In the given circuit, a charge of +80μC is given to the upper plate of the 4μF capacitor. Then in the steady state, the charge on the upper plate of the 3μF capacitor is
(A)+32μC
(B)+40μC
(C)+48μC✓ Correct
(D)+80μC
Answer: (C)
Solution
In steady state, the voltage across the 2μF and 3μF capacitors must be equal. Let q be the charge on the 3μF capacitor. Then the charge on the 2μF capacitor is 80−q. Setting the voltages equal: 280−q=3q. Solving: 3(80−q)=2q, giving 240=5q, so q=48μC.
JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option