Simple RL question with laplace

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SUMMARY

The discussion centers on solving a circuit problem involving a resistor and inductor (RL) using Laplace transformation. The initial voltage is 50 V, resistance is 10 ohms, and inductance is 0.2 H. The user initially calculated the current incorrectly as 250 A at t=0, but later corrected it to 0 A by applying the correct voltage source representation as 50u(t) volts in the s-domain. The final expression for current in the time domain is I(t) = 5 - 5e^(-50t).

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Homework Statement


I have a circuit with resistor and inductor (RL) that is placed serial and a power supply.
the voltage is constant 50 V at t=0 , R=10 ohm, and L=0.2 H.
Find the I using laplace transformation

Homework Equations



-

The Attempt at a Solution


my attempt is transform everything to S-domain and with nodal analysis
so the equation go like this
50/(10+0.2S)=I(s)
so, 250/(S+50)=I(s)
then i transform it to t-domain
I(t)=250 e^(-50t)
at t=0
I(0)=250 e^0
so I(0)= 250
is this right?
i have V at 50 Volt, but the current become 250 A
am i doing it right?
thanks for the reply
 
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If you recall how inductors behave, you wouldn't expect the current to instantaneously go from zero to 250 Amps at time t = 0 when the voltage is switched on. So clearly something has gone awry in your workings.

Your voltage source should "turn on" at time t = 0, so it's really 50u(t) volts. In the s-domain that becomes 50/s ;)
 
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ooh, so the equation go like this?

I(s)=250/s(s+50)
I(s)=5/s - 5/(s+50)
I(t)=5 - 5 e^ -50 t
then because t=0
then I(0) = 5-5 = 0
so because it haven't been started so the current would be 0
okay, thanks
 

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