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}} $

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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}} $



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