I'm understanding everything until its gets to the part where it states to "Equations (1) AND (2) may be solved simultaneously by multiplying Equation (2) by 3 and adding the result to Equation (1)::"
Why 3, Where are we getting the 3 from ?
To write an equation for the first mesh, select a convenient starting point such as point a and go around the mesh clockwise, following the direction of I1. Since the first circuit element encountered is the generator, and since the tracing direction passes through the ernf from the negative to the positive terminal, the first term of our equation should be written 8. A drop of +0.1I1 occurs from points b to c, and another drop of +0.05I1 occurs between points d and e.
We next encounter the positive terminal of the battery which should be accounted for in the equation by the term +6.4. On emerging from the battery we arrive at point f which is the same as point a and our traversal of Mesh 1 appears to be complete. At this point, we could very easily make the mistake of writing the equation for the first mesh as follows:
8 + 0.1I1 + 0.05I1 + 6.4 = 0
However, an extremely important point has been overlooked! Although current I1 flows downward between points d and e, this is not the only current flowing between these points. Current I2, in the adjacent mesh, also flows between these same points and I2 flows upward from e to d contrary to the direction given I1. Since the drop caused by I2 is in opposition to the drop caused by I1, this serves to reduce the total drop caused by I1 and must be indicated in the equation of the first by a minus sign. The complete equation for Mesh 1 then becomes:
8 + 0. 1I1 + 0.05I1  0.05I2 + 6.4 = 0
Or by collecting terms and transposing
0.15I1  0.05I2 = 1.6 (1)
Turning to Mesh 2, start at point f and traverse the loop fdgh, following the direction I2, writing the signs of the emfs according to the signs of the terminals first encountered and considering all drops caused by I2 to be positive. This time, any drops caused by I1 carry a negative sign since they are in opposition to the drops caused by I2
The complete equation for Mesh 2 is then,
6.4 + 0.05I2  0.05I1 + 0.5I2 = 0
collecting terms and transposing
0.05I1 + 0.55I2 = 6.4 (2)
Equations (1) and (2) may be solved simultaneously by multiplying Equation (2) by 3 and adding the result to Equation (1):
0.15I1  0.05I2 = 1.6 (1)
0.15I1 + 1.65I2 = 19.2 (2)
1.60I2 = 20.8
I2 = 13 amps
Why 3, Where are we getting the 3 from ?
To write an equation for the first mesh, select a convenient starting point such as point a and go around the mesh clockwise, following the direction of I1. Since the first circuit element encountered is the generator, and since the tracing direction passes through the ernf from the negative to the positive terminal, the first term of our equation should be written 8. A drop of +0.1I1 occurs from points b to c, and another drop of +0.05I1 occurs between points d and e.
We next encounter the positive terminal of the battery which should be accounted for in the equation by the term +6.4. On emerging from the battery we arrive at point f which is the same as point a and our traversal of Mesh 1 appears to be complete. At this point, we could very easily make the mistake of writing the equation for the first mesh as follows:
8 + 0.1I1 + 0.05I1 + 6.4 = 0
However, an extremely important point has been overlooked! Although current I1 flows downward between points d and e, this is not the only current flowing between these points. Current I2, in the adjacent mesh, also flows between these same points and I2 flows upward from e to d contrary to the direction given I1. Since the drop caused by I2 is in opposition to the drop caused by I1, this serves to reduce the total drop caused by I1 and must be indicated in the equation of the first by a minus sign. The complete equation for Mesh 1 then becomes:
8 + 0. 1I1 + 0.05I1  0.05I2 + 6.4 = 0
Or by collecting terms and transposing
0.15I1  0.05I2 = 1.6 (1)
Turning to Mesh 2, start at point f and traverse the loop fdgh, following the direction I2, writing the signs of the emfs according to the signs of the terminals first encountered and considering all drops caused by I2 to be positive. This time, any drops caused by I1 carry a negative sign since they are in opposition to the drops caused by I2
The complete equation for Mesh 2 is then,
6.4 + 0.05I2  0.05I1 + 0.5I2 = 0
collecting terms and transposing
0.05I1 + 0.55I2 = 6.4 (2)
Equations (1) and (2) may be solved simultaneously by multiplying Equation (2) by 3 and adding the result to Equation (1):
0.15I1  0.05I2 = 1.6 (1)
0.15I1 + 1.65I2 = 19.2 (2)
1.60I2 = 20.8
I2 = 13 amps
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