Simple RL question with laplace

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The discussion revolves around calculating the current in an RL circuit using Laplace transformation. The initial attempt incorrectly suggested a current of 250 A at t=0, which contradicts the behavior of inductors that do not allow instantaneous changes in current. After clarification, it was noted that the voltage source should be represented as 50u(t) volts, leading to a corrected expression for I(s). The final result indicates that the current I(t) starts at 0 A at t=0, aligning with the expected behavior of the circuit. This highlights the importance of accurately modeling circuit components in time-domain analysis.
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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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