Obtain the energy stored in each capacitor under dc conditio

Obtain the energy stored in each capacitor under dc conditions

Solution

Here,

as the capacitor will behave as open circuit

current in 2 kOhm

I(2) = 6 * (3)/(2 + 4 + 3)

I(2) = 2 mA

Now , current through 3 kOhm , I(3) = 6 - 2 = 4 mA

energy stored in 2 mF = 0.5 * C * V^2

energy stored in 2 mF = 0.5 * 2 * 10^-3 * (2 * 2)^2

energy stored in 2 mF = 0.016 J

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energy stored in 4 mF = 0.5 * 4 *10^-3 * (4 * 2)^2

energy stored in 4 mF = 0.128 J