Question

Circuit help please!

Answer 1

Sorry - I don't understand. I cannot help you.

Answer 2

Okay, I'll try.

First of all, where are you stuck?

Answer 3

which law should I apply?

Answer 4

the setup and all

Answer 5

What are i1 and i2?

Answer 6

current names

Answer 7

How about v and r? r is resistor.... v is? battery?

Answer 8

it's okay don't worry about it

Answer 9

Am I correct? They are resisters and batteries, right?

Answer 10

@abb0t You remember this stuff?

Answer 11

Remember what?

Answer 12

How to turn circuits into linear equations

Answer 13

I know there is something about the current going in and out of each node adding to zero.

Answer 14

i think it depends if they are in series or parallel

Answer 15

@sheena101
Can you post your attempt..coz I don't know what to explain here..

Answer 16

Then the voltage drop in each loop equal to the voltage rise

Answer 17

Maybe @nincompoop may be able to help. I know he took a physics course recent. He may be more familiar with how to explain.

Answer 18

Also I can't see the link anymore :'(

Answer 19

Wish I had draw tool

Answer 20

@sheena101
Can you now solve for ix ??

Answer 21

Looks like I am talking to myself...*dissolves in silence*

Answer 22

@sheena101

Answer 23

She's busy with chat box..

Answer 24

do you guys get it?

Answer 25

@LastDayWork gets it

Answer 26

He's trying to figure out which part you could do.

Answer 27

i got this so far :

v1+i2R1-R1i1-i1R2+v2-ix

Answer 28

is that right?

Answer 29

When i1 and i2 are zero; why do you even bother to use them ??

Answer 30

LastDayWork, how do you know they are 0?

Answer 31

There is no current between two points which are at the same potential...See my diagram..

Answer 32

same potential? what do you mean?

Answer 33

wouldn't all the voltage go through i1 as the path of least resistance?

Answer 34

Okay..I can see your point now..battery v1 is shot-circuited..

Answer 35

So what does that mean here? We ignore that battery?

Answer 36

It means ix = 0 (current is lazy - like me ;)
TBH, I wasn't expecting this ^^ answer..

Answer 37

you use kirchoff's first law

Answer 38

The answer would be different (more practical) if we are given the internal resistance of the battery and ammeter.

Answer 39

Assuming that the battery has some internal resistance (and hence is not shot-circuited); we can calculate ix by applying Kirchhoff -
(V2) + (ix)(R2) = 0

Answer 40

wait there has to be more though

Answer 41

What "more" are you expecting ??

Answer 42

use r1,v1,i1,i2

Answer 43

ix will not depend on them..

Answer 44

Q2) Use Kirchhoff ; the answer should be obvious..
Q3) What will be the value of ix in Q1 if the battery and ammeter were ideal :P

Answer 45

I agree with LastDayWork here.

Answer 46

I would say ix = i1 and i2 = 0

Answer 47

"...ix = i1..." why ??

Answer 48

I mean ix = -i1

Answer 49

I'll accept I was wrong about i1=0
But I can't see any relation between i1 and ix..

Answer 50

Well, thinking about it in terms of the electrons... they all have to go through that wire. I dunno that is my thinking

Answer 51

And I think none of the current goes in opposite direction if ix because of short circuit?

Answer 52

In terms of electron, current will distribute into ix and i1 at the junction;
where i1 tends to infinity
and ix tends to zero

Answer 53

also the direction of ix will come out to be opposite of what it is shown in the question..

Answer 54

Looking over the problem more, I think in this case, actually that the current for \(I_1\) and \(I_2\) are controlled. That is, they aren't a direct result of the battery and resistors, but something else has fixed them to be some unspecified number. Given whatever value they are manipulated into being, we have to solve the circuit.

Answer 55

In other words, \(i_1\) and \(i_2\) are current sources. Here's an image to play around with (since we can draw reply to it).