Energy stored in the steady state circuit

AI Thread Summary
The discussion centers on a circuit problem involving energy storage, where the user initially calculated the stored energy as 4.4J. Other participants requested clarification on the approach used and confirmed that applying Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) was appropriate. It was pointed out that the user miscalculated the energy formula, specifically the term 0.5 LI^2, leading to an incorrect result by a factor of ten. After correcting the calculation, the user arrived at the correct energy value of 4.37 nJ. The conversation emphasizes the importance of careful calculations in circuit analysis.
hoangpham4696
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TL;DR Summary: Hi everyone. I have this circuit problem. I know that vL=0 and iL is just a constant number at t-> infiniti. I have attempted this problem and got energy stored = 4.4J. Please confirm it with me if my approach to the problem is correct. Thank you all.

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hoangpham4696 said:
Please confirm it with me if my approach to the problem is correct.
What approach? Please show the work you used to come to that result. Thank you.
 
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Here is my approach
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Yes! A good cookbook application of KVL & KCL.
Later, you will learn to do this much more quickly by reducing the complexity of the network with some observations. For example, as you said for steady state ##v_l=v_b=0##, so you can replace both of those with a short circuit. That voltage will be zero regardless of the current flow.
 
DaveE said:
Yes! A good cookbook application of KVL & KCL.
Later, you will learn to do this much more quickly by reducing the complexity of the network with some observations. For example, as you said for steady state ##v_l=v_b=0##, so you can replace both of those with a short circuit. That voltage will be zero regardless of the current flow.
Thank you so much. I appreciate you checking it for me.
 
hoangpham4696 said:
Thank you so much. I appreciate you checking it for me.
Oops! I didn't check the final result. You miscalculated ##0.5 LI^2##. You're off by powers of 10. Try that again.
 
DaveE said:
Oops! I didn't check the final result. You miscalculated ##0.5 LI^2##. You're off by powers of 10. Try that again.
Ok I see. It should be in nJ. So 4.37 nJ. Thank you
 
hoangpham4696 said:
Ok I see. It should be in nJ. So 4.37 nJ. Thank you
Yep. Good work!
 
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