# RL circuit problem, Examp7.4, Irwin

#### PG1995

Joined Apr 15, 2011
832
Hi

Please have a look on the attachment. Where am I going wrong? There is a minor error in my final answer. Could you please help me? Thank you.

Regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi

Please have a look on the attachment. Where am I going wrong? There is a minor error in my final answer. Could you please help me? Thank you.

Regards
PG
Your mistake is that you assumed that the pre-existing coil current is equal to the starting supply current. This is not correct.

You correctly calculated the preexisting coil current is 8/3, and you are correctly noting that this coil current can not change instantaneously.

However, when the switch is thrown, it is not correct to say that the supply current is 8/3 as it was before the switch was thrown. You need to find a new supply current valid after the switch is thrown. The switch puts an effective 4 ohm resistance across the coil and the voltage on the coil will create an additional current through the supply. You have to solve this circuit to find the true starting supply current. Note that the starting supply current equals the starting coil current plus the starting current through the resistance in parallel with the coil.

Everything else you did is correct.

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#### PG1995

Joined Apr 15, 2011
832
Hi Steve

I'm still unable to get the correct answer. Yesterday, I was trying to solve the problem hurriedly and didn't have much time to ask follow-on question but I still very much appreciate your response to my request. Now I have solved it little carefully. You see I'm still getting "12", not "24", which I should according to the given answer. Please help me with it. Thank you.

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi Steve

I'm still unable to get the correct answer. Yesterday, I was trying to solve the problem hurriedly and didn't have much time to ask follow-on question but I still very much appreciate your response to my request. Now I have solved it little carefully. You see I'm still getting "12", not "24", which I should according to the given answer. Please help me with it. Thank you.

Best wishes
PG
Well, even though you solved it "a little carefully" this time, you still solved it the same way as previously. There is a famous quote which says, "insanity is doing the same thing over and over and expecting different results".

I take it you didn't understand my first answer, but that answer is the key to finding your mistake. So let me try to say it in a different way.

Most of what you did is correct, but there is a subtle flaw. Even though the current in the coil does not change when the switch is thrown, the current from the supply (which generates the v(t) across the 4 ohm resistor) does change instantaneously when the switch is thrown. The coil does act like a current source at that instant in time, but the starting current from the supply must be recalculated after the switch is thrown. The supply current is 8/3 A at t=0-, where the 0 minus means at time zero just before the switch makes contact. The thing you have to solve for is i(t=0+) which is the current at time zero just after the switch makes contact. The two currents are not the same. You have correctly drawn the effective circuit valid for t=0+, so just solve for that current and use it in place of the 8/3 A value in your equation. Then multiply this current time 4 Ohms and you have your voltage. I did it out myself and obtained the exact answer you are looking for, so I know it works out.

Now, let's ask why the supply current changes instantaneously when the switch is thrown. When the switch is thrown, the previously determined steady state condition is no longer valid and you have a dynamic condition again. Before the switch is thrown, the 2 Ohm plus the 1 Ohm resistor add to make a 3 ohm resistor which is in parallel with the 6 ohm resistor to make an effective 2 ohm resistor in parallel with the coil. After the switch is thrown, the 2 and 1 Ohm resistors are gone and the 12 Ohm resistor is active again. This 12 Ohm resistor is now in parallel with the 6 Ohm resistor, which now makes an effective 4 Ohm resistor in parallel with the inductor. So the effective circuit you have to solve has an 8/3 A current source, a 24 V voltage source, and two 4 Ohm resistors. Now solve for the current coming from the voltage supply, and then you have your current i(t=0+).

#### PG1995

Joined Apr 15, 2011
832
Well, even though you solved it "a little carefully" this time, you still solved it the same way as previously. There is a famous quote which says, "insanity is doing the same thing over and over and expecting different results".

Thanks a lot, Steve.

This was the second RL circuit that I solved. In the past I have solved RC circuits and they were straightforward and didn't led to such 'subtle' problem. So, I think while solving for RL circuits I have to be little careful. The answer I found is incorrect when put t=∞ and I needed to add 3. I understand it now. I'm much grateful.

With best regards
PG

#### steveb

Joined Jul 3, 2008
2,436

Thanks a lot, Steve.

This was the second RL circuit that I solved. In the past I have solved RC circuits and they were straightforward and didn't led to such 'subtle' problem. So, I think while solving for RL circuits I have to be little careful. The answer I found is incorrect when put t=∞ and I needed to add 3. I understand it now. I'm much grateful.

With best regards
PG
Good work. I follow your point of view now. I misinterpreted one of your steps originally, so I can see how my answer would be hard to follow.

#### PG1995

Joined Apr 15, 2011
832
I follow your point of view now. I misinterpreted one of your steps originally, so I can see how my answer would be hard to follow.
Hi again,

Some days ago when I was revising the material, I found out that I had forgotten the reason for getting the wrong answer. I have tried to remember it but don't really understand.

If you don't mind, please help me with it. I understand that you would need to look at the problem from the beginning once again. I'm grateful for your kindness. Thanks.

Best regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
I'm confused by what you wrote out. I see i(t), I(t) and I_T, but it's not clear to me what these are referring to. Is i(t) the coild current or the current in the 4 ohm resistor?

Please label the schematic and identify what i(t) is.

#### PG1995

Joined Apr 15, 2011
832
I'm confused by what you wrote out. I see i(t), I(t) and I_T, but it's not clear to me what these are referring to. Is i(t) the coild current or the current in the 4 ohm resistor?

Please label the schematic and identify what i(t) is.
Hi

Is it any clearer now? If not, please do let me know. Thank you.

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi

Is it any clearer now? If not, please do let me know. Thank you.

Best wishes
PG
I cant' see that attachment for some reason.

#### PG1995

Joined Apr 15, 2011
832
Sorry. I don't know what happened.

#### steveb

Joined Jul 3, 2008
2,436
So, it looks to me that the coil current is correct, but the method of calculating the current in the four ohm resistor is not. The circuit is not quite the same as a current source driving parallel resisters.

Use v=L di/dt to get the voltage drop on the coil. Then 24 minus v will be the voltage drop on the four ohm resister. Then, divide that by four to get the current in the four ohm resister.

#### PG1995

Joined Apr 15, 2011
832
So, it looks to me that the coil current is correct, but the method of calculating the current in the four ohm resistor is not. The circuit is not quite the same as a current source driving parallel resisters.

Use v=L di/dt to get the voltage drop on the coil. Then 24 minus v will be the voltage drop on the four ohm resister. Then, divide that by four to get the current in the four ohm resister.
Thank you.

I have looked at it again to check if there was some error in finding the current through the 4 ohm resistor using current divider rule. It looks correct. I understand that it's not a 'regular' current source rather an exponential current source but I still don't find any flaw in the method. I think my method should work. The error comes into play when I use CDR. Why is so? Could you please help me with it?

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
Thank you.

I have looked at it again to check if there was some error in finding the current through the 4 ohm resistor using current divider rule. It looks correct. I understand that it's not a 'regular' current source rather an exponential current source but I still don't find any flaw in the method. I think my method should work. The error comes into play when I use CDR. Why is so? Could you please help me with it?

Best wishes
PG

I would have to spend some time to figure out if the basic method you are trying to employ can be made to work. But, my usual approach in doubtful cases is to avoid that path and take a path that I'm 100% confident is correct.

Since you are getting the wrong answer, your approach, even if fundamentally correct, has not been correctly implemented. I would say that for your approach to work you might need to consider the source impedance of your current source. This source impedance is also in parallel with your resistors. Try including that, and see if it works. However, I still think there may be additional flaws in the approach because the voltage source in not being considered. Once, all important factors are included, you can probably get the viewpoint to work, but this takes time to figure out, to be sure.

#### PG1995

Joined Apr 15, 2011
832
Thank you.

This is where you helped me to figure out why my straightforward and regular approach fails.

I don't know why you say that the effect of the voltage source isn't factored in. When the transient response dies away, i.e. at t = ∞, i(t)=6 - 10/3 * e^(-t/2) = 6A, we are left with only 6A and this current is due to the voltage source.

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
I don't know why you say that the effect of the voltage source isn't factored in.
I meant not factored in to the final step where you use an equivalent circuit.

Consider the attached drawing. If I claim to have a current source (with a finite source resistance Rs) driving a load path that includes a resistor Ro and a voltage source, it seems to be a mistake to not include the voltage source in the calculations. It also seems to be a mistake to ignore the source impedance of the current source.

Analyze this simple case and verify that you get a different answer in each case. This looks essentially the same as what you are doing, which is why I mentioned that the voltage source isn't factored in. You'll have to look at this to see if my judgement is correct. Am I interpreting what you did correctly?

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#### PG1995

Joined Apr 15, 2011
832
Thank you very much, Steve.

If you help me with the query in the attachment, then perhaps I would be able to understand where I'm going wrong. Thanks.

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
Thank you very much, Steve.

If you help me with the query in the attachment, then perhaps I would be able to understand where I'm going wrong. Thanks.

Best wishes
PG

I don't know the answer to your question. I'm not understanding Fig. 3 at all, so I can't anwser a question about it. By not answering my question above, you aren't allowing me to try and understand what you are doing. I asked whether the diagrams I provided in any way reflects what you are trying to do. If it does, then I feel that your equivalent circuit (Fig. 3) is flawed. If I'm right that it is flawed, then this is exaclty where you are going wrong. You are making an incorrect equivalent circuit.

#### PG1995

Joined Apr 15, 2011
832
Hi Steve

Actually I wasn't able to understand the diagram you gave. But now I have modified the equivalent circuit diagram a little and included the voltage source. Is it any good and can you now tell me which branch has i(t) running through it? Please have a look. Thanks.

Best regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi Steve

Actually I wasn't able to understand the diagram you gave. But now I have modified the equivalent circuit diagram a little and included the voltage source. Is it any good and can you now tell me which branch has i(t) running through it? Please have a look. Thanks.

Best regards
PG
We are kind of doing this backwards. You are the one proposing that you have an equivalent circuit that will give you the correct answer. My feeling is that the circuit is not correct. If you propose an equivalent circuit, the burden is on you to prove that it is equivalent and to transform the relevant variables and parameters.

But, let's put that point aside for a moment.

I presume that your intention is to have i(t) from the coil be the current source in your equivalent circuit. Otherwise, I don't know where else you would put it. However, I feel the circuit is wrong, even with the voltage source added because you have not included the fact that the current i(t) is flowing through an inductor that will generate a particular voltage across its terminals.

This is what I was trying to show in my diagrams above. The bottom line is that if I calculate the answer from your circuit, I get the wrong answer. It's possible I'm making a mistake, but you also have not been able to get the correct answer by that method. What more proof do you need that it is wrong, or at least more trouble than it is worth to fix.

I say, why go through all of this effort. Use the real circuit. It is simple enough to analyze. I already mentioned how easy it is to calculate the current I4, once you know i(t), and gave you the algorithm.