Electrostatics Question 582
Question: Two condensers $ C _{1} $ and $ C _{2} $ in a circuit are joined . The potential of point $ A $ is $ V _{1} $ and that of $ B $ is $ V _{2} $ . The potential of point $ D $ will be [MP PMT 1997]
Options:
A) $ \frac{1}{2}(V _{1}+V _{2}) $
B) $ \frac{C _{2}V _{1}+C _{1}V _{2}}{C _{1}+C _{2}} $
C) $ \frac{C _{1}V _{1}+C _{2}V _{2}}{C _{1}+C _{2}} $
D) $ \frac{C _{2}V _{1}-C _{1}V _{2}}{C _{1}+C _{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Charge on $ C _{1} $ = charge on $ C _{2} $
therefore $ C _{1}(V _{A}-V _{D})=C _{2}(V _{D}-V _{B}) $
therefore $ C _{1}(V _{1}-V _{D})=C _{2}(V _{D}-V _{2}) $
therefore $ V _{D}=\frac{C _{1}V _{1}+C _{2}V _{2}}{C _{1}+C _{2}} $
BETA
