Hello Ramiel,

The pictures have to less information to get them enhanced, as I did before.
Do you have a flatbed scanner?
My printer here at home has one.

Bertus

If i wrote them in a bigger font will you be able to give me feedback?

 

The firs

the first picture is only meants for the circuit
Look at the other pictures for the work

If the first picture is only meant to provide the circuit, then crop the other stuff out. Don't make the reader have to guess what is and what is not important to trying to follow what you are trying to convey. Remember, you are asking strangers for free help -- YOU need to make it easier for them to WANT to do that.

 

If i wrote them in a bigger font will you be able to give me feedback?

It's not the size nearly so much as it is the contract and the sharpness (contrast, primarily).

 

Hello,

When you would have written the papers with a black pen instead of a pencil, they already would be better readable.

Bertus

 

@bertus @WBahn
That is the best I can.
The only thing i can not find is i2(0)
Help!!

 

Just look at the solutions and you will see that your assumption that those equations work is wrong!

But let's assume they aren't.

You are claiming

i1=i*L1/(L1+L2)
and:
i2=i*L2/(L1+L2)

A direct consequence of this is that, at all time,

i2(t) = i1(t)·(L2/L1)

Since L1 = 6 H and L2 = 3 H, that means that

i2(t) = 2·i1(t)

What is i1(0)?

It is given that i1(t) = 0.6 A e^(t/0.5 s)

That means that i1(t=0) = 0.6 A.

Thus, you're claiming that i2(t=0) = 1.2 A

But we are also given that i(t=0) = 1.4 A.

Since

i(t) = i1(t) + i2(t)

i2(t = 0) = 1.4 A - 0.6 A = 0.8 A

Therefore, contrary to what you are assuming, those equations do not hold for this problem.

The reason why not is very simple and I have stated it repeatedly. I see no utility in stating it yet again -- go back and DO THE MATH!

Hi,

Did i type that wrong?

I meant this:
i1=i*L2/(L1+L2)
and:
i2=i*L1/(L1+L2)

 

Hi,

Did i type that wrong?

I meant this:
i1=i*L2/(L1+L2)
and:
i2=i*L1/(L1+L2)

This method holds for specific situations. For example if you tried it here, it wont work

 

Hello,

When you would have written the papers with a black pen instead of a pencil, they already would be better readable.

Bertus

I hope you can take a look at post 45 and give me some advice on how to find i2(0)

 

It's not the size nearly so much as it is the contract and the sharpness (contrast, primarily).

I hope you can check post 45 and give me feedback on how to get i2(0)

 

I

if i an able to find either i(0) or i2(0), i would be able to find the other using i= i1+ i2
But i am not able to find either

Hello again,

I had a hard time reading it too. It is best to type these things.

Let me just say this without saying to much right away...

If you have one inductor driven by a voltage source (like a unit step) the current ramps up to a certain level at time t1 call this iL1.
Now if it was a different inductor value the current would ramp up to a different level at time t1 call this iL2.
Now if these two inductors (L1 and L2) are driven by two identical unit step sources and the first inductor reaches the level iL1 in time t1,
then the other inductor must reach the other value iL2 in time t1.
The question then is what is the relationship between iL1 and iL2.
The relationship is simply the ratio of the two inductor values.
So if L1 was 1H and L2 was 2H, and L1 ramped up to 1 amp in time t1, then L2 would ramp up to 1/2 amp in the same time t1.
That is a consequence of v=L*di/dt and solving for di we get:
di=v*dt/L
and so you can see that the larger the inductor the less di will be because L is in the denominator.

Now knowing this, see if you can finish the example. You do have to think backwards a little because you know the current in one inductor and the other inductor in parallel must have been exposed to the same excitation and so you should be able to calculate that current too, given the discussion above.

 

This method holds for specific situations. For example if you tried it here, it wont work

Hi,

Really? Then how did i get the right result using that idea? Is the book wrong?

 

Hi,

Really? Then how did i get the right result using that idea? Is the book wrong?

Try it with the question in post 47, and the answer using your method will be wrong since it doesnt always hold to be true

 

Hi,

Did i type that wrong?

I meant this:
i1=i*L2/(L1+L2)
and:
i2=i*L1/(L1+L2)

Doesn't matter. These equations do NOT hold for this problem because this problem does not abide by the assumptions that are required to be true in order for those equations to apply!

DO THE MATH!

Or, at the very least, explain why the provided solutions to the problem do not satisfy the relationships that you keep insisting hold. If the solutions are wrong, then show that they are wrong. If they are not wrong, then how can you keep insisting that the current divides as you say it does here?

Since doing the math seems to be asking way too much, at least follow along as I do it, step by step.

You have an inductor, L1, that has a voltage v(t) applied across it. What is the current i1(t)?

For an inductor in general:

v(t) = L di(t)/dt

so

i(t) = ∫v(t)/L

i1(t) = i1(0) + (1/L1)·[∫v(τ) dτ evaluated from 0 to t]

You have another inductor, L2, that has this same voltage applied across it. What is the current i2(t)?

i2(t) = i2(0) + (1/L2)·[∫v(τ) dτ evaluated from 0 to t]

Solving for the integral, we get

[∫v(τ) dτ evaluated from 0 to t] = L2(i2(t) - i2(0))

Plugging this back into the expression for i1(t), we get

i1(t) = i1(0) + (1/L1)·[L2(i2(t) - i2(0))]

i1(t) = i1(0) + (L2/L1)·[i2(t) - i2(0)]

i1(t) = (L2/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ]

Similarly, we have for i2(t)

i2(t) = (L1/L2)·i1(t) + [i2(0) - (L1/L2)·i1(0) ]

Since

i(t) = i1(t) + i2(t)

i(t) = (L2/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ] + i2(t)

i(t) = ((L1 + L2)/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ]

So i2(t) in terms of i(t) is

i2(t) = [L1/(L1 + L2)]{i(t) - [i1(0) - (L2/L1)·i2(0) ] }

i2(t) = [L1/(L1 + L2)]·i(t) + [L2·i2(0) - L1·i1(0)]/(L1 + L2)

Can you NOW finally see what I have been saying over and over?

You are insisting on assuming that the initial currents in the inductors are either zero or, at the very least, are in the same proportion as the inductances.

This is simply NOT TRUE for this problem. The initial conditions are very explicitly given and they simply are NOT in this proportion!

 

Doesn't matter. These equations do NOT hold for this problem because this problem does not abide by the assumptions that are required to be true in order for those equations to apply!

DO THE MATH!

Or, at the very least, explain why the provided solutions to the problem do not satisfy the relationships that you keep insisting hold. If the solutions are wrong, then show that they are wrong. If they are not wrong, then how can you keep insisting that the current divides as you say it does here?

Since doing the math seems to be asking way too much, at least follow along as I do it, step by step.

You have an inductor, L1, that has a voltage v(t) applied across it. What is the current i1(t)?

For an inductor in general:

v(t) = L di(t)/dt

so

i(t) = ∫v(t)/L

i1(t) = i1(0) + (1/L1)·[∫v(τ) dτ evaluated from 0 to t]

You have another inductor, L2, that has this same voltage applied across it. What is the current i2(t)?

i2(t) = i2(0) + (1/L2)·[∫v(τ) dτ evaluated from 0 to t]

Solving for the integral, we get

[∫v(τ) dτ evaluated from 0 to t] = L2(i2(t) - i2(0))

Plugging this back into the expression for i1(t), we get

i1(t) = i1(0) + (1/L1)·[L2(i2(t) - i2(0))]

i1(t) = i1(0) + (L2/L1)·[i2(t) - i2(0)]

i1(t) = (L2/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ]

Similarly, we have for i2(t)

i2(t) = (L1/L2)·i1(t) + [i2(0) - (L1/L2)·i1(0) ]

Since

i(t) = i1(t) + i2(t)

i(t) = (L2/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ] + i2(t)

i(t) = ((L1 + L2)/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ]

So i2(t) in terms of i(t) is

i2(t) = [L1/(L1 + L2)]{i(t) - [i1(0) - (L2/L1)·i2(0) ] }

i2(t) = [L1/(L1 + L2)]·i(t) + [L2·i2(0) - L1·i1(0)]/(L1 + L2)

Can you NOW finally see what I have been saying over and over?

You are insisting on assuming that the initial currents in the inductors are either zero or, at the very least, are in the same proportion as the inductances.

This is simply NOT TRUE for this problem. The initial conditions are very explicitly given and they simply are NOT in this proportion!

1) I really hope you took some time to see my attempt in post 45 as i wrote it clearly
2) i understand the math and i am working by the math and not the division rule
3) are you refering to the question i posted when giving the explanation because the voltage across l1 and l1 is not v(t)

 

Hi,

Really? Then how did i get the right result using that idea? Is the book wrong?

Please show how you got the book's result using that idea, since the book's result is completely at odds with your claim.



Show how that idea leads to:

i2(t) = (-0.4 + 1.2e^(-2t/sec)) A

If the book's answer is correct, then

i(t) = (-0.4 + 1.8e^(-2t/sec)) A

Since i1(t) is given as

i1(t) = 0.6e^(-2t/sec)) A

And since you claim that

i2(t) = i(t) * L1/(L1+L2)

Please show how

i2(t) = (-0.4 + 1.2e^(-2t/sec)) A = [(-0.4 + 1.8e^(-2t/sec)) A ]* [L1/(L1+L2)]

 

1) I really hope you took some time to see my attempt in post 45 as i wrote it clearly
2) i understand the math and i am working by the math and not the division rule
3) are you refering to the question i posted when giving the explanation because the voltage across l1 and l1 is not v(t)

1) It is much clearer -- I'll try to take a look at it when I have some time (more than I do right now).
2) Good
3) I'm talking about the original problem. This is why we strongly recommend NOT talking about multiple problems in the same thread -- it leads to confusion and chaos.

 

Please show how you got the book's result using that idea, since the book's result is completely at odds with your claim.

View attachment 166888

Show how that idea leads to:

i2(t) = (-0.4 + 1.2e^(-2t/sec)) A

If the book's answer is correct, then

i(t) = (-0.4 + 1.8e^(-2t/sec)) A

Since i1(t) is given as

i1(t) = 0.6e^(-2t/sec)) A

And since you claim that

i2(t) = i(t) * L1/(L1+L2)

Please show how

i2(t) = (-0.4 + 1.2e^(-2t/sec)) A = [(-0.4 + 1.8e^(-2t/sec)) A ]* [L1/(L1+L2)]

The book’s answer is right. I solved it. If you want a detailed answer i would do it.
I first tried the way @MrAl mentioned using division rule, but it is not right.

 

Doesn't matter. These equations do NOT hold for this problem because this problem does not abide by the assumptions that are required to be true in order for those equations to apply!

DO THE MATH!

Or, at the very least, explain why the provided solutions to the problem do not satisfy the relationships that you keep insisting hold. If the solutions are wrong, then show that they are wrong. If they are not wrong, then how can you keep insisting that the current divides as you say it does here?

Since doing the math seems to be asking way too much, at least follow along as I do it, step by step.

You have an inductor, L1, that has a voltage v(t) applied across it. What is the current i1(t)?

For an inductor in general:

v(t) = L di(t)/dt

so

i(t) = ∫v(t)/L

i1(t) = i1(0) + (1/L1)·[∫v(τ) dτ evaluated from 0 to t]

You have another inductor, L2, that has this same voltage applied across it. What is the current i2(t)?

i2(t) = i2(0) + (1/L2)·[∫v(τ) dτ evaluated from 0 to t]

Solving for the integral, we get

[∫v(τ) dτ evaluated from 0 to t] = L2(i2(t) - i2(0))

Plugging this back into the expression for i1(t), we get

i1(t) = i1(0) + (1/L1)·[L2(i2(t) - i2(0))]

i1(t) = i1(0) + (L2/L1)·[i2(t) - i2(0)]

i1(t) = (L2/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ]

Similarly, we have for i2(t)

i2(t) = (L1/L2)·i1(t) + [i2(0) - (L1/L2)·i1(0) ]

Since

i(t) = i1(t) + i2(t)

i(t) = (L2/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ] + i2(t)

i(t) = ((L1 + L2)/L1)·i2(t) + [i1(0) - (L2/L1)·i2(0) ]

So i2(t) in terms of i(t) is

i2(t) = [L1/(L1 + L2)]{i(t) - [i1(0) - (L2/L1)·i2(0) ] }

i2(t) = [L1/(L1 + L2)]·i(t) + [L2·i2(0) - L1·i1(0)]/(L1 + L2)

Can you NOW finally see what I have been saying over and over?

You are insisting on assuming that the initial currents in the inductors are either zero or, at the very least, are in the same proportion as the inductances.

This is simply NOT TRUE for this problem. The initial conditions are very explicitly given and they simply are NOT in this proportion!

Hi again,

Hey thanks for the detailed info.
We can talk about this in PM if you prefer. Maybe i have misunderstood the starting conditions?

WAIT, didnt we move on to the next exercise?


Here are the fundamental reasons for my calculation:
What i read was that everything was zero to begin with, and that means that the current values were attained by the driving source.
That means that the SAME voltage source MUST have appeared across BOTH inductors in order to attain that initial current. Thus the initial current in BOTH inductors can be calculated. Since they both must follow v=L*di/dt, the INITIAL currents must follow that also.
When i use this reasoning, i get the 'book' answers.

So are you saying that the book answers are wrong too?

[LATER]
I'll go over this again asap and see what i did that was different.

 

Please show how you got the book's result using that idea, since the book's result is completely at odds with your claim.

View attachment 166888

Show how that idea leads to:

i2(t) = (-0.4 + 1.2e^(-2t/sec)) A

If the book's answer is correct, then

i(t) = (-0.4 + 1.8e^(-2t/sec)) A

Since i1(t) is given as

i1(t) = 0.6e^(-2t/sec)) A

And since you claim that

i2(t) = i(t) * L1/(L1+L2)

Please show how

i2(t) = (-0.4 + 1.2e^(-2t/sec)) A = [(-0.4 + 1.8e^(-2t/sec)) A ]* [L1/(L1+L2)]

Hi,

Yeah i moved on to the next problem.
For that problem take a look at L2/L1 for example but i'd have to go back to that one.

 

The book’s answer is right. I solved it. If you want a detailed answer i would do it.
I first tried the way @MrAl mentioned using division rule, but it is not right.

Hi,

Ok i'll have to go back to that problem and look at it again.
I must have done something different than you guys did.