Calculated as follows
Eq1: i1 + i2 = i3
From Node ‘B’ we go from - to + over the 10v battery giving +10, then because we do not have the voltage we use ohm’s law over the Voltage drop of R1 so that is 4i1, then we get a voltage drop over R3 so that is 8i3 ...
Eq2: 10 -4i1 -8i3 = 0
From Node ‘B’ again we go against the current so this will be a plus, over R3 so 8i3 R2 is also against the current so 6i2 and through the 4v battery will be a -4...
Eq3: 8i3 + 6i2 - 4 = 0
Rewrite with i3 equivalent
Eq1: i1 + i2 = i3
Eq2: 10 -4i1 -8i3 = 0
Eq3: 8i3 + 6i2 - 4 = 0
Rewrite to simplify:
Eq2b: -4i1 - 8i1 - 8i2 = -12i1 - 8i2 = -10
Eq3b: 8i1 + 8i2 + 6i2 = 8i1 + 14i2 = 4
We now have the makings of a simultaneous equation.
Multiply Eq2b x 2 and Eq3b x 3 to get
-24i1 - 16i2 = -20
Add them 24i1 + 42i2 = 12
26i2 = -8
So now we are left with 26i2 = -8
Divide both sides by 26 to find i2
i2 = -8 / 26 = 0.308
The value we got for i2 can now be used to substitute i2 values in the original equations.
Use Eq3 to solve for i2
8i3 + 6i2 - 4 = 0
8i3 + 6 (-0.308) -4 = 0
8i3 = 6(0.308) + 4
i3 = 5.848 / 8 = 0.731
We now have 2 of the missing currents
i2 = -0.308
i3 = 0.731
With simple maths manipulation we get i1
i1 + i2 = i3
i1 = i3 - i2
i1 = 0.731 + 0.308
i1 = 1.039
i3 = We now have all 3 currents
i1 = 1.039
i2 = -0.308
i3 = 0.731
.
These can now be used to solve for voltage drops over each resistor
Voltage drop on R1 = 4i1 = 4 (1.039) = 4.15v
Voltage drop on R2 = 6i2 = 6 (0.308) = 1.85v
Voltage drop on R2 = 8i3 = 8 (0.731) = 5.85v
