Using Kirchhoff's Laws to Find Currents, Source Current and Power Dissipated

[tiny-tim]

So the right loop ie the 80Ω and 40Ω i3 and i2
=80i3 - 40i2 = 0 OR i2 = 2i3

Middle is the 20 and 40 ie i1 and i2. So
i1 = 2i2 40i2 - 20i1 = 0

yes and yes

so i₁ = 2i₂ = 4i₃

The left? Is there not voltage already there on the left as 10 volts is Vs.
So I5 = 2i1
10i5 - 20i1 = 0?

no

the arrows for i₅ and i₁ are both clockwise, so the IR is positive for both

also, you need to put the 10 V on the RHS …

10i₅ + 20i₁ = 10

(or is it -10? i can never remember which way round it is! )

first, any questions about that?

second, now try KVL for the outside loop (the one with V₅ i₅ and i₃) ​

 

[cschear87]

yes and yes

so i₁ = 2i₂ = 4i₃


no

the arrows for i₅ and i₁ are both clockwise, so the IR is positive for both

also, you need to put the 10 V on the RHS …

10i₅ + 20i₁ = 10

(or is it -10? i can never remember which way round it is! )

first, any questions about that?

second, now try KVL for the outside loop (the one with V₅ i₅ and i₃) ​

I thought that about the voltage but wasn't sure. So the right and middle loops have no voltage immediately in them so they have to balance out. But because on the left there's 10 volts going directly into that loop, the equation has to = 10? Is that right?
The outside loop is...
10i5 + 80i3 = 10?

 

[tiny-tim]

yup!

so we have i₁ = 2i₂ = 4i₃

10i₅ + 20i₁ = 10

10i₅ + 80i₃ = 10

(and if we do the other two loops, we would get:

10i₅ + 40i₂ = 10

20i₁ = 80i₃)

note that the outside loop gives us no new information …

from 10i₅ + 20i₁ = 10 and i₁ = 2i₂ = 4i₃,

we can get 10i₅ + 80i₃ = 10 anyway (without KVL) …

the moral is that you only need as many KVL equations as there are independent loops (in this diagram, there are 6 loops altogether, but only three are independent … so in this case you need exactly 3 KVL equations, and any more would be superfluous …

in this case, we chose 2 really easy loops, with V = 0, and only one loop with V ≠ 0, to keep it as simple as possible! )​

ok, now that should give you the answer to the first part of the question: what are the currents i₁ i₂ i₃ and i₅

any questions?

 

[cschear87]

yup!

so we have i₁ = 2i₂ = 4i₃

10i₅ + 20i₁ = 10

10i₅ + 80i₃ = 10

(and if we do the other two loops, we would get:

10i₅ + 40i₂ = 10

20i₁ = 80i₃)

note that the outside loop gives us no new information …

from 10i₅ + 20i₁ = 10 and i₁ = 2i₂ = 4i₃,

we can get 10i₅ + 80i₃ = 10 anyway (without KVL) …

the moral is that you only need as many KVL equations as there are independent loops (in this diagram, there are 6 loops altogether, but only three are independent … so in this case you need exactly 3 KVL equations, and any more would be superfluous …

in this case, we chose 2 really easy loops, with V = 0, and only one loop with V ≠ 0, to keep it as simple as possible! )​

ok, now that should give you the answer to the first part of the question: what are the currents i₁ i₂ i₃ and i₅

any questions?

So now that I have the equations and have some understanding, what exactly does that tell me about the currents? And how do I confirm this with power dissipated?

 

[tiny-tim]

(btw, there's no need to keep repeating the entire previous post! )

So now that I have the equations and have some understanding, what exactly does that tell me about the currents?

those equations tell you what the currents are, don't they?

And how do I confirm this with power dissipated?

write out the power equations

 

[cschear87]

So when I write out that
i1 = 40i2 + 20i1 = 10
i2 = 80i3 - 40i2 = 0
i3 = 40i2 - 80i3 = 0
i5 = 10i5 + 20i1 = 10?

I think I confused myself with the different equations...

 

[tiny-tim]

I think I confused myself with the different equations...

yes!

first, you should never write that "introductory" equals-sign at the start of your equations (the "i1 =" etc), you should write …

40i2 + 20i1 = 10
80i3 - 40i2 = 0
40i2 - 80i3 = 0
10i5 + 20i1 = 10​

however, the first equation is wrong … where did you get it from?

and the second and third equations are the same, aren't they?

try again! ​

 

[cschear87]

yes!

first, you should never write that "introductory" equals-sign at the start of your equations (the "i1 =" etc), you should write …

40i2 + 20i1 = 10
80i3 - 40i2 = 0
40i2 - 80i3 = 0
10i5 + 20i1 = 10​

however, the first equation is wrong … where did you get it from?

and the second and third equations are the same, aren't they?

try again! ​

OK, I'll try again...
Right Loop: 40i2 - 80i3 = 0
Middle Loop: 20i1 - 40i2 = 10
Left Loop: 10i5 + 20i1 = 10
Outside Loop: 10i5 + 80i3 =10?

Power Dissipated is
P=V^2/R I think. Is that the right formula?
Voltage for each resistor would depend on which loop it's for?

 

[tiny-tim]

OK, I'll try again...
Right Loop: 40i2 - 80i3 = 0
Middle Loop: 20i1 - 40i2 = 10
Left Loop: 10i5 + 20i1 = 10
Outside Loop: 10i5 + 80i3 =10?

yes, but

i] the middle loop should end "= 0"
ii] there's no point in doing the outside loop (once you've done the first three) …*i thought you understood that there's only three independent loops, and so only three independent loop equations?

anyway, you now still need to find what i₁ i₂ and i₃ actually are

Power Dissipated is
P=V^2/R I think. Is that the right formula?
Voltage for each resistor would depend on which loop it's for?

P = V²/R is correct, but so is P = I²R

as you say, V would be different for different resistors, so there's no point in using the first equation

use the second equation instead (because you know what I is for each resistor, from the first part of the question)

 

[cschear87]

yes, but

i] the middle loop should end "= 0"
ii] there's no point in doing the outside loop (once you've done the first three) …*i thought you understood that there's only three independent loops, and so only three independent loop equations?

anyway, you now still need to find what i₁ i₂ and i₃ actually are


P = V²/R is correct, but so is P = I²R

as you say, V would be different for different resistors, so there's no point in using the first equation

use the second equation instead (because you know what I is for each resistor, from the first part of the question)

Yes I think I understand.
So the middle loop is
20i1 - 40i2 = 0

 

[tiny-tim]

that's better!

ok, then the 3 KVL equations (one for each loop) are …

20i₁ - 40i₂ = 0
40i₂ - 80i₃ = 0
10i₅ + 20i₁ = 10​

and the KCL equation is …

i₅ = i₁ + i₂ + i₃​

… so what are the actual values of i₁ i₂ and i₃ ?

 

[cschear87]

I'm not sure how the KVL equations tell me i1, i2, and i3. OR the source current.

 

[tiny-tim]

you have 4 equations (in my last post), and 4 unknowns

so you can solve them, can't you? ​

 

[cschear87]

you have 4 equations (in my last post), and 4 unknowns

so you can solve them, can't you? ​

Would you multiply i1, i2, i3 by the r in each equation?
Ex: 20i1 - 40i2 = 0
20 (2) - 40 (1) = 0?

 

[tiny-tim]

that's not multiplying them, that's getting rid of them!

you have to find them, not eliminate them! ​

 

[cschear87]

I'm just really frustrated that this isn't just popping into my head, second nature. I just can't seem to understand this 100%. Grr

 

[aralbrec]

I'm just really frustrated that this isn't just popping into my head, second nature. I just can't seem to understand this 100%. Grr

ok, then the 3 KVL equations (one for each loop) are …

20i₁ - 40i₂ = 0
40i₂ - 80i₃ = 0
10i₅ + 20i₁ = 10​

and the KCL equation is …

i₅ = i₁ + i₂ + i₃​

… so what are the actual values of i₁ i₂ and i₃ ?

The physics has led to a set of equations that must be satisfied and now you have transitioned to math which will tell you how to solve the equations.

What you have is called a system of linear equations. It's a system because you have more than one equation that must be satisfied at the same time. It's linear because the equations are sums of constants times variables. No square roots, squares, sines of variables appear. Because this type of problem is so common, a branch of mathematics called linear algebra evolved to solve it but there's no need to go too deeply into that. You'll see that solving a system of linear equations is almost common sense.

To solve systems like this, it is almost always easiest to use Gaussian elimination. Scroll down to the section "Solve the following system of equations using Gaussian elimination." on http://www.purplemath.com/modules/systlin6.htm for an overview. Writing down all the +/- signs and unknowns repetitively is a pain so we usually use a matrix shortcut, see http://www.sosmath.com/matrix/system1/system1.html . Sorry I have a hard time finding a decent link, surprising as this is a pretty elementary subject.

Gaussian elimination, a method for solving systems of linear equations, follows three rules:

* Given a set of equations written down in order, you can change the order of the equations without changing the solution (this is called interchanging of rows).

* You can multiply an equation by a constant without changing the solutions (x+y=2 is the same equation as 2x+2y=4).

* You can add two equations together to get another equation consistent with the system (if you have " x - y = 6" and "x + y = 10" then you can add them together to get "2x = 16"). Note that you are not finding more information by doing this so you want to replace one of those two original equations with this new one.

You need as many equations as unknowns to find a unique solution. In your problem you have four independent equations and four unknowns (i1,i2,i3,is).

Take a look at those webpages and see if you can get anywhere with your system of equations.

 

[aralbrec]

I think you didn't catch the concept of resistance either so maybe this will help.

Resistors are devices that follow Ohm's Law: V=IR. To get a current I to flow through them, a voltage V must appear across its terminals according to this formula. Or you could rearrange the equations as I=V/R and say to get a current I to flow through them, you need to apply a voltage V across them. The resistance, R, is a property of the device and will depend on the device material and geometry.

One of your KVL equations was this one:

10i5 + 20i1 = 10

This was the loop involving V5, 10 ohm resistor, 20 ohm resistor. Nature has currents flowing through these resistors and you marked on the diagram those currents with variables i5, i1, etc. We don't know what those currents are yet. But in order for a current i5 to flow through the 10 ohm resistor, eg, a voltage of 10*i5 volts must appear across the 10 ohm resistor and in order for a current i1 to flow through the 20 ohm resistor, a voltage of i1*20 must appear across the 20 ohm resistor. These voltages are energy changes experienced by the moving charges (making up the current) and KVL says the sum of energy changes around a loop must be zero.

I can write that loop like so:

V5 - 10i5 - 20i1 = 0

It's important to keep the signs straight. We must follow one direction around the loop in order to distinguish energy gain from energy loss consistently.

I started at the bottom left corner under the battery V5 and moved clockwise. The battery adds energy in the direction I travel because I am moving from the -ve side (the shorter line at lower voltage) to the +ve side (the longer line at higher voltage). So I record V5 in my equation as positive (adding energy). Then I encounter the 10 ohm resistor. A direction for i5 is assumed that, if true, requires the left side of the 10 ohm resistor to be at higher voltage than the right side. So moving in the direction I am from left to right, I am losing energy. So I record the energy loss 10i5 in the equation as negative. Next we pass through the 20 ohm resistor. Again, the assumed direction of i1 requires that the top of the 20 ohm resistor is at higher voltage so there is another energy loss in across that resistor and I record it in the equation as -20i1. We're back to where we started, so the sum of those energy changes around that closed loop must be zero and we have the equation above.

Another way to write it was the way you had it written down:

10i5 + 20i1 = V5

Some people prefer to write the equation as Energy Loss = Energy Gain, which is equivalent.

 

[jshoe]

am i missing i missing something here why are you making this so complicated. just find total resistance of the circuit to find total current. then use current divider to find current in parallel...done... you guys are doing mesh analysis

 

[tiny-tim]

welcome to pf!

hi jshoe! welcome to pf!

am i missing i missing something here …

yes, you're missing that cschear87's professors want him to learn how to use kirchhoff's laws

it's easiest to learn on very simple cases, and of course those cases can often be solved more quickly in some other way, but doing that doesn't help in the long run!

(and btw, mesh analysis isn't the same as kirchhoff … it involves assigning a current to each loop, so that each wire has two currents going through it … and it can't be used if there are "crossovers")

 

[cschear87]

So based on the circuit and resistors etc, we've gotten equations for KVL. How do those equations translate into the other equation i5 = i1 + i2 + i3? I have another tutorial on this tonight but I really want to make a bit more progress.
For my own review I know that:
Right Loop: 40i2 -80i3 = 0
Middle Loop: 20i1 - 40i2 = 0
Left Loop: 10i5 + 20i1 = 10

I mean, is it simple algebra so each equation comes to 0, 0 and 10? Ie, i3 would be 1, i2 would be 2, i1 would be 4?

 

[tiny-tim]

hi cschear87!

Right Loop: 40i2 -80i3 = 0
Middle Loop: 20i1 - 40i2 = 0
…
I mean, is it simple algebra so each equation comes to 0, 0 and 10? Ie, i3 would be 1, i2 would be 2, i1 would be 4?

i'm not really following you

those two equations tell us the ratios: i₁ = 2i₂ = 4i₃

Left Loop: 10i5 + 20i1 = 10
So based on the circuit and resistors etc, we've gotten equations for KVL. How do those equations translate into the other equation i5 = i1 + i2 + i3?

the first 2 equations, in 3 unknowns, give you only the ratios between those 3 unknowns (because 3 = 2 + 1)

the last 2 equations (the equations with 1₅) give you a total of 4 equations in 4 unknowns, which enable you to solve them completely (because 4 = 4)

I have another tutorial on this tonight but I really want to make a bit more progress.

you need to ask for help on solving simultaneous equations