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

[cschear87]

Homework Statement

In the circuit shown in Figure Q8 below, if Vs = 10V, use Kirchoff’s Laws to determine the currents i1, i2, i3 and the source current is. Calculate the power dissipated by the resistors in this circuit. Confirm that the power dissipated by the resistors is the same as the power supplied by the power sources.

Homework Equations

kirchhoff's current law
kirchhoff's voltage law

i1=i2+i3

The Attempt at a Solution

I've genuinely tried to get the answers to this but I just don't understand it. I want to understand the answer and how the answer was reached. I have another diagram which I want to do on my own from the help given here.

 

[tiny-tim]

welcome to pf!

hi cschear87! welcome to pf!

you can get the ratios of i₁ i₂ and i₃ by doing KVL round each of the two loops at the right (separately)

and you can get the ratio of i₅ and i₁ by doing KVL round the loop at the left

(and of course KCL gives you i₅ = i₁ + i₂ + i₃)

show us what you get ​

 

[gneill]

Homework Statement

In the circuit shown in Figure Q8 below, if Vs = 10V, use Kirchoff’s Laws to determine the currents i1, i2, i3 and the source current is. Calculate the power dissipated by the resistors in this circuit. Confirm that the power dissipated by the resistors is the same as the power supplied by the power sources.

Homework Equations

kirchhoff's current law
kirchhoff's voltage law

i1=i2+i3

The Attempt at a Solution

I've genuinely tried to get the answers to this but I just don't understand it. I want to understand the answer and how the answer was reached. I have another diagram which I want to do on my own from the help given here.

Can you show one of your attempts? How have you tried to apply Kirchoff's laws?

 

[cschear87]

hi cschear87! welcome to pf!

you can get the ratios of i₁ i₂ and i₃ by doing KVL round each of the two loops at the right (separately)

and you can get the ratio of i₅ and i₁ by doing KVL round the loop at the left

(and of course KCL gives you i₅ = i₁ + i₂ + i₃)

show us what you get ​

Any chance you could phrase that as though I was say a 5 year old. lol. I'm just starting this class and feel a bit at the deep end.
Before this post I tried
1/i1 + 1/i2 + 1/i3 - 1/20 + 1/40 + 1/80 = 7/80 = 11Ω
Total R. = 140Ω
V/R = 11/140 = .08 (A?) -- .08 x 11 = .86 -- .86 x 140
=.006

I'm really really bad at math so it was my ABSOLUTE best guess and the book from college was written by the university and thus not great. I got through my binary questions and I'm stuck on algebra *cries*

 

[tiny-tim]

hi cschear87!

… use Kirchoff’s Laws to determine the currents i1, i2, i3 and the source current is.

1/i1 + 1/i2 + 1/i3 - 1/20 + 1/40 + 1/80 = 7/80 = 11Ω
Total R. = 140Ω
V/R = 11/140 = .08 (A?) -- .08 x 11 = .86 -- .86 x 140
=.006

no, those aren't using Kirchoff’s Laws

Kirchhoff’s Laws are the circuit (or loop) law (KVL), and the node (or junction) law (KCL)

look them up in wikipedia or the PF Library

(and then start by applying KVL to the right-hand loop, the one with only i₂ and i₃)

 

[cschear87]

hi cschear87!



no, those aren't using Kirchoff’s Laws

Kirchhoff’s Laws are the circuit (or loop) law (KVL), and the node (or junction) law (KCL)

look them up in wikipedia or the PF Library

(and then start by applying KVL to the right-hand loop, the one with only i₂ and i₃)

So does i1 + i2 + i3 = 10v?
I think I understand that voltage in = voltage out. So should the currents add up to 10v?

 

[mink_man]

I = current

and the voltage stays the same in a parallel circuit

 

[tiny-tim]

So does i1 + i2 + i3 = 10v?
I think I understand that voltage in = voltage out. So should the currents add up to 10v?

sorry, cschear87, but this is nonsense

how can currents equal a voltage?

and it isn't "voltage in = voltage out"

it's "current in = current out"​

cschear87, you clearly have no grasp of how Kirchhoff's laws work

you need to read your book (or lecture notes) again, carefully, until you understand them

(or wikipedia, or other websites)

 

[cschear87]

sorry, cschear87, but this is nonsense

how can currents equal a voltage?

and it isn't "voltage in = voltage out"

it's "current in = current out"​

cschear87, you clearly have no grasp of how Kirchhoff's laws work

you need to read your book (or lecture notes) again, carefully, until you understand them

(or wikipedia, or other websites)

That's what I was saying. I don't understand it. And I've read the notes dozens of times. All two pages in the book I'm given about it. I've just started this course and I'm trying to understand but have limited literature for this subject (there isn't a textbook for this class just notes provided by university). This isn't an excuse, I just don't understand this. This 10Q assignment I've gotten through the first 7 parts without help. I just don't understand based on the material provided. That's why I came here. I thought someone could help. It doesn't matter how many times I read that "the total current flowing into any junction equals the total current flowing out of it" I don't understand. It also doesn't matter how many times I read that "in any closed loop in a network, the algebraic sum of the voltage drops (PDs) around the loop is equal to the emf acting in that loop"... makes no sense. Looks like greek to me. Nevermind

 

[tiny-tim]

It doesn't matter how many times I read that "the total current flowing into any junction equals the total current flowing out of it" I don't understand.

ok, i believe you …

but that doesn't help me explain it to you …

if you don't follow it from the books, how am i going to make any difference?

all i can suggest is that you think of electric current as being like the current (of water) in a river …

when the river divides into three channel (as in the diagram), the total water flowing into the juction has to equal the total water flowing out of the same junction​

... then re-read your book on the junction law (node law, KCL), and see if it makes any more sense

It also doesn't matter how many times I read that "in any closed loop in a network, the algebraic sum of the voltage drops (PDs) around the loop is equal to the emf acting in that loop"... makes no sense.

sorry, but i have no suggestion for this, other than to point out that voltage is just another name for energy-per-charge, and KVL is basically an energy equation

 

[aralbrec]

It doesn't matter how many times I read that "the total current flowing into any junction equals the total current flowing out of it" I don't understand. It also doesn't matter how many times I read that "in any closed loop in a network, the algebraic sum of the voltage drops (PDs) around the loop is equal to the emf acting in that loop"... makes no sense. Looks like greek to me. Nevermind

You will get it, maybe see a few different explanations. Tinytim's is above, here is another.

KCL:

Keep in mind what current is -- it is moving charges, inside a wire it is moving electrons. At a point where wires connect, each wire is carrying some moving electrons. If the number of electrons coming into that point did not equal the number of electrons leaving, you would have electrons accumulating there. This is not allowed**. So the number of electrons coming in must equal the number of electrons leaving. If you divide that by time you can speak of the rate at which electrons arrive must equal the rate that electrons leave. This becomes KCL: the sum of incoming currents = the sum of outgoing currents.

*** If electrons were allowed to accumulate at a point, charge would build up and generate an electric field that would oppose and eventually stop the currents from flowing. We call those devices capacitors so when we want to model that we would put a capacitor in the circuit. A simple wire cannot perform this function. A capacitor is actually a broken wire with large plates connected at either end. Electrons do not flow across the broken wire but charges accumulate on the plates.KVL:

As mentioned, this is an energy equation. But maybe you can understand it intuitively. If I take an charged particle at some point in the wire, it will have a certain amount of energy. If I then move this charged particle along the wire, its energy will change because it experiences a force (the electric field) that is summarized in a quantity known as a voltage. But if I keep moving this particle around the wire and come back to the point I started at, it should have the same energy as when I started.

KVL means the change in energy as I travel around a closed loop is zero. The energy changes are measured as voltage drops across each circuit element (the wire itself is assumed ideal with no resistance and therefore no energy drop; if it were not ideal it would be modeled as a resistor on an ideal wire).

 

[cschear87]

You will get it, maybe see a few different explanations. Tinytim's is above, here is another.

KCL:

Keep in mind what current is -- it is moving charges, inside a wire it is moving electrons. At a point where wires connect, each wire is carrying some moving electrons. If the number of electrons coming into that point did not equal the number of electrons leaving, you would have electrons accumulating there. This is not allowed**. So the number of electrons coming in must equal the number of electrons leaving. If you divide that by time you can speak of the rate at which electrons arrive must equal the rate that electrons leave. This becomes KCL: the sum of incoming currents = the sum of outgoing currents.

*** If electrons were allowed to accumulate at a point, charge would build up and generate an electric field that would oppose and eventually stop the currents from flowing. We call those devices capacitors so when we want to model that we would put a capacitor in the circuit. A simple wire cannot perform this function. A capacitor is actually a broken wire with large plates connected at either end. Electrons do not flow across the broken wire but charges accumulate on the plates.


KVL:

As mentioned, this is an energy equation. But maybe you can understand it intuitively. If I take an charged particle at some point in the wire, it will have a certain amount of energy. If I then move this charged particle along the wire, its energy will change because it experiences a force (the electric field) that is summarized in a quantity known as a voltage. But if I keep moving this particle around the wire and come back to the point I started at, it should have the same energy as when I started.

KVL means the change in energy as I travel around a closed loop is zero. The energy changes are measured as voltage drops across each circuit element (the wire itself is assumed ideal with no resistance and therefore no energy drop; if it were not ideal it would be modeled as a resistor on an ideal wire).

OK, I think I understand a bit better now. CURRENTS in = CURRENTS out and voltage drops when contact is made with each circuit element. That seems to make sense. How do I put that information into an equation based on my diagram?

 

[tiny-tim]

start by applying KVL to the right-hand loop, the one with only i₂ and i₃ …

what do you get? ​

 

[cschear87]

start by applying KVL to the right-hand loop, the one with only i₂ and i₃ …

what do you get? ​

For KVL, the sum of the voltages = 0. Right?
For KCL current in = current out. So depending on the arrows in the diagram, and how they connect to the nodes, that would, I guess, dictate the equation. If generally i1 = i2 + i3, then iS = i1 + i2 + i3?

 

[tiny-tim]

For KVL, the sum of the voltages = 0. Right?

(let's leave KCL for the moment, it's a bit more complicated than it looks)

Yes, the sum of the voltage drops has to be zero, because the emf (in that loop) is zero.

ie, the RHS of the KVL equation is zero.

So, in terms of I and R, what is the LHS?

 

[cschear87]

(let's leave KCL for the moment, it's a bit more complicated than it looks)

Yes, the sum of the voltage drops has to be zero, because the emf (in that loop) is zero.

ie, the RHS of the KVL equation is zero.

So, in terms of I and R, what is the LHS?

So is the current Voltage (I)/Resistance (R)?
Voltage is 10V and would resistance be 20, 40 and 80?
(LHS? Not far enough to know abbreviations yet :( )

 

[tiny-tim]

So is the current Voltage (I)/Resistance (R)?
Voltage is 10V and would resistance be 20, 40 and 80?
(LHS? Not far enough to know abbreviations yet :( )

LHS is left-hand-side, and RHS is right-hand-side (of an equation)!

ok, i'll do this one for you …

KVL for the right hand loop:

the RHS is 0

(because there's no emf in the loop)​

the LHS is 80i₃ - 40i₂

(because the voltage drop across each resistor is IR, and because if we go round clockwise then i₃ is positive and i₂ is negative)​

so the whole KVL equation is 80i₃ - 40i₂ = 0,

ie i₂ = 2i₃ ​

(btw, if we'd gone anti-clockwise round the loop, it'd have been -80i₃ + 40i₂ = 0, which still gives i₂ = 2i₃, doesn't it?

so it doesn't matter which way we go round ! )

any questions about all that?

if not, now you try the middle loop (the one with i₁ and i₂)

 

[cschear87]

LHS is left-hand-side, and RHS is right-hand-side (of an equation)!

ok, i'll do this one for you …

KVL for the right hand loop:

the RHS is 0

(because there's no emf in the loop)​

the LHS is 80i₃ - 40i₂

(because the voltage drop across each resistor is IR, and because if we go round clockwise then i₃ is positive and i₂ is negative)​

so the whole KVL equation is 80i₃ - 40i₂ = 0,

ie i₂ = 2i₃ ​

(btw, if we'd gone anti-clockwise round the loop, it'd have been -80i₃ + 40i₂ = 0, which still gives i₂ = 2i₃, doesn't it?

so it doesn't matter which way we go round ! )

any questions about all that?

if not, now you try the middle loop (the one with i₁ and i₂)

2i1 - i2?
= 2(20) - 40 = 0

 

[tiny-tim]

that doesn't make sense

write it in the standard form, i₁R₁ + (or -) i₂R₂ = 0

 

[cschear87]

that doesn't make sense

write it in the standard form, i₁R₁ + (or -) i₂R₂ = 0

I think where I'm confused is what exactly is R. There's no R in the diagram which is partially why I'm confused I think. And I thought that I had to use the current law to find the current for the first part of the question. I'm confusing myself And you're very good to try and help. This course is all online with my university so I have very little interaction with other students and professors unfortunately which is why I haven't gotten an answer back about this from them. Online tutorial next week though.

 

[tiny-tim]

I think where I'm confused is what exactly is R.

R is the resistance (easy to remember!)

(V is for voltage, C is for capacitance, I is for something German, i think )

so in this case R is either 20 or 40

 

[cschear87]

R is the resistance (easy to remember!)

(V is for voltage, C is for capacitance, I is for something German, i think )

so in this case R is either 20 or 40

So, 2(20i1) - 40(i2) = 0?

 

[tiny-tim]

20i₁ - 40i₂ = 0

(where did your "2" come from? )

so i₁ = 2i₂​

ok, now try the left-hand loop (the one with V₅ i₅ and i₁)

 

[cschear87]

20i₁ - 40i₂ = 0

(where did your "2" come from? )

so i₁ = 2i₂​

ok, now try the left-hand loop (the one with V₅ i₅ and i₁)

Thought it had to = 0?
So r5 is 10
So 2i5 = i1? - 2(10i5) = 20i1
or
2 (10) = 20

 

[tiny-tim]

now you're completely losing me

Thought it had to = 0?

you thought what had to = 0?

and can we go back a little … where did that "2" come from? (i see you have another one a couple of lines down)

let's sort those two things out before we look at the rest of what you did

 

[cschear87]

now you're completely losing me


you thought what had to = 0?

and can we go back a little … where did that "2" come from? (i see you have another one a couple of lines down)

let's sort those two things out before we look at the rest of what you did

I thought the equation had to =0. The 2 (both times) were because I thought the equation had to come to 0

 

[tiny-tim]

I thought the equation had to =0.

it depends how you write it

for KVL, i usually put V on one side, and the IRs on the other side

but of course you can put them all on the same side (being careful about the signs), and 0 on the other

The 2 (both times) were because I thought the equation had to come to 0

let's see, you originally wrote 2(20i1) - 40(i2) = 0

but the correct equation was 20(i1) - 40(i2) = 0 (IR + IR = V) …

why did you think there had to be a 2 there? (i need to know to stop you doing it again! )

 

[cschear87]

it depends how you write it

for KVL, i usually put V on one side, and the IRs on the other side

but of course you can put them all on the same side (being careful about the signs), and 0 on the other


let's see, you originally wrote 2(20i1) - 40(i2) = 0

but the correct equation was 20(i1) - 40(i2) = 0 (IR + IR = V) …

why did you think there had to be a 2 there? (i need to know to stop you doing it again! )

lol, Well, I thought that 2 x 20 =40 and therefore 40-40=0. It all comes back to thinking things = 0. So is this to determine the current or the voltage?

 

[tiny-tim]

lol, Well, I thought that 2 x 20 =40 and therefore 40-40=0. It all comes back to thinking things = 0.

oh now i see!

ok, you're trying to balance the two Rs on their own,

ie you have 20Ω and 40Ω, and you know you have to multiply each one by something to get 0

but that's what the current is for …

you multiply each R by its I and nothing else

you must follow the formula! (for KVL)​

try again

So is this to determine the current or the voltage?

KVL is sometimes to determine the current, and sometimes to determine the voltage

in this case, you know the voltage (it's 0 in that loop), but you don't know either of the two currents …

so this KVL gives you one equation to help you find the two unknown currents

(obviously, you always need two equations to find two unknowns, so you'll need another equation from somewhere else)

 

[cschear87]

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

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

 

[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")