# Circuit with serie-parallel inductors

#### N0t_S0_Sm4rt

Joined Apr 6, 2019
10
Hi!

I need a help with calculating/solving for VL4 (voltage across inductor L4) in circuit attached as file.
I know L1, L2, L3, L4, V and frequency.
Anytime I try to solve it by myself, I'm stuck at point where I solved total current (I which should be 2,49A by my calculations). But what now?

Little bit of background: I got this circuit at exam, ended up with F, I need to repair my grade, otherwise I won't pass...
Any help appreciated!

#### Attachments

• 42.2 KB Views: 24

#### WBahn

Joined Mar 31, 2012
25,309
You need to show your best attempt to answer the question. That's the only way we can possibly see where you are going wrong so that we can help you overcome it.

#### N0t_S0_Sm4rt

Joined Apr 6, 2019
10
You need to show your best attempt to answer the question. That's the only way we can possibly see where you are going wrong so that we can help you overcome it.
Well, my best attempt so far was:
1) Solve L234 parallel - L2 * L3 / L2 + L3 ... L23 * L4 / L23 + L4\
2) Solve XL total = angular frequency * L
3) Solve for I current total = V total / XL total
I just don't know what should I do now, in order to solve VL4.... I think, I need to know IL4 and then do the Ohm's law, IL4 * XL4, but the problem is, what size or how to calculate IL4...

#### LesJones

Joined Jan 8, 2017
2,533
Can you calculate the voltage across L4 ?

Les.

#### Dodgydave

Joined Jun 22, 2012
8,880
Solve the total current like you said, then you can work out the Voltage across the parallel inductors, and use Ohms law for each parallel inductor to give you their own current.

#### WBahn

Joined Mar 31, 2012
25,309
Well, my best attempt so far was:
1) Solve L234 parallel - L2 * L3 / L2 + L3 ... L23 * L4 / L23 + L4\
2) Solve XL total = angular frequency * L
3) Solve for I current total = V total / XL total
I just don't know what should I do now, in order to solve VL4.... I think, I need to know IL4 and then do the Ohm's law, IL4 * XL4, but the problem is, what size or how to calculate IL4...
I have no idea what

L2 * L3 / L2 + L3 ... L23 * L4 / L23 + L4\

means.

I can guess, but I shouldn't have to guess about how to evaluate a mathematical expression, should I?

Note that the first part of this is

L2 * L3 / L2 + L3 = ((L2 * L3) / L2) + L3 = 2*L3

You need to pay attention to order of operations. This WILL bite you.

I think what you are trying to say is

L234 = L2 || L3 || L4 = L23 || L4
L23 = L2 || L3 = (L2 * L3) / (L2 + L3)
L234 = L23 || L4 = (L23 * L4) / (L23 + L4)

If you have an equivalent circuit with L1 and L234, what is the relationship between the voltage across L234 and the voltage across L4 in the original circuit?

#### N0t_S0_Sm4rt

Joined Apr 6, 2019
10
I have no idea what

L2 * L3 / L2 + L3 ... L23 * L4 / L23 + L4\

means.

I can guess, but I shouldn't have to guess about how to evaluate a mathematical expression, should I?

Note that the first part of this is

L2 * L3 / L2 + L3 = ((L2 * L3) / L2) + L3 = 2*L3

You need to pay attention to order of operations. This WILL bite you.

I think what you are trying to say is

L234 = L2 || L3 || L4 = L23 || L4
L23 = L2 || L3 = (L2 * L3) / (L2 + L3)
L234 = L23 || L4 = (L23 * L4) / (L23 + L4)

If you have an equivalent circuit with L1 and L234, what is the relationship between the voltage across L234 and the voltage across L4 in the original circuit?
Yeah, that was exactly what I was trying to say...
I don't know? Have you seen the attached file? All I was told is described here, and then just asked to solve for VL4.
Just my guess - since voltage total is 50V, isn't voltage across L234 just: Voltage total - Voltage L1
Well, I moved a bit forward, since I know VL234 (voltage across L234). But still, this doesn't solve the problem .
I think, my problem is I don't know to divide the current between L2, L3 and L4.
=>
However I guess, if I would know how to divide the current between them, I should be able to divide the voltage across VL234 to each of them...
By the way thank you for help, I know it's tough with me.

#### ericgibbs

Joined Jan 29, 2010
9,575
hi,
What numerical value do you calculate for the Total circuit impedance.?....

Current thru the total circuit.? ...

What do you calculate the voltage to be across the paralleled inductor section.?...

Lets see some numbers.

E

#### N0t_S0_Sm4rt

Joined Apr 6, 2019
10
hi,
What numerical value do you calculate for the Total circuit impedance.?....

Current thru the total circuit.? ...

What do you calculate the voltage to be across the paralleled inductor section.?...

Lets see some numbers.

E
Total circuit impedance -yikes, haven't been paying attention in school, sorry, didn't think of this...
Voltage across the parallel inductor section = V total - VL1 = 25V
- I used different values for this measurement, as L1 = 5H, L2 = 20 H, L3 = 10 H, L4 = 10 H, V total = 50V , f = 800 Hz

#### N0t_S0_Sm4rt

Joined Apr 6, 2019
10
Oh jeez. It seems like, I forgot the meaning of "parallel" , because voltage for parallel section is the same among all the parts... Well, thank you guys for help! I better read all my old notes from the beginning

Last edited:

#### MisterBill2

Joined Jan 23, 2018
5,250
You could also do this by super-position. The current through L1 is the sum of the currents through L2, L3, and L4. Only 2 calculations because L3=L4. And inductors in series add just like resistors in series. So then add the three currents and find the voltage across L1. Subtract that from "V" and that is the voltage across the other three in parallel. No complicated math to do it this way. And it is a good application of Kirchov's voltage law.

#### WBahn

Joined Mar 31, 2012
25,309
You could also do this by super-position. The current through L1 is the sum of the currents through L2, L3, and L4. Only 2 calculations because L3=L4. And inductors in series add just like resistors in series. So then add the three currents and find the voltage across L1. Subtract that from "V" and that is the voltage across the other three in parallel. No complicated math to do it this way. And it is a good application of Kirchov's voltage law.
You seem to be saying that to find the current in, say, L2 you can ignore L3 and L4 and treat L1 and L2 as though they are in series. Is that really what you are trying to say?

#### MisterBill2

Joined Jan 23, 2018
5,250
You seem to be saying that to find the current in, say, L2 you can ignore L3 and L4 and treat L1 and L2 as though they are in series. Is that really what you are trying to say?
Yes, but not forgetting to add the three currents flowing in R1.

#### WBahn

Joined Mar 31, 2012
25,309
Yes, but not forgetting to add the three currents flowing in R1.
This won't work -- and this isn't what superposition is all about.

Let's consider a simple resistor circuit in which we have R1 in series with the parallel combination of R2 and R3. This resistive network is then driven by a 60 V supply and we want to know the voltage across R2.

For simplicity we will say that R1 is 1000 Ω and R2 and R3 are both 2000 Ω.

Clearly the parallel combination of R2 and R3 is 1000 Ω and so the total current flowing from the supply is 30 mA and the voltage across R2 is 10 V. The currents are I(R1) = 30 mA, I(R2) = I(R3) = 15 mA.

Now let's do it they way you are suggesting.

Ignore R2 and we have R1 in series with R3 for a total of 3000 Ω for a total current of 20 mA.

When we ignore R3 we have R1 in series with R2 for a total of, again, 3000 Ω and a current of 20 mA.

You are now saying that the total current in R1 is 40 mA and the total current in R2 and R3 are each 20 mA. But this makes the voltage across R1 40 V and the voltage across R2 and R3 also 40 V.

The situation gets far worse if R2 and R3 are not the same, since now you will get that the voltages across these two parallel elements are not the same.

Superposition says that, in a linear circuit, the currents and voltages in the circuit are linear combinations of the voltage and currents due to each of the independent supplies individually (where the output of the other supplies is set to zero).

#### MisterBill2

Joined Jan 23, 2018
5,250
Each of the three parallel resistors has the same voltage across them, and each of those three has a current flowing through it, and R1 So the total current through R1 must equal the sum of the currents through all three of the parallel resistors. I am only talking about currents, not voltages. The current through R1 equals the current flowing through R2 plus the current flowing through R3 plus the current flowing through R4. The voltage drop would be different for each, but my solution is not using those voltage drops, only the sums of the currents.
But since L3=L4=0.01H each, only one reactance needs to be calculated , the parallel XL is half that, in parallel with XL2. So the parallel is XL2 x XL34/ XL2+XL34 .
So then the voltage across L4=voltage across L3= voltage across L2 =V times XLparallel3/ XL1+XLparallel3.
And that answer will be the same as arrived at from the other method.

#### WBahn

Joined Mar 31, 2012
25,309
Each of the three parallel resistors has the same voltage across them, and each of those three has a current flowing through it, and R1 So the total current through R1 must equal the sum of the currents through all three of the parallel resistors. I am only talking about currents, not voltages. The current through R1 equals the current flowing through R2 plus the current flowing through R3 plus the current flowing through R4. The voltage drop would be different for each, but my solution is not using those voltage drops, only the sums of the currents.
But since L3=L4=0.01H each, only one reactance needs to be calculated , the parallel XL is half that, in parallel with XL2. So the parallel is XL2 x XL34/ XL2+XL34 .
So then the voltage across L4=voltage across L3= voltage across L2 =V times XLparallel3/ XL1+XLparallel3.
And that answer will be the same as arrived at from the other method.
What you are describing here does not involve superposition at all. Assuming that you correct your errors associated with order of operations, you simply have the normal way of analyzing a series-parallel circuit.