Current Electricity Charging Discharging Of Capacitors Question 646
Question: The maximum power drawn out of the cell from a source is given by (where r is internal resistance)
[DCE 2002]
Options:
A) $ E^{2}/2r $
B) $ E^{2}/4r $
C) $ E^{2}/r $
D) $ E^{2}/3r $
Show Answer
Answer:
Correct Answer: B
Solution:
$ i=\frac{E}{r+R} $
Therefore $ P=i^{2},R $
Therefore $ P=\frac{E^{2}R}{{{(r+R)}^{2}}} $ Power is maximum when r = R
Therefore $ {P _{\max }}=E^{2}/4r $
BETA
