Answer
$W_{2F} = 11.756 J$
$W_{1F} = 0 J$
$W_{3F} = 17.632J$
$W_{2H} = 0.184 J $
Work Step by Step
Setting this problem up is the hardest part, so start by redrawing the circuit with the capacitors as open circuits and the inductors as short circuits.
The goal is to find the current through the short circuit and the open circuit voltages using the simplified circuit of resistors.
Find the voltage over the 2F capacitor using a voltage divider by simplifying the 8ohm and 4 ohm resistors in parallel.
$V_{2F} = 6 * 2.6667/4.6667 = 3.429V$
This voltage is the same as the 3F capacitor because they are in parallel. (The 6 ohm resistor has 0 current flowing in it, so it has no voltage drop/rise.
Notice that the 1F capacitor is shorted by the inductor so it has 0V across its terminals
Use current divider to find the short circuit current, after calculating total current in the circuit.
$I_{total} = 6 / 4.6667 = 1.286A$
$I_{2H} = 1.286 * 2.6667/8 = 0.429 A$
Now use these values in the energy equations for inductors and capacitors to calculate energy.
$W_{2F} = 1/2 * 2 * (3.429)^2 = 11.756 J$
$W_{1F} = 0 J$
$W_{3F} = 1/2 * 3 *(3.429)^2 = 17.632J$
$W_{2H} = 1/2 * 2 *(0.429)^2 = 0.184 J $
