Current Electricity Charging Discharging Of Capacitors Question 591
Question: When a resistance of 2ohm is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell is
[MP PMT 1996]
Options:
A) $ 0.5,ohm $
B) $ 1.0,ohm $
C) $ 1.5,ohm $
D) $ 2.0,ohm $
Show Answer
Answer:
Correct Answer: D
Solution:
Let the e.m.f. of cell be E and internal resistance be r. Then $ 0.5=\frac{E}{(r+2)} $ and $ 0.25=\frac{E}{(r+5)} $ On dividing, $ 2=\frac{5+r}{2+r} $
Therefore $ r=1\Omega $
BETA
