Current Electricity Charging Discharging Of Capacitors Question 601
Question: n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is
[DPMT 2002]
Options:
A) $ \frac{nE}{R+nr} $
B) $ \frac{nE}{nR+r} $
C) $ \frac{E}{R+nr} $
D) $ \frac{nE}{R+r} $
Show Answer
Answer:
Correct Answer: A
Solution:
Total e.m.f. = nE, Total resistance R + nr
Therefore $ i=\frac{nE}{R+nr} $
BETA
