Here is the problem

I need to find the Stored energy of the inductor and capacitor.

Attempt at problem:
Now looking at the circuit I know that the inductor is treated as a short and the capacitor is treated as an open circuit. In which I can redraw the circuit.

Now here is where I get hung up on, I'm not sure where to go from this point.
Would I use mesh Analysis in order to find out the Vx. I'm just thinking there is something I'm not seeing to make this problem easier and I'm making it more complicated than it is.

 

You've got two 10Ω resistors. Let's call the vertically oriented one R1, and the one with Vx across it, R2. Let the bottom of the 4A current source be the reference node.

What is the current through R2 (this is also the current through the inductor) and what is the current through the 14Ω resistor? Then what is the current through R1, and the voltage across R1?

 

You've got two 10Ω resistors. Let's call the vertically oriented one R1, and the one with Vx across it, R2. Let the bottom of the 4A current source be the reference node.

What is the current through R2 (this is also the current through the inductor) and what is the current through the 14Ω resistor? Then what is the current through R1, and the voltage across R1?

Much thanks for the quick response. I'll try this after I do some other problems with RL Circuits. Going over stuff for a test next week.

 

Ok I figured it out using KCL.

Considering I have a 4A of current flowing on the right branch, we need the same upon returning back to the current source. So 4*10=40V = Vx
so power in the inductor is = 1/2*.5H*(4)^2 = 4 Joules

Then since Vx, is found, use that in the dependent current source, use KCL to see how much current goes through the vertical 10 resistor, I = .4*40 + 4 = 20A, 20*10 = -200 V, then energy in the conductor is = 1/2(8*10^-4 F)*(200)^2 = 16 Joules.

Much thanks Electrician.

 

Ok I figured it out using KCL.

Considering I have a 4A of current flowing on the right branch, we need the same upon returning back to the current source. So 4*10=40V = Vx
so power in the inductor is = 1/2*.5H*(4)^2 = 4 Joules

Then since Vx, is found, use that in the dependent current source, use KCL to see how much current goes through the vertical 10 resistor, I = .4*40 + 4 = 20A, 20*10 = -200 V, then energy in the conductor is = 1/2(8*10^-4 F)*(200)^2 = 16 Joules.

Much thanks Electrician.

Looks good except it should be "then energy in the capacitor is = 1/2(8*10^-4 F)*(200)^2 = 16 Joules.
"