Current Electricity Charging Discharging Of Capacitors Question 636
Question: When a resistor of 11 W is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead, when a resistor of 5 W is connected to the same electric cell in series, the current increases by 0.4 A. The internal resistance of the cell is
[EAMCET 2001]
Options:
A) 1.5 W
B) 2 W
C) 2.5 W
D) 3.5 W
Show Answer
Answer:
Correct Answer: C
Solution:
By using $ i=\frac{E}{R+r} $
$ \Rightarrow 0.5=\frac{E}{11+r} $
Therefore $ E=5.5+0.5r $ ..?(i) and $ 0.9=\frac{E}{5+r} $
Therefore $ E=4.5+0.9r $ ..?(ii) On solving these equation, we have $ r=2.5\Omega $
BETA
