KCET 2017 Physics Solved Paper

The effective capacitance of system of two given capacitors is
=‌

1

2µF

+‌

1

4µF

=‌

(2+1)

4µF

=‌

3

4

⇒C=‌

4

3

µF
Now, using Q = CV

Therefore, electric charge = 8 μC
Now, energy stored =‌

1

2

CV2=‌

1

2

×‌

4

3

×10−6×6×6
=2×10−6×2×6=24×10−6J
Therefore, energy stored = 24 μJ