Comments on Profile Post by MangoGuy

I got the I thingies by using junction rule.
Then,use loop rule to get 4 eqns from 4 loops.You have 4 variables,so solve for them.

hmm so you continue it here ahahaha

actually just googling kirchof problem and solution you will get many example

I took left terminal to be 25V and right one to be 0V. That is the better way,as you get away from +ve and -ve thingy.In the end,the signs f unknown voltages will be taken care off by the eqn.

thanks dude, but what's confusing me the most is the part in the middle . how do I solve it ?. Do I add both the positive and the negative voltage in the equation or only one?

And what's the big loop?

@asriu I believe that having someone guide you personally is a better way. I am not sure whether pancake is the best at self learning. I might be wrong!! Plus..I like to complete things~~

@Mr Pancakes the big loop is made of 2R,17R,7R,6R,1R and 2R,including the 25V terminal.

I know but I need this for tomorrow and our professor didn't explain this type of problems, only the most basic ones

When u cant decide the sign of potential,forget the potential and focus on the currents.Assign a direction to them.If you get wrong direction,then the current value will be negative.Simple as that.

hmm oh well if you say so~
I was encounter this problem long time ago~
hmm too bad I only remember first step is to break down the circuit into easier diagram

@Mr Pancakes Idk..This is intermediate level. If your prof taught you to only use Potentials,then no problem.Essentially,there are 4 junctions with unknown potential.Call them V something something. Assume everything is positive and then apply the rules to get 4 eqns.If any of the V turns out to be negative,then there you go~~~ That is the opposite sign.

.In the middle part , would the formula have something like E1+E2(or v1+v2 depending on how you mark it) +I1R1... ?

@asriu that is actually a good method to solve,except it is not easy for this particular problem.I remember teaching similar problem 6 months back.It is not easy to restructure circuits after set number of junctions,and when the circuit components are numbers without patterns.Breaking this into smaller loops is the best method instead of changing thee shape of circuit.I have a feeling that is what you meant.

@Mr Pancakes yes. the I1R1 is negative sign,because resistance opposes current flow. So E1+E2-I1R1

This is the only method we learned (ie. E1+R1I1+...=0) so I dont know how to break them down

So both of them are positive ( E1 and E2) in this case. what would it be like if i had 2 sources next to r3 and r6 would they still add up or would it be E-E?

@Mr Pancakes you can still break them down.For left most loop,I think it looks like: E1-3(I3)+E2-2(I2)-2(I1)=25 as external voltage applied is 25.

what about the middle one?

and thanks for helping

@Mr Pancakes We ASSUME them to be positive,even though it is wrong.As a result,in the final answer one of them comes with a negative sign.Like you thought E1 and E2 and they are 2V and -3V in reality.Then solving the eqn gives you E1=2V and E2=-3V. If you ASSUMED them to be E1 and -E2,then you get E1=2 and E2=3V meaning -E2=-3V.

@Mr Pancakes I can see ur problem here.Even if you incorrectly assign the sign for a potential,as long as it is a variable the equation will give you the value of the variable which gives correct answer when paired with the wrong sign.Basically,believe in mathematical equations.

If that got you clean,then I will go to bed.Good Luck!!

Thanks. Good night