# Mixed RL circuit in series and parallel

Discussion in 'Homework Help' started by fido_circuit, Aug 3, 2016.

1. ### fido_circuit Thread Starter New Member

Aug 3, 2016
6
0
Hello

I am trying to solve the circuit below:

Are these equations above correct according to the diagram? If yes how do I find solve I1?

2. ### DGElder Member

Apr 3, 2016
348
87
They are correct. Let's see your best effort at solving them.

3. ### WBahn Moderator

Mar 31, 2012
18,094
4,920
It's impossible to say whether they are correct according to the diagram because your diagram doesn't define many of the terms used by the equations. For instance, your equations use u, i1, i2, and i3 yet these are never defined on your diagram. While we can make pretty good guesses what you mean by these terms, that is exactly all they are -- good guesses. And while it is one thing to guess that the i3 is the current in R3, that is only part of the definition -- in what direction is i3 defined?

Engineering is not about guessing. You need to clearly define all of the variables and parameters you use.

4. ### fido_circuit Thread Starter New Member

Aug 3, 2016
6
0
Here is a new diagram:

The direction of the current is from plus to minus.

I tried to solve these equation:

Is this the right way to solve the equations?

Aug 3, 2016
6
0
i_m=i_3

6. ### DGElder Member

Apr 3, 2016
348
87
You still have not unambiguously identified the direction of the currents. Quit dragging around Im, you only need the two loop current variables, and two differential equations. Im is not an independent variable it is just the difference of the two loop currents.

Last edited: Aug 4, 2016
7. ### fido_circuit Thread Starter New Member

Aug 3, 2016
6
0
The directions of the current is from V1 through R1 and L1 and again through R2 and L2.

But the my two equations are dependent on Im. How do I have to write my two equations without Im?

8. ### WBahn Moderator

Mar 31, 2012
18,094
4,920

For currents, the best way to show polarity is with an arrow showing the direction the current is flowing. Doing so on a computer-drawn schematic can be difficult, but if you can add text annotation you can do a pretty good job. For horizontal branches use "--->" and "<---". For vertical branches stack pipe symbols, '|', and use '^' and 'v' for the arrow heads.

9. ### WBahn Moderator

Mar 31, 2012
18,094
4,920
You can use that last equation, Im = I1 - I2, to eliminate it two have two diffy-Qs in two unknowns.

10. ### fido_circuit Thread Starter New Member

Aug 3, 2016
6
0
Okay thanks for the advice regarding current directions

Is the following equations correct?
0=u(t)-r1*i1(t)-l1*di1(t)/dt-rm*(i1(t)-i2(t))

0=rm(i1(t)-i2(t))-r2*i2(t)-l2*di2(t)/dt​

The last equation (im=i1-i2) has been substituted in the two equations

11. ### MrAl Distinguished Member

Jun 17, 2014
2,573
523
Hello,

They look right but you need to put the multiplication sign after the 'rm' in the second equation.
Also, if you have "R3" in the schematic then you should have R3 in the equations, not 'rm'.
You can also write these a little neater if you get rid of all the references to 't' because time is understood. This however is subject to your instructors wishes.
0=u-R1*i1-L1*di1-R3*(i1-i2)
0=R3*(i1-i2)-R2*i2-L2*di2

But one question, are you really solving for an actual pulse or just a step input?

Last edited: Aug 5, 2016
12. ### fido_circuit Thread Starter New Member

Aug 3, 2016
6
0
I want to solve these equation which ables me to see how the step response are for i1 and i2 are when u(0)=10V.
If R3 instead was an L3 inductor is the procedure the same? Substitute Im=i1-i2 into the two equation like we did earlier?

0=u-R1*i1-L1*di1-L3*(di1-di2)
0=R3*(di1-di2)-R2*i2-L2*di2

The reason why I ask is because then I am able to understand the concept of substitution with differential equations.

13. ### MrAl Distinguished Member

Jun 17, 2014
2,573
523
Hi,

That looks right, except for the second equation where there is still an R3 in there, so i think you meant to write this:
0=u-R1*i1-L1*di1-L3*(di1-di2)
0=L3*(di1-di2)-R2*i2-L2*di2

You really dont have to write three equations though to start with. You can just draw two clockwise loops inside each inner part of the circuit. So i1 would flow through R1, L1, and R3 in that order, and i2 would flow through R3, R2, and L2 in that order. That gives you the two equations.
If you want to do it with three equations though that's up to you. Then just reduce to two again.