Energy stored in the steady state circuit

In summary, a steady state circuit refers to an electrical circuit that has reached a condition where all voltages and currents remain constant over time, following the initial transients after a change in the circuit. In this state, the behavior of the circuit can be analyzed using linear circuit theory, allowing for the application of techniques such as Ohm's law and Kirchhoff's laws. Steady state conditions are crucial for understanding the long-term operation of circuits in various applications, including AC and DC systems.
  • #1
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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  • #3
Here is my approach
IMG_0399.JPG
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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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