Mesh/Kirchhoffs theorem.

In the second figure below, how was 11.2v found?

The voltage drop from B1 to B2 is 28 - 7 = 21 volts

This is split into a drop across a 4 ohm resistor and a 1 ohm resistor in series with it.  This means the 21 volt  drop is divided in proportion to these two resistances.  

Let's call the voltage in between the two resistors V1.  Then (28 - V1)/(28 - 7) = R1/(R1 + R2)

(28 - V1)/21 = 4/(4 + 1) 

28 - V1 = 21*4/5

V1 = 28 - 21*4/5

$${\mathtt{V1}} = {\mathtt{28}}{\mathtt{\,-\,}}{\frac{{\mathtt{21}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{5}}}} \Rightarrow {\mathtt{V1}} = {\mathtt{11.2}}$$

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Why is it a i2=-i2 in the image. is it just because clockwise is positve. I got it. Thanks

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For the above post, using the matrix (mesh theorem) I have to do it like a matrix? or can i just do a vector cross product of the top line, then the bootm line, and please give an example if yes. How do I solve this 3 unknowns by not using simultaneous and perhaps by using Thevenins theorem instead? 

Stu, if you are unhappy about solving with matrices, you can always solve the two equations for I1 and I2 in the following way:

10 = 50*I1 - 40*I2      ...(1)

-20 = -40*I1 + 60*I2  ...(2)

Divide through both equations by 10

1 = 5*I1 - 4*I2         ...(3)

-2 = -4*I1 + 6*I2     ...(4)

Multiply (3) by 4 and (4) by 5

4 = 20*I1 - 16*I2        ...(5)

-10 = -20*I1 + 30*I2  ...(6)

Add (5) and (6) term by term

4 - 10 = 0 -16*I2 + 30*I2   Simplify

-6 =  14*I2

Divide both sides by 14

I2 = -6/14 = -3/7  

Put this back into (3)

1 = 5*I1 - 4*(-3/7) Simplify

1 = 5*I1 + 12/7

Subtract 12/7 from both sides

1- 12/7 = 5*I1  Simplify

-5/7 = 5*I1

Divide both sides by 5

I1 = -1/7

Or in terms of decimals

$${\mathtt{I1}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{7}}}} \Rightarrow {\mathtt{I1}} = -{\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$

$${\mathtt{I2}} = {\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{7}}}} \Rightarrow {\mathtt{I2}} = -{\mathtt{0.428\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

However, I think this is a poorly expressed question, as the definition of the currents changes within the question.  It would have been better to call the clockwise looping currents I3 and I4, say, to avoid confusion with the currents marked on the lines (although I1 is the same in both cases, I2 is not).

Why was 10 in the divide both sides by 10 step chosen? 

I didn't understand this step. Please explain it- I looked at t for 5 minutes. Maybe I should look longer?

Multiply (3) by 4 and (4) by 5

4 = 20*I1 - 16*I2        ...(5)

-10 = -20*I1 + 30*I2  ...(6)

does this work for all silmultaneous equatoins, only those with 2 unknowns or only where one unknown will cancle? If it works for all i am very grateful. $_$

In the second figure below, how was 11.2v found? Can I solve it through P.D and which values? What about how I would do it through Kirchhoff's current theorem? I tried the latter and got a error. Additionally, what step was taken to get the 2 ohm as a load, as well why or what is load defined as ( i mean with respect to theorems and what not- how is it used? Thanks. 

Why was 10 in the divide both sides by 10 step chosen? 

It's usually better to work with small numbers rather than large ones if possible, and it was clear that all the constants were multiples of 10 (they all ended in zero), so I divided by 10.  This step wasn't essential, the process works without doing this, but it just made the equations easier to view in my opinion.

I didn't understand this step. Please explain it- I looked at t for 5 minutes. Maybe I should look longer?

Multiply (3) by 4 and (4) by 5

4 = 20*I1 - 16*I2        ...(5)

-10 = -20*I1 + 30*I2  ...(6)

 The purpose of this step is to make the magnitudes of the coefficients of the I1 terms the same in both equations.  This means that the two equations could then be added together term by term and the I1 terms would disappear, leaving I2 as the only unknown. 

An alternative would have been to multiply (3) by 6 and (4) by 4.  In this case the magnitudes of the coefficients of the I2 term would have been the same, and I2 could have been eliminated, leaving I1 as the only unknown.

does this work for all silmultaneous equatoins, only those with 2 unknowns or only where one unknown will cancle? If it works for all i am very grateful. $_$

It will work for more than two equations, but only one term at a time will be eliminated.  Therefore the process must be repeated several times until you get down to a single unknown.

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I see, however the 1 ohm resistor is not in the loop with the i2=21/4. Also, the 1 ohm resistor is in parallel with the 1 ohm resistor. So I still don't understand according to the logic explained. 

Additionally, the current is flowwing (ccw), would it be correct to say this is not for conventional current? Does this also work for the conventional current, would the dirrection or anything else need to be changed?