# simple LR circuit

Discussion in 'Homework Help' started by Musab, Oct 23, 2009.

1. ### Musab Thread Starter Member

Sep 20, 2008
25
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For a simple LR circuit where a DC supply is connected in series with a resistor and an inductor. How is the graph of Vin vs Vout (across the inductor) is going to look?

I think Vout is going to be zero for any value of Vin if the inductance is small. But I am not sure if this is going to change if the inductance is increased.

Thanks

2. ### Thav Member

Oct 13, 2009
82
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Do you have any experience with Laplace transforms or differential equations? That's how you'll find the precise answer.

You are correct in that the steady state (not transient!) Vout (I assume you are measuring across the inductor) will always be zero, no matter what the inductance value is.

Wikipedia has a good page with graphs that you can look at, and maybe we can answer more detailed questions. http://en.wikipedia.org/wiki/RL_circuit

3. ### Accipiter New Member

Sep 20, 2009
9
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Remember, too, that although ideally the steady-state DC voltage drop across any inductor will be zero, this is only true for an inductor with zero resistance. Any practical inductor actually has some resistance (although small). So, there will be a (small) voltage drop across a practical inductor.

Mar 24, 2008
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5. ### Musab Thread Starter Member

Sep 20, 2008
25
0
Thanks

I derived the equation of Vout across the inductor using Laplace transform, $Vout = Vin . e^{-RT/L}$

But how am I suppose to conclude that Vout will be zero for any value of Vin ?

Last edited: Oct 23, 2009
6. ### RimfireJim Member

Apr 7, 2008
22
1
Vout will be zero only for the steady state condition, which is when time goes to infinity. Take the limit of the right side of your equation for t->infinity, and you'll see that the result goes to zero regardless of finite values of R and L.