Answer:
the branch currents are as follows:
top left: I2 = 0.625 A
middle left: I1 = 2.500 A
bottom left: I1-I2 = 1.875 A
top center: I2+I3 = 2.500 A
bottom center: I2+I3-I1 = 0 A
right: I3 = 1.875 A
Explanation:
You can write the KVL equations:
Top left loop:
I2(4) +(I2 +I3)(2) +I1(1) = 10
Bottom left loop:
(I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10
Right loop:
(I2+I3)(2) +(I2+I3-I1)(2) = 5
In matrix form, the equations are ...
![\left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%266%262%5C%5C7%26-6%26-2%5C%5C-2%264%264%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7DI_1%5C%5CI_2%5C%5CI_3%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D10%5C%5C10%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
These equations have the solution ...
![\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7DI_1%5C%5CI_2%5C%5CI_3%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2.500%5C%5C0.625%5C%5C1.875%5Cend%7Barray%7D%5Cright%5D)
This means the branch currents are as follows:
top left: I2 = 0.625 A
middle left: I1 = 2.500 A
bottom left: I1-I2 = 1.875 A
top center: I2+I3 = 2.500 A
bottom center: I2+I3-I1 = 0 A
right: I3 = 1.875 A
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This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.
When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.
Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.
Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.
(I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)
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It helps to be familiar with the formulas for resistors in series and parallel.
