Analysis of an RL Circuit: Voltage, Current and Time Dependence

AI Thread Summary
The analysis of the RL circuit indicates that immediately after the switch is closed, the current in resistor R1 is at its maximum (E/R1), while the current in R2 is zero due to the inductor's initial state. As time progresses, the current in R1 does not decay to zero because it remains connected to the battery, while the current in R2 increases exponentially until it reaches a steady state. The inductor behaves like a short circuit in the long-term, allowing for current flow through both resistors in parallel with the battery. The reasoning regarding the initial current in R2 being zero is confirmed as correct. Overall, the circuit analysis reveals interesting dynamics between the resistors and the inductor over time.
Mola
Messages
22
Reaction score
0
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
 

Attachments

Physics news on Phys.org
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
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...

Similar threads

Back
Top