Polarity of capacitor/inductor in RLC circuit

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SUMMARY

The discussion focuses on determining the polarities of components in an RLC circuit when a switch transitions from position A to position B. At time t = 0, the inductor carries a current of 1A, establishing its upper terminal as positive and lower terminal as negative. The resistor and capacitor, connected in series, exhibit opposite polarities to the inductor, with the lower terminal of the resistor being positive and the upper terminal negative, while the right terminal of the capacitor is positive and the left terminal negative. This polarity assignment is crucial for understanding the behavior of the circuit during the switch transition.

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Abdulwahab Hajar
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Homework Statement


In the figure given, the switch was at position A for a long time, at time =0 the switch is turned to position B...
My question is how do we assume the polarities for both the resistor and the capacitor after the switch is in position B

Homework Equations


??

The Attempt at a Solution


I know that a time t = 0, the current at the inductor is 1A and will be moving in a clockwise direction, and as current moves from higher potential to lower potential meaning that the the upper side of the inductor will be its positive terminal and the lower side will be its negative...
What about the resistor and capacitor??
Since in they are connected in series won't they have the opposite sign of the element which leads/ follows them...
As in the lower terminal of the resistor will be positive
the upper terminal of the resistor will be negative
the right terminal of the capacitor will be positive
the left terminal of the capacitor will be negative
Is this correct??
Thank you
 

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How did you determine the direction of the current in the inductor for time t=0?
 
Assuming that it reached steady state when the switch was in position A, the current through the inductor was 1A and going downwards
 
Abdulwahab Hajar said:
Assuming that it reached steady state when the switch was in position A, the current through the inductor was 1A and going downwards
Right. So when the switch moves to position b, what direction will it be going?
 
Abdulwahab Hajar said:
I know that a time t = 0, the current at the inductor is 1A and will be moving in a clockwise direction, and as current moves from higher potential to lower potential meaning that the the upper side of the inductor will be its positive terminal and the lower side will be its negative...
Okay, the part I have difficulty with in the above is the assignment of a polarity to the inductor for time t = 0 (or really, t = 0-). At steady state there will be no potential across the inductor; it will behave as a short circuit.

At time t = 0+, the inductor will develop a potential across it in order to maintain the current flow with the same magnitude and direction it had the instant before the switch commutation. So you can assign a polarity to that potential accordingly.
What about the resistor and capacitor??
Since in they are connected in series won't they have the opposite sign of the element which leads/ follows them...
Usually they can be assigned by the direction of the current flow: The "+" end of the component will be the terminal that the current enters. Things can be a bit trickier with energy-storing components such as inductors and capacitors, since they can be the source for a current or hold a potential previously stored there. In this case the inductor is going to develop a potential that will drive the current, so you should be able to figure out which end of that potential difference has to be positive in order to do so. The capacitor is initially uncharged so it will end up building a potential difference with a certain polarity depending upon the current starting at t=0.

Don't confuse these "dynamic" polarity assignments with any pre-existing labels on the circuit diagram that are there to tell you how to interpret various potentials in the circuit.

As it turns out, the following of potential polarity assignments are fine:
As in the lower terminal of the resistor will be positive
the upper terminal of the resistor will be negative
the right terminal of the capacitor will be positive
the left terminal of the capacitor will be negative
Is this correct??
Thank you
 
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gneill said:
Okay, the part I have difficulty with in the above is the assignment of a polarity to the inductor for time t = 0 (or really, t = 0-). At steady state there will be no potential across the inductor; it will behave as a short circuit.

At time t = 0+, the inductor will develop a potential across it in order to maintain the current flow with the same magnitude and direction it had the instant before the switch commutation. So you can assign a polarity to that potential accordingly.

Usually they can be assigned by the direction of the current flow: The "+" end of the component will be the terminal that the current enters. Things can be a bit trickier with energy-storing components such as inductors and capacitors, since they can be the source for a current or hold a potential previously stored there. In this case the inductor is going to develop a potential that will drive the current, so you should be able to figure out which end of that potential difference has to be positive in order to do so. The capacitor is initially uncharged so it will end up building a potential difference with a certain polarity depending upon the current starting at t=0.

Don't confuse these "dynamic" polarity assignments with any pre-existing labels on the circuit diagram that are there to tell you how to interpret various potentials in the circuit.

As it turns out, the following of potential polarity assignments are fine:
Thank you sir much appreciated
 

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