Find the amount by which the total energy stored in the capacitor (in μJ) in will increase in the circuit shown in the figure after switch K is closed. Consider all capacitors are of capacitance 3 μF and battery voltage is 10 V.

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Detailed Solution

We can redraw the circuit with K is open and when K is closed as shown in figures below.Initial and final energies stored in capacitors areUi=12CV2⇒Uf=1253CV2Change in stored energy is given asΔU=Uf−Ui=125C3−CV2⇒ΔU=122C3V2=13CV2⇒ΔU=13×3×10−6×102=100μJ

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