What Happens to the Current in an LR Circuit After the Switch is Opened?

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When the switch in an LR circuit is opened, the current through the inductor remains constant immediately after the switch is opened due to the inductor's property of resisting changes in current. The initial current, Io, flows through the remaining circuit components, which include the resistance of the inductor and the external resistor in parallel. The current in the resistor will change instantaneously, while the inductor maintains its current until it begins to decay. The decay of the current can be modeled as an exponential decay with a time constant of L/(r+R). This understanding allows for the calculation of current distribution in the circuit after the switch is opened.
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Homework Statement


Please refer to the diagram which i have attached for better understanding...

A solenoid of inductance L with resistance r is connected in parallel to a resistance R. A battery of emf E and of negligible resistance is connected across the parallel combination. At time t=0 switch is opened, calculate current through the solenoid after the switch is opened.

Homework Equations





The Attempt at a Solution



This i guess will not pose too much difficulty for people who have done this topic for sometime but i am pretty new to it...
Anyway as long as switch was closed the inductor played no role...(di/dt = 0)
So we just take into consideration the resistance of the inductor and the external resistance in parallel and get current through each of them in the simple manner...

Now i am not getting exactly what will happen after the switch is opened. After the switch is opened means immediately after right? So just after it is opened what will be the distribution of current in the upper part of the circuit(the lower part has zero current). I think we have to get an equation using Kirchoff'd Law and then integrate it and put t=0.

But i am not being able to frame this equation because i think the current in each just after opening the switch will be same as before then maybe the currrent through one decreases and the other increases then maybe they become uniform and decrease uniformly to become zero after a long time...This is what i think and i guess am totally wrong here...So please help me out and correct my thinking...Waiting for a reply soon...Thank you
 

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Hi Abhishekdas. When the switch opens, the current through the inductor wants to remain as it was before (inductors provide a sort of current inertia). The current through other components isn't so constrained.

So if the current through the inductor before the switch opens is Io, it will still be Io the instant after the switch opens. That means it will flow via the loop that exists with r and R in it. Now you've got a standard RL type circuit with an initial current. What do you know about the general behavior of currents and voltages in RL and RC type circuits?
 
hi gneill...thanks for the reply...

So you mean to say that whatever be the initial value of current in the resistor it will change instantaneously but the current in the inductor wont?(coz of it resists change in current)

So now does it becomes a sum of decay of current where you simply have to take initial current as the value of current in the Inductor and the time constant as L/(r+R)?
 
Yessir. That's it.
 
Got it...Thanks a lot man...
 

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