
A) 0.5
B) 5.0
C) 2.0
D) 8.0
Show Answer
Answer:
Correct Answer: A
Solution:
$ E=V+ir $ After short-circuiting, $ V=0; $
Therefore $ r=\frac{E}{i}=\frac{2}{4}=0.5\Omega $
BETA
