Engineering First order circuits, Inductors

AI Thread Summary
The discussion revolves around analyzing a security alarm circuit involving an inductor after a switch is opened. The inductor is initially charged with a current of 2A, and when the switch opens, it creates two separate circuits. Participants clarify that the inductor behaves like a current source, maintaining the current direction initially. The voltage across a resistor can be calculated by multiplying the inductor current, but this only holds true at t=0, as the current will decrease over time. Using Kirchhoff's Voltage Law (KVL) is recommended to derive the differential equation needed to solve for the circuit's behavior post-switch opening.
popo902
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Homework Statement


A security alarm for an office building door is modeled by the circuit [below]. The switch represents the door interlock, and v is the alarm indicator voltage. Find v(t) for t>0 for the circuit [below]. The switch has been closed for a long time at t=0-

http://i36.photobucket.com/albums/e47/jo860/img014.jpg

Homework Equations



V = L di/dt

The Attempt at a Solution



ok so, i see that it has been closed for a long time, which must mean the inductor is "charged" and at steady state at t=0-, t=0+, and i think also t=0 since current cannot change instantaneously
I used mesh analysis and found the current "stored" in the inductor to be 2A

then i see that when the switch is opened, it splits the circuit into two circuits...right?
so then only the left side is where i work.
then the inductor would sort of be like a finite current source pumping current through the 9 then 3 ohm resistor--the opposite way the previous current was going(is this correct?)

I see it like if i find the current through the inductor at t>0, then i just multiply that by 9 to get the voltage across 9
could someone maybe give me a hint to what I'm missing or if I'm wrong?
would it right to use KVL?
 
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popo902 said:

Homework Statement


A security alarm for an office building door is modeled by the circuit [below]. The switch represents the door interlock, and v is the alarm indicator voltage. Find v(t) for t>0 for the circuit [below]. The switch has been closed for a long time at t=0-

http://i36.photobucket.com/albums/e47/jo860/img014.jpg

Homework Equations



V = L di/dt

The Attempt at a Solution



ok so, i see that it has been closed for a long time, which must mean the inductor is "charged" and at steady state at t=0-, t=0+, and i think also t=0 since current cannot change instantaneously
I used mesh analysis and found the current "stored" in the inductor to be 2A

then i see that when the switch is opened, it splits the circuit into two circuits...right?
so then only the left side is where i work.
then the inductor would sort of be like a finite current source pumping current through the 9 then 3 ohm resistor--the opposite way the previous current was going(is this correct?)
No, the current through the inductor continues in the same direction.
I see it like if i find the current through the inductor at t>0, then i just multiply that by 9 to get the voltage across 9
This is true only at t = 0. After that the current and consequently the voltage through the resistor will diminish.
could someone maybe give me a hint to what I'm missing or if I'm wrong?
would it right to use KVL?
You can use KVL to write the differential equation for the mesh. After that, you solve the equation with the initial condition i(0) = 2.
 

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