Current Electricity Charging Discharging Of Capacitors Question 596
Question: A new flashlight cell of e.m.f. 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of resistance $ 0.04,\Omega $ . The internal resistance of cell is
[MP PET 1994]
Options:
A) $ 0.04,\Omega $
B) $ 0.06,\Omega $
C) $ 0.10,\Omega $
D) $ 10,\Omega $
Show Answer
Answer:
Correct Answer: B
Solution:
Let the internal resistance of cell be r, then $ i=\frac{E}{R+r} $
Therefore $ 15=\frac{1.5}{0.04+r} $
Therefore $ r=0.06\Omega $
BETA
