Finding the inductance of an circuit knowing the energy stored

[Jd303]

If the total energy stored in the circuit below is 190 mJ, what is the value of L?
IS = 2 A
R1 = 250 Ω : R2 = 38 Ω
C = 41 µF
Give your answer to the nearest whole number, in mH (I have attached the diagram)

I first try to find the equivalent impedence knowing that the impedence of an inductor is jωL and the impedence of a capacitor is -j/(ωC)

I try to do this so as I can find Vs and hence using the formula:
E = (CV^2 + LI^2)/2
I would be able to find the value of L

However I have too many unknowns to be able to find the equivalent impedance, so how do i go about this question? Any help would be greatly appreciated I am really stuck. Thanks!

 

[Lancelot59]

If the total energy stored in the circuit below is 190 mJ, what is the value of L?
IS = 2 A
R1 = 250 Ω : R2 = 38 Ω
C = 41 µF
Give your answer to the nearest whole number, in mH (I have attached the diagram)

I first try to find the equivalent impedence knowing that the impedence of an inductor is jωL and the impedence of a capacitor is -j/(ωC)

I try to do this so as I can find Vs and hence using the formula:
E = (CV^2 + LI^2)/2
I would be able to find the value of L

However I have too many unknowns to be able to find the equivalent impedance, so how do i go about this question? Any help would be greatly appreciated I am really stuck. Thanks!

Notice how your current is unchanging. Your circuit is at a steady state.

Using this fact you can simplify your circuit. Recall that at a steady state, capacitors act like open circuits, and inductors act as wires. Since you have a known current entering the circuit, you can use Kirchoffs Current Law to determine the current through the inductor at a steady state.

 

[Jd303]

Thanks for your help! Sorry to be a bit slow, but I still can't obtain the correct answer.
As I am looking at it with the inductor replaced by a wire, and the capacitor replaced with an open circuit I am left with a simple circuit with the two resistors in parallel.

I then go to find the current going through R2 such that:

Ix = (R1/(R1 + R2))*Is

I then use the formula E = (LI^2)/2

To find the value of L

Can anyone please explain the mistake I have made, thanks.!

 

[Lancelot59]

Thanks for your help! Sorry to be a bit slow, but I still can't obtain the correct answer.
As I am looking at it with the inductor replaced by a wire, and the capacitor replaced with an open circuit I am left with a simple circuit with the two resistors in parallel.

I then go to find the current going through R2 such that:

Ix = (R1/(R1 + R2))*Is

I then use the formula E = (LI^2)/2

To find the value of L

Can anyone please explain the mistake I have made, thanks.!

I think you're forgetting that the capacitor stores energy as well.

 

[Jd303]

Sorry lazy mistake.

In that case I get this:

E = (CV^2 + LI^2)/2
Therefore:

L = (2E - CV^2)/I^2

I(through inductor) = (R/(R1 + R2))*Is

V = (R1*R2/(R1+R2))*Is

Doing this I still am left with an incorrect answer

Where am I going wrong?

 

[Jd303]

Thanks for your help! corrected my mistakes was just a calculator error. cheers!