Current Electricity Charging Discharging Of Capacitors Question 649
Question: Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are $ R _{1} $ and $ R _{2},(R _{2}>R _{1}) $ . If the potential difference across the source having internal resistance $ R _{2} $ is zero, then
[AIEEE 2005]
Options:
A) $ R=R _{1}R _{2}/(R _{1}+R _{2}) $
B) $ R=R _{1}R _{2}/(R _{2}-R _{1}) $
C) $ R=R _{2}\times (R _{1}+R _{2})/(R _{2}-R _{1}) $
D) $ R=R _{2}-R _{1} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ i=\frac{2E}{R+R _{1}+R _{2}} $ From cell (2) $ E=V+iR _{2}=0+iR _{2} $
Therefore $ E=\frac{2E}{R+R _{1}+R _{2}}\times R _{2} $
Therefore $ R=R _{2}-R _{1} $
BETA
