Calculate Current, Power, EMS..

I have to calculate 1. current in each branch, 2. voltameter result 3. electromotive force (EMS) mode and 4. power bilance.
And need to use four methods:
Node Voltage Analysis (Node potential method)
Superposition Method
Loop Current Analysis
Kirchoff's Rules Analysis

I hope you can rede something or give me advices. What would be the right way to start?
Then I could try to calculate the tasks with your help..
Please.

 

Yes, R01, R02, R03 are internal resistances.
But why there are four loops? Maybe I don't understand something, but I guess, that, when switch is OFF, then we don't have to use R02, R2 and E2, and of course we don't have to use voltmeter.
And then there could be two loops like this:

Or am I wrong?
And when switch is ON, then there could be three loops, because, as you mentioned, we don't have to use voltmeter.
I'm confused. When I will know, how exactly there are loops, then I could try to start calculations.

 

Heh, I don't know what, but I did something wrog.
I tried Nr.32 using Kirchoff's rules analysis/method.

I got system of:
I5+I1-I4-I3-I6=0
I6-I5-I1=0
-E3=I3R3 - I6R6 + I4R4
E1 = I6R6 + I5R5 + I1R1

A, B - nodes. What I did wrong?
And also do I have to count R1+R01, R2+R02 and R3+R03 to get R on whole branch, because they are on chain?
Also, to get result of voltmetre, I have to count E1+U1+U01?
Tomorrow I will try other methods, but I want to know, what I did wrong using Kirchoff's rules.

 

I count R03+R3 = 0.1 + 3.9 = 4, same as R01+R1 = 4 (let it be R3 and R1, but just without R01 and R03)
Then:
I6R6+I3R3+I4R4 = E3
-I1R1-I5R5-I6R6=-E1
I6-I1-I3=0
As I5=I1 and I3=I4 (as you metioned), then is system of:
10I1+3I6=40
9I3+3I6=28
I6-I1-I3=0

When I solve it, I get I1=I5= 2.7A; I3=I4= 1.7 A; I6= 4.4 A
Is it wright as your calculations with your algebra system?

Then I tried to solve this task with Node Voltage Analysis (Node potential method).
R(upper) = R3+R03+R4=9, R(lower) = R1+R01+R5 = 10 and R6 = 3
U(ab)= (E1*(1/Rlower) + E3*(1/Rupper))/((1/Rlower)+(1/Rupper)+(1/R6)) = (40*(1/10) + 28*(1/9))/((1/10)+(1/9)+(1/3)) = 13.1 V
I6 = (U(ab))/R6 = 13.1/3 = 4.6 A
But, when I tried to solve other I1, I3, I4, I5 ... results were differet as they had been solved by Kirchoff's Rules before. Only I6, as you can see, match. What's wrong again?

Also, can you chek other task? This ir much easier, I just want to be sure.

E = E1 + E2 = 19 + 10.1 = 21.9V
R56 = 177
R23 = R2 + R3 = 814 + 254 = 1068
R2356 = 151.1
R(total) = R2356 + R1 + R4 = 151.8 + 121 + 521 = 793.8
I = E/R(total) = 29.1/793.8 = 0.037A
I = I1 = I4 = I2356
I2356= I23 + I5 + I6
U1 = I1 * R1 = 0.037 * 121 = 4.5 V
U4 = I4 * R4 = 0.037 * 521 = 19.3 V
U2356 = I2356 * R2356 = 0.037 * 151.8 = 5.6 V
I2 = I3;
U23 = U2356
U2+U3 = U5 = U6
I5 = U4/R5 = 5.6/326 = 0.017A
I6 = U6/R6 = 5.6/387 = 0.014A
I23 = I2356 - I5-I6 = 0.037-0.017-0.014 = 0.006A
U2 = I2 * R2 = 0.006*814 = 4.9V
U3= I3 * R3 = 0.006 * 254 = 1.5 V

Thanks for helping!

 

Please, can you tell your solution with nodal method or at least give me a few hints? Actually, I have calculated this task with Kirchoff's Rules method (as you saw my answers), but there is needed Nodal method as well to be sure with my solution, but I don't know how exactly I can get it with nodal method.

 

Thanks!
But I also have to calculate currents with Loop Current Analysis for that loop, when switch is ON:

Here I have example, but still I'm not sure, how to write these equations correctly:


I guess it can take just a minute for you.
I tried something like in example:
E3+E2=(R02+R2+R3+R03)*Ia - (R02+R2)*Ib
E1=(R6+R5+R1+R02)*Ic + R6*Ib
-E2=(R6+R4+R02+R2)*Ib - (R02+R2)*Ia + R6*Ic

Can you help to write equations correctly?

 

30+50=(0.2+3.8+2.9+0.2)*Ia - (0.2+3.8)*Ib
40=(4+5+3.9+0.1)*Ic + (0.2+3.8)*Ib
-50=(4+4+3.8+0.2)*Ib-(0.2+3.8)*Ia + 4*Ic
I solve that system and get:
Ia = 10.127A
Ib = -2.024 A (I guess, there is opposite current direction)
Ic = 3.7 A
Then:
I3=I03=Ia = 10.127A
I5=I1=I01=Ic = 3.7A
I6 = -Ib-Ic = -(-0.024)-3.7 = -1.676A
I4 = -Ib = -(-2.024) = 2.024A
I02 = I2= -Ib+Ia = -(-2.024)+10.127 = 12.151A

Is it wright?
Can you chek it and show solution using Kirchhoff's current law (KCL) method?
Please.

And also - to determine, what is result of voltmeter, I have to calculate U01+U1+E1?

 

Actually, there is no need for other method if Loop Current Analysis equations were right. [as you mentioned]. I just wanted to be sure, but I will believe and hope, that there is all OK.
But when we reconnect voltmeter, we get that diagram or not?

I guess that bottom corners are nodes, because there is votmeter or I'm wrong?

I just started to think, that if now I'm wrong, that all these solutions when switch is OFF are wrong in this case:

because there is no bottom nodes.
What's the truth? I have got lack of understanding.

 

Despite that I believe, that solution in post #14 is wright. I tried to solve it with KCL method.


I1*(R5+R01+R1) + I6*R6 = E1
-I6*R6 + I4*R4 + I2*(R02+R2) = E2
-I2*(R2+R02) + I3*(R03+R3) = -E2-E3
A] -I2 + I4 - I3=0
B] -I4 - I6 + I1 =0
C] -I1 + I6 + I2=0

I didn't include I01, I02, I03, I5, because
I1 = I01 = I5
I02 = I2
I3 = I03

But I guess something is missing to solve that system. Can you look at it, what's there is wrong? For me it's easier to try to use this KLC method, that's why I don't want use nodal solution, but of course we can try if it's necessary and you shouldn't suggest, what's wrong in KLC equations.

For the same time I want to ask question for other task.
I have got this solution with nodal method.


When I calculate, I get:
fi1 = -10.14 V
fi2 = 3.39 V
fi3 = 20.37 V
Actually, you don't have to recalculate. It's no matter about numbers. Can you tell, how exactlly I can get I1, I2, I3, I4, I5 and I6, using R, E and fi?