Current Electricity Charging Discharging Of Capacitors Question 612
Question: A current of two amperes is flowing through a cell of e.m.f. 5 volts and internal resistance 0.5 ohm from negative to positive electrode. If the potential of negative electrode is 10V, the potential of positive electrode will be
[MP PMT 1997]
Options:
A) 5 V
B) 14 V
C) 15 V
D) 16 V
Show Answer
Answer:
Correct Answer: B
Solution:
$ V _{2}-V _{1}=E-ir=5-2\times 0.5=4volt $
Therefore $ V _{2}=4+V _{1}=4+10=14,volt $
BETA
