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[SOLVED] Circuit analysis of Source free RL Circuits

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Eshal

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Hello all,

I have two questions.

1) Below is the image of the solution for a question. Kindly tell me where does vo=1V come from in figure 7.14(a)?
Capture.PNG

2) We know that inductor acts as a short-circuited to the DC source.
In below figure 7.18, if switch is closed for long time then inductor will be short. Then what would be resulting circuit. Kindly draw it and tell me how.
Capture.PNG

Best Regards,
Princess
 

_Eduardo_

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(1)
It's a trick to calculate the Thevenin resistance.
Feeding the circuit with an auxiliary source, find the current and then makes
\[R_{th} = \frac{V_{aux}}{I}\]


But solving the circuit in that way is the worst idea that a human can have.

Note:
\[I_l + I_s = I_l + \frac{V_l}{2} = \frac{3 I_l - V_l}{4} \;\;\longrightarrow\;\; I_l = -3 V_l\]

and \[V_l = L I_l' \;\;\longrightarrow\;\; I_l' = -\frac{2}{3} I_l\]

whose well-known solution is \[I_l = I_0 e^{-\frac{2}{3}t}= 10 e^{-\frac{2}{3}t}\]
 
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Eshal

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But why 1V of source? Why not other volts like 2V, 3V, 4V, etc..

Also In this example I have dependent source. So author describe two methods. But in the previous example 7.2 (I didn't show here but is in the book) there is no any dependent voltage source and author didn't provide the method 1 there. It means only dependent sources if exist then method 1 is applicable otherwise not.
 

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Why 1V? Because it's a nice number.

If you have chosen vo = 10V --> Io = 30A
and Rth = 10/30 = 1/3

You could also have chosen a current source 100A, calculate the voltage Vo, and Rth=Vo/Io would be the same.


But in the previous example 7.2 (I didn't show here but is in the book) there is no any dependent voltage source and author didn't provide the method 1 there. It means only dependent sources if exist then method 1 is applicable otherwise not.

The previous example could also have done so. But if there are a more simple way why not use it?
 
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Got it.

And what about my second question?

Can you draw the circuit when inductor is short?
 

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You can draw it.
Just remove the R5Ω and replace L by a wire.
 
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Ohhh Thanks a lot. However, always be here to help me. ;-)

Best Regards,
Princess
 

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