Energy stored in the steady state circuit

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SUMMARY

The forum discussion centers on calculating the energy stored in a steady state circuit, where the voltage across the inductor (vL) is zero and the current (iL) is constant as time approaches infinity. The initial calculation of energy stored was reported as 4.4J, but it was later corrected to 4.37 nJ due to a miscalculation involving the formula 0.5 LI^2. The discussion emphasizes the application of Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) in simplifying circuit analysis.

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  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with energy storage in inductors
  • Basic circuit analysis techniques
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  • Study the principles of steady state analysis in electrical circuits
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hoangpham4696
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Thread moved from the technical forums to the schoolwork forums
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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Welcome to PF.

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
IMG_0399.JPG
 
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
 
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hoangpham4696 said:
Ok I see. It should be in nJ. So 4.37 nJ. Thank you
Yep. Good work!
 
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