Current Electricity Charging Discharging Of Capacitors Question 615
Question: When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 A. When the resistance is increased to 5 ohm, the current is 0.25 A. The e.m.f. of the cell is
[MP PET 1999, 2000; Pb. PMT 2002; MP PMT 2000]
Options:
A) 1.0 V
B) 1.5 V
C) 2.0 V
D) 2.5 V
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ i,=( \frac{E}{R+r} ), $ we get $ 0.5=\frac{E}{2+r} $ ……(i) $ 0.25=\frac{E}{5+r} $ …..(ii) Dividing (i) by (ii), we get $ 2=\frac{5+r}{2+r} $
Therefore $ r=1\Omega $ \ $ 0.5=\frac{E}{2+1} $
Therefore $ E=1.5V $
BETA
