Find the voltage of an inductor in an LR circuit

[crashdirty86]

Homework Statement

at t=0 switch S is closed. Just after the switch is closed what are i1, i2, the potential difference across V2 across resistor 2, and Vl across the inductor? A long time after the switch is closed, what are i1, i2, V2, and Vl?

Emf = 12V, R1= 15 ohms, R2 = 25 ohms, and L = 5.0 H

Now switch S is opened at t'=0. Just after the switch is opened, what are i1, i2, V2, and Vl?

Homework Equations

V=iR
Vl= -Ldi/dt

The Attempt at a Solution

So just after t=0 no current passes through the inductor, so the Vl=12V and V2=0V. Using ohms law, the current in Resistor 1 is i=V/R => i1=0.8A

A long time after the switch is closed the inductor now acts like a wire, and R1 & R2 are in paralell, therefore, i1=0.8 A and i2=0.48 A V2= 12V and Vl= 0V.

When the switch is opened at t'=0 R1 & R2 are now in series with the inductor. So i1=i2 => i=V/(R1+R2)= 0.3A. The voltage on the inductor will be 12V to oppose the change in magnetic flux of the circuit. And V2=iR2=7.5V.

So I am pretty positive about the first two parts of this question. It is really the opening of the switch at t'=0 that I am stuck on. Am I right to say that the Voltage through the inductor is 12V?

 

[crashdirty86]

anyone want to attempt to help?

 

[gneill]

Your first two parts look okay.

For the the time when the switch re-opens, you've got a problem. There's no voltage source V in the remaining circuit, so it shouldn't play a role in your calculations. Instead, what do you know about the current through an inductor with regards to sudden changes?

 

[crashdirty86]

The current will change continuously, so the current through the inductor should be 0.3 A i do believe.

 

[gneill]

The current will change continuously, so the current through the inductor should be 0.3 A i do believe.

Nope. Inductors resist a change in current. That's why you wrote that the initial current through the inductor was zero when the switch first closed.

This time, at the instant when the switch opens at t'=0, the current must remain the same as immediately before the switch opened.

 

[crashdirty86]

Nope. Inductors resist a change in current. That's why you wrote that the initial current through the inductor was zero when the switch first closed.

This time, at the instant when the switch opens at t'=0, the current must remain the same as immediately before the switch opened.

Okay, so after a long time the inductor acts like a wire which I can then find the current acting on resistor 2. Once the switch is opened, then that current through two will be the current through the inductor which I can then use to find its voltage?

 

[gneill]

Okay, so after a long time the inductor acts like a wire which I can then find the current acting on resistor 2. Once the switch is opened, then that current through two will be the current through the inductor which I can then use to find its voltage?

Yes, the initial current around the loop will be the same as the inductor current immediately prior to the switch opening. Use KVL around the loop to determine all the voltages.

 

[crashdirty86]

Yes, the initial current around the loop will be the same as the inductor current immediately prior to the switch opening. Use KVL around the loop to determine all the voltages.

So the current through the inductor when the switch is opened should be 0.48 A. Giving a voltage of 2.4 V on the inductor, 12 V on resistor 2 and 7.2 volts on resistor 1? This seems odd... I feel like I have broken the law of conservation on energy?

 

[SammyS]

So the current through the inductor when the switch is opened should be 0.48 A.

That's correct for the current throught the inductor.

Giving a voltage of 2.4 V on the inductor, 12 V on resistor 2 and 7.2 volts on resistor 1? This seems odd... I feel like I have broken the law of conservation on energy?

If the current through the inductor is 0.48 A, then what is the current through R₁ and R₂ ?

 

[crashdirty86]

That's correct for the current throught the inductor.
If the current through the inductor is 0.48 A, then what is the current through R₁ and R₂ ?

[ IMG]https://www.physicsforums.com/attachment.php?attachmentid=58314&d=1367173841[/PLAIN]

Well if the current through the inductor is 0.48 A and the switch is open, then we have a series of elements which means that the current through resistors 1 & 2 would also be 0.48 A.

 

[gneill]

So the current through the inductor when the switch is opened should be 0.48 A. Giving a voltage of 2.4 V on the inductor, 12 V on resistor 2 and 7.2 volts on resistor 1? This seems odd... I feel like I have broken the law of conservation on energy?

Nope. The circuit is drawing on the energy stored in the inductor's magnetic field. That potential energy was put there while the switch was closed and the current through the inductor built up.

You need to check the polarities of the potential drops that you are summing up; Which way does the current flow through each resistor when the switch opens?

 

[SammyS]

Well if the current through the inductor is 0.48 A and the switch is open, then we have a series of elements which means that the current through resistors 1 & 2 would also be 0.48 A.

That's correct.

What is the voltage drop across each resistor?

After answering that, What's the voltage across the inductor?

 

[crashdirty86]

So this is a late response, but I do have the concept down for finding the voltage on the inductor when the switch is opened at t'=0. Since the current inside an inductor must change continuously, then the current flowing through the inductor must be the same as the current flowing through resistor two before the switch was opened for t'=0. This would be the 0.48A. To provide this continuous flow of current the inductor creates a momentary spike in its voltage as the release of energy stored in the magnetic field of the inductor depletes. Therefore, the voltage in the inductor at t'=0 is 19.2V. I found this odd because it seems like the laws of physics were broken, but since the inductor has to change continuously, conservation of charge must be instated and the voltage then spikes. Thanks for everyone's help, it allowed me to score a 98 on my exam.