# [Circuit theory] Help me solve this mathematically

Discussion in 'Homework Help' started by ZeroTorrent, Apr 5, 2012.

1. ### ZeroTorrent Thread Starter New Member

Mar 28, 2012
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Hi, the problem is attached and i'm trying to solve it and despite me doing circuit theory for the past weeks, this problem has me stumped.

I want to calculate the current I over R1 and R2.

Regarding R2 using the loop current method.

Why won't U3 give any current to R2? This has me stumped.

Taking the U2 in a loop counter-clock I'd get:

U2-I2*R2 + U3*R2 = 0

Is it so easy that the current on the negative side of U3 is zero because we have it grounded there? So:

U2-I2*R2 +0*R2 = 0 ?

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2. ### vvkannan Active Member

Aug 9, 2008
138
11
As U2 is parallel to R2 the voltage across R2 is U2 and you can divide U2/R2 to get the current through it .Since U2 is larger than U3 it forces current through R3,R4 and that is equal to the difference between the U2 and U3 divided the R2+R3

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3. ### ZeroTorrent Thread Starter New Member

Mar 28, 2012
10
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Hmm i'm not quite sure I understand. Could you show me using current loop method, please? To me nothing about circuit theory is intuitive and it never will be, I can only know or see what i'm doing through pure math.

4. ### vvkannan Active Member

Aug 9, 2008
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5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Since U2 is the the highest voltage in this circuit.
The situation will look like this

U2 = I2 * R2
U2 = U1 + I1*R1
U2 = U3 + I*(R3 + R4)

Because all these components are connected in parallel

So we have

I2 = U2/R2

And so on

I1 = ?

I = ?

And look at the picture that I attached ( post 17 and 19)

And also can I ask you about you country of origin ? I am curious because you use letter "U" for voltage instead of "V".
Just like me in Poland.

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Mar 28, 2012
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Thank you!

7. ### studiot AAC Fanatic!

Nov 9, 2007
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You may be confused here.

Kirchoff's Methods do not involve a loop current. That is reserved for Maxwell's Mesh Method.

So either use K's laws and node currents, as others have offered.

or

Use the mesh method and loop currents. You should note that loop currents are fictitious and must be combined to obtain real currents in actual components.

Don't mix them up

However the mesh method is directly applicable to AC theory using impedances/admittances and can be directly written in standard matrix format so that it can be assembled for input to a computer. This is why it is a particularly important method.

I have appended the beginning of the mesh solution. If you wish to continue see if you can solve the simultaneous equations (easy) before we proceed.

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8. ### WBahn Moderator

Mar 31, 2012
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It is better to say "the current through R1 and R2". Generally, you want to talk about the current through a device and the voltage across a device.

What makes you say that it won't? I'm assuming that someone told you that, perhaps as a hint. Even if they did (and even if it was the instructor), don't assume that it doesn't until you can show mathematically that it doesn't.

You want to get in the habit of always checking units. In this case, your first term has units of volts, meaning that all of the other terms being added (or subtracted) along with it MUST have units of volts. The second term has units of (amps)(ohms) which if volts, so life is good so far. But the third term has units of (volts)(ohms), which may or may not have physical meaning, but you know it is NOT volts. Hence you KNOW this equation is WRONG and there is no point proceeding until it is corrected.

Most (not all, but most) errors you make will affect the units in some way. Hence, ensuring that the units work out correctly will permit you to catch the majority of your errors. It won't guarantee a correct answer, but if the units don't work you are guaranteed an incorrect answer.

You have no basis for this assertion; in other words, no, it isn't so easy.

A point you need to keep in mind is that loop-current analysis (more commonly referred to as mesh current analysis, at least in the U.S.) is nothing more than a formalized application of KVL that permits you to quickly write down a set of simultaneous linear equations to solve the currents in a circuit. As such, you almost always need ALL of the equations to solve ANY of them, otherwise, you will have more unknowns than you have equations. There exist rare exceptions to this rule and, as it happens, this circuit is one of them. The fact that the U2 supply forces a constant voltage across the segment shared by two of the loop currents separates the circuit to the left of it from the circuit to the right of it. That's an important point for you to see and understand: changing things to the left of U2 will have no impact on what happens to the right of it and vice versa. Do you see that?

This circuit is actually better analyzed using Node Voltage Analysis, but I suspect the person that prepared it knows that and is using it specifically to let you get experience working problems for which other techniques are better suited so that you can start developing a feel for when to use one method over another.

9. ### studiot AAC Fanatic!

Nov 9, 2007
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Good morning WBahn.

Mesh analysis is not a form of KVL and should not be attributed to him.

10. ### panic mode Senior Member

Oct 10, 2011
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it's nothing but formalized Ohm's law ;p

be systematic, mark nodes and loops and write equations.
for example mark "A" the top side of U2, mark "B" the left side od R5, mark "C" the right side of R5.

make all currents through resistors arbitrary, for example top to bottom (for vertical resistors) and left to right (for horiz.). supposedly all currents have same index as resistor they flow through.

I1 then must be top to bottom of R1. voltage difference is Va-U1 which is same as U2-U1, hence
I1=(U2-U1)/R1

next one is
I2=(U2-GND)/R2

where GND=0V so we get I2=U2/R2

I3=(Va-Vb)/R3

I4=(Vb-U3)/R4

I5=0 because loop is open and Vc=Vb so
I5=(Vc-Vb)/R5=0/R5=0

This means that I3=I4 (KCL at "B").

11. ### studiot AAC Fanatic!

Nov 9, 2007
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What is ???????

Oct 10, 2011
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13. ### studiot AAC Fanatic!

Nov 9, 2007
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Maxwells Mesh Method is neither a direct application of Ohms or Kirchoffs laws.

It is a mathematical trick to make the arithmetic easy and systematic, based on the fact that in linear circuit analysis we can employ a change of basis (variable) to put the arithmetic in the most convenient form.

This change of variable involves the introduction of fictitious currents, known as the loop currents. None of these loop current are the actual currents flowing except in special cases.

14. ### panic mode Senior Member

Oct 10, 2011
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erm, it was supposed to be a joke (hint - smiley face...). apparently it didn't work on you...

15. ### studiot AAC Fanatic!

Nov 9, 2007
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I'm sorry I still don't see the joke (but keep trying, I like a laugh even at my expense).

The point I am trying to make here is that the OP referred to loop currents.

This can only mean the particlar method I referred to and I am concerned that the OP is mixing up several different methods, which is why I have not posted a detailed description of the loop current method.

I will certainly post it if asked, but I do not wish to add further to the confusion at this stage.

Meanwhile several of the other methods have been proposed and I have referred the OP to these if he wants them.

16. ### panic mode Senior Member

Oct 10, 2011
1,662
472
your comment was absolutely correct (mesh and K-loops are different methods), i just found it cute how clean, almost pedantic response was on distinction between the two, along with giving credit to the correct scientist. i was sure that crediting Ohm for everything (and giving no credit to Kirchoff and Maxwell) would make tease obvious and not provoke similar response. i just found it tempting to find out (sorry).

please forgive me, i'll try to straighten up now...

17. ### studiot AAC Fanatic!

Nov 9, 2007
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522

filler filler filler

18. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Am I missing something here? Why are you multiplying U3 b R2?

If I saw this problem, I would probably lean towards using the superposition theorum, which kills all sources but one at a time, and solves the circuit for that source. The process needs to be repeated with each of the other sources as well. Then, at the end, the results are combined to give you the final value.

19. ### studiot AAC Fanatic!

Nov 9, 2007
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Do you not think superposition would be quite onerous for a large complicated mesh with many sources?

(Edit Wbahn not Jony sorry) discussed your other question in his post 8.

Last edited: Apr 7, 2012
20. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,373
1,354
Yes, I suppose you're right about that....

Oops, sorry. I missed that.