The currents i1 i2 i4 i5 i6 in the circuit that is shown can

The currents, i_1, i_2, i_4, i_5, i_6, in the circuit that is shown can be determined from the solution of the following system of equations. (Obtained by applying Kirch-hoff\'s law.) 8i_1 - 4i_2 i_4 = 20, - 4i_1 + 11.5i_2 - 2.5i_3 - 5i_5 = -12 -2.5i_2 + 4.5i_3 - 2i_6 = 14, -i_1 + 3i_4 - 2i_5 = 8 - 5i_2 - 2i_4 + 8.5i_5 - 1.5i_6 = -30, - 2i_3 - 1.5i_5 + 8_i_6 = 0 Solve the system of the equations for the unknown currents.

Solution

{ x1 = -3.891976061384913, x2 = -9.099637919066469, x3 = -8.744310826782733, x4 = -5.992945988030693, x5 = -11.04343095135358, x6 = -4.25672101007448 }.

x=i
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