RL Circuit

  • Thread starter Mola
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Main Question or Discussion Point

I am trying to analyse an RL circuit, particulartly the dependence of current in 2 resistors with time, in an RL circuit. This is what I think but I would really love some people to tell me if I am getting it wrong or if there is anything they want to add to it.

***Just after the the switched at t = 0, the voltage in R1 is the same as the voltage in R2 because they are in parallel. Therefore the current in R1 is: I = E/R1 which is the maximum current. But the current in R2 is zero because current in the inductor is zero at t = 0.

Question:
Do I have a good reasoning for the current in the resistor R2 being zero at t = 0?

***A long time after the switched is closed: I think as time goes to infiniti, the current in R1 drops down from maximum (E/R1) to zero.
The current in R2 starts growing exponentially as time increases. And at some point the current in R2 goes steady or reach equilibrium(not growing and not decaying).

Question: Is it true that the current in R1 decays to zero as time goes to infiniti? I am not sure the current in R1 decays though because the resistor does not consume current.

I would like some help analysing this circuit. It looks pretty interesting.

Thanks, Momodou
 

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  • #2
kuruman
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The answer to the first question is yes, your reasoning is correct.
The answer to the question is no. The current in R1 does not decay to zero because it is still connected across the battery. A long time after the switch is closed, the current in the branch with the inductor is not changing. This means that the inductor acts as a straight wire (short). The current in each branch is obtained by considering R1 and R2 in parallel with the battery.
(I changed the original figure format from BMP to PNG and uploaded it in case anyone looks at this post)

drawing1.png
 

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