Current Electricity Charging Discharging Of Capacitors Question 620
Question: Two resistances $ R _{1} $ and $ R _{2} $ are joined as shown in the figure to two batteries of e.m.f. $ E _{1} $ and $ E _{2} $ . If $ E _{2} $ is short-circuited, the current through $ R _{1} $ is
[NDA 1995]
Options:
A) $ E _{1}/R _{1} $
B) $ E _{2}/R _{1} $
C) $ E _{2}/R _{2} $
D) $ E _{1}/(R _{2}+R _{1}) $
Show Answer
Answer:
Correct Answer: A
Solution:
After short circuiting, $ R _{2} $ becomes meaningless.
BETA
