Current Electricity Charging Discharging Of Capacitors Question 635
Question: Two batteries A and B each of e.m.f. 2 V are connected in series to an external resistance R = 1 ohm. If the internal resistance of battery A is 1.9 ohms and that of B is 0.9 ohm, what is the potential difference between the terminals of battery A
[MP PET 2001]
Options:
A) 2 V
B) 3.8 V
C) Zero
D) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
$ i=\frac{2+2}{1+1.9+0.9}=\frac{4}{3.8}A $ For cell A $ E=V+ir $
Therefore $ V=2-\frac{4}{3.8}\times 1.9=0 $ .
BETA
