Electrostatics Question 578

Question: In the circuit shown here C1=6μF, C2=3μF and batteryB=20V . The switch S1 is first closed. It is then opened and afterwards S2 is closed. What is the charge finally on C2

Options:

A) 120μC

B) 80μC

C) 40μC

D) 20μC

Show Answer

Answer:

Correct Answer: C

Solution:

Common potential V=6×20+3×0(6+3)=1209Volt

So, charge on 3μF capacitor

Q2=3×106×1209=40μC



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