Finding Currents in a Circuit: Kirchhoff's Law

[Blu3eyes]

Homework Statement

Find the currents
[PLAIN]http://img695.imageshack.us/img695/7042/78217840.png

Homework Equations

Using Kirchhoff's Law, Loop, Junction rule

The Attempt at a Solution

The problem here is the inside circuit .I try to solve it as one emf, one resistance, but honestly I do not know how to start.
Any suggestions would be greatly appreciated.

 

[Delphi51]

I don't think you can replace the inner loop with a simple equivalent circuit.
Do the sum of potentials around the three loops.
I5 = I2 + I3.
I4 relates simply to the other I's, so you have only 3 unknowns.

 

[Blu3eyes]

I don't think you can replace the inner loop with a simple equivalent circuit.
Do the sum of potentials around the three loops.
I5 = I2 + I3.
I4 relates simply to the other I's, so you have only 3 unknowns.

Can you please be a little more specific?
I understand I5=I23=I2 + I3 (series circuit -> same current)

Do the sum of potentials around the three loops.

But how should I deal with the inner circuit?
Emf2 and emf3 are in parallel -> same Voltage
V terminal: Vt2 = Vt3 ?

 

[Delphi51]

I don't understand "Vt2 = Vt3".
Define I2 and I3 both to the right. Do we still say the current comes out of the big end on the battery symbol?
Then -V2 + I2*R2 - I3*R3 + V3 = 0
where V2 is the potential on emf2 and V3 is the potential across emf3.

 

[Blu3eyes]

I don't understand "Vt2 = Vt3".
Define I2 and I3 both to the right. Do we still say the current comes out of the big end on the battery symbol?
Then -V2 + I2*R2 - I3*R3 + V3 = 0
where V2 is the potential on emf2 and V3 is the potential across emf3.

Sorry I am not familiar with conventional current, but I am kinda getting the idea, (I believe I could do it with electron current.)
So, I would say I5= -V2 + I2*R2 - I3*R3 + V3. and then have 3 equations with 3 unknowns I1, I5, and I4.
Solve for these first and then come back to the inner circuit.

 

[gneill]

Do you have numerical values for the components, or are you expected to solve for all the currents symbolically?

 

[Delphi51]

would say I5= -V2 + I2*R2 - I3*R3 + V3

units of current on the left, voltage on the right - can't be correct

Why not I5 = I2 + I3?

For the inner loop, the idea is to write that the sum of the potentials around the loop is zero. By all means use electron current directions. Be sure you are clear on which direction you define each current as going so you get the right signs on all the terms.

 

[Blu3eyes]

Do you have numerical values for the components, or are you expected to solve for all the currents symbolically?

Yes, I do have numerical values for resistors and emfs.
[tex]\epsilon1[/tex]=8V,[tex]\epsilon2[/tex]=12V,[tex]\epsilon3[/tex]=24V,[tex]\epsilon4[/tex]=16V

R1=150[tex]\Omega[/tex], R2=250[tex]\Omega[/tex], R3=300[tex]\Omega[/tex], R4=600[tex]\Omega[/tex], R5=350[tex]\Omega[/tex]

 

[gneill]

You have a choice of methods to solve this circuit. For example, you could transform the inner loop to a Thevenin equivalent, as you were hinting at earlier. Then you would have two loop equations to write to solve for the rest.

The problem with this method is that your stated objective is to find all the currents through all the components. If you 'transform away' some of the components, they will no longer be 'accessible' for analysis. I suppose you could solve for the net voltage across the in-circuit Theveninized loop and then convert back to the original components and analyze that portion separately. Sounds like a lot of extra work though.

Instead, just write three KVL loop equations, one for each loop. Three equation, three unknowns. Use the loop currents to find the individual component currents.

 

[Blu3eyes]

Here is how I did with junction, loop rules:
[PLAIN]http://img101.imageshack.us/img101/7042/78217840.png
Junction C:
I₁ + I₂ + I₃ +I₄ + I₅=0
I₁ + I₂ + I₃ +I₄ + I₂ + I₃ = 0 (I₅= I₂ + I₃)
I₁ +2I₂ +2I₃ +I₄=0


Loop ABCHGF:
8 - 150I₁ + 250I₂ -12 -350I₅=0
-150I₁ -100I₂ -350I₃=4


Loop FJICDE:
-350I₅ -24 -300I₃ +600I₄ +16=0
-350I₂ -650I₃ +600I₄=8

Loop GHIJ:
12 -250I₂ +300I₃ +24=0
-250I₂ +300I₃ = -36



I came up with 4 equations, 4 unknowns. Not sure how you did with 3 equations, 3 unknowns.

 

[gneill]

But I5 = I2 + I3. Currents don't materialize out of nowhere. Your junction C equation seems to be in trouble.

 

[Blu3eyes]

But I5 = I2 + I3. Currents don't materialize out of nowhere. Your junction C equation seems to be in trouble.

I don't quite get you there

 

[gneill]

Three loops, as shown in the figure. In order to determine the individual currents through the various components, determine which of the loop currents pass through that component and sum them (according to direction).

[tex] \left(\begin{array}{ccc}R1 + R2 + R5&-R2&-R5\\-R2&R2 + R3&-R3\\-R5&-R3&R3 + R4 + R5\end{array}\right) \left(\begin{array}{c}i1&i2&i3\end{array} \right) = \left(\begin{array}{c}-E1 + E2&-E2 - E3&E3 - E4\end{array}\right)[/tex]

Yields

[tex] \left(\begin{array}{c}i1&i2&i3\end{array}\right) = \left(\begin{array}{c}-41.979&-100.63&-29.505\end{array}\right) mA [/tex]

 

[gneill]

I don't quite get you there

You've written that at junction C the sum of the currents is

I1 + I2 + I3 +I4 + I5=0

But clearly I5 = I2 + I3 from your schematic.

 

[Blu3eyes]

On the second and third equations, is it:
-R2 "+" R2 + R3 - R3
-R5 - R3 "+" R3 +R4 +R5

And I really don't understand how you came up with these equation.
1st equation: where did the extra -R2 and -R5 come from?
To finish the loop , shouldn't (R1 + R2 +R5)i1= -E1 +E2 is enough to finish the loop?
The same with 2nd and 3rd equations.

 

[gneill]

On the second and third equations, is it:
-R2 "+" R2 + R3 - R3
-R5 - R3 "+" R3 +R4 +R5

I cast the equations into matrix form. It's a 3 x 3 resistance matrix multiplied by a vector of currents, set equal to a voltage vector. In short, R*I = V. To extract the individual equations, multiply the resistance matrix by the current vector. The first equation would be

(R1 + R2 + R5)*i1 - R2*i2 - R5*i3 = -E1 + E2

 

[gneill]

On the second and third equations, is it:
-R2 "+" R2 + R3 - R3
-R5 - R3 "+" R3 +R4 +R5

And I really don't understand how you came up with these equation.
1st equation: where did the extra -R2 and -R5 come from?
To finish the loop , shouldn't (R1 + R2 +R5)i1= -E1 +E2 is enough to finish the loop?
The same with 2nd and 3rd equations.

Take the first equation, which represents the first loop for current i1. If you look at the current loops that I drew, i2 and i3 both contribute, passing through R2 and R5 respectively.

 

[Blu3eyes]

Can I assume that you used conventional current on the resistors?
Currents come in with (+) and out with (-)

 

[gneill]

Can I assume that you used conventional current on the resistors?
Currents come in with (+) and out with (-)

I always use conventional current. It makes it sooooo much easier to converse with colleagues that way. It's the latest thing. In fact, it's been the latest thing for a couple of hundred years now!

 

[Blu3eyes]

I always use conventional current. It makes it sooooo much easier to converse with colleagues that way. It's the latest thing. In fact, it's been the latest thing for a couple of hundred years now!

Unfortunately, my physics course is taught with electron current. Yes, I was introduced conventional current but it was like 10 minutes lecture. Assignments are done using electron current, lectures are same.

 

[Blu3eyes]

The equations make sense to me now.
So to figure out the individual currents:
I1=i1
I4=i3
I5=i1+i3
I2=i1+i2
I3=i2+i3

 

[gneill]

The equations make sense to me now.
So to figure out the individual currents:
I1=i1
I4=i3
I5=i1+i3
I2=i1+i2
I3=i2+i3

If you look at the figure, the currents i1, i2, and i3 may run in different direction through any given component. So be sure to 'add' appropriately. Pick a direction that you want to assign to the current through the component, and add the current that goes in that direction and subtract the current that goes in the opposite direction.

So, suppose I decided that I wanted to publish the value of the current through R2 in the "right to left" direction through the resistor. Then I would say that the current is i1 - i2.

Good?

 

[Blu3eyes]

If you look at the figure, the currents i1, i2, and i3 may run in different direction through any given component. So be sure to 'add' appropriately. Pick a direction that you want to assign to the current through the component, and add the current that goes in that direction and subtract the current that goes in the opposite direction.

So, suppose I decided that I wanted to publish the value of the current through R2 in the "right to left" direction through the resistor. Then I would say that the current is i1 - i2.

Good?

I'm still not sure. Does the battery do anything with the +,- here?

 

[gneill]

I'm still not sure. Does the battery do anything with the +,- here?

Nope. All the influences of the sources were taken into account when the equations were determined.

 

[Blu3eyes]

Like you said, pick R2, direction from right to left, R2 would be (+)R2(-)
i1, which is clockwise, hits the (-) of R2. On the other hand, i2 hits the (+) of R2.
i1 goes the opposite direction with which I chose on R2. i2 goes the same direction with R2. If I was right, it should be i2-i1

 

[gneill]

Like you said, pick R2, direction from right to left, R2 would be (+)R2(-)
i1, which is clockwise, hits the (-) of R2. On the other hand, i2 hits the (+) of R2.
i1 goes the opposite direction with which I chose on R2. i2 goes the same direction with R2. If I was right, it should be i2-i1

Sorry, I don't understand your nomenclature.

If you want to quote a (conventional) current I passing through resistor R2 that is flowing from right to left, then it will be I = i1 - i2.

The two loop currents i1 and i2 (and i3 as a matter of fact) are both clockwise currents. i1 flows through R2 from right to left. i2 flows through i2 from left to right.

Now, when the equations are solved the values for the currents i1, i2, and i3, all turned out to be negative. No problem, that just means that the assumed directions (clockwise ) was not what they turned out to be. This is not a problem. The actual currents are still determined by the simple expedient of the appropriate sums. No conversion or interpretation is required.

So, let's once again take the current through R2 as an example. I've calculated that i1 = -41.979mA, and i2 = -100.62 mA. i1 flows right to left through R2, and i2 flows left to right. That makes the "right to left" current though R2 [-41.979 - (-100.62)]mA = 58.65 mA. R2 is passing 58.65 mA, from right to left, of conventional current.