**Mesh Analysis**

Solve the following two equations for *I*_{2}:

(2+*j*2)*I*_{1} - 2*I*_{2} = 8

-2 *I*_{1} + (4-*j*2)*I*_{2} = 6

Life is short! I put this into MathCAD and found

*I*_{2} = 3.4+*j*0.2 A = 3.41 A \/ -3.4 deg

**Nodal Analysis**

Solve the following two equations for *V*_{1} and *V*_{2}:

Eq. 1: *j**V*_{1} + *V*_{2} = 8

Eq. 2: *V*_{1} + (-1+*j*)*V*_{2} = -*j*6

Multiplying Eq. 1 times *j* and then adding to Eq. 2 yields:

(-1+*j*2)*V*_{2} = *j*2,

thus

*V*_{2} = 0.8-*j*0.4

(after conversion of the two complex numbers to polar form, division, then conversion back).

Back substitution into Eq. 1 yields:

*V*_{1} = 0.4-*j*7.2

so that

*V*_{1}-*V*_{2} = -0.4-*j*6.8

Finally,

*I*_{2} = (*V*_{1}-*V*_{2})/(-*j*2) = (-0.4-*j*6.8)/(-*j*2) = 3.4-*j*0.2

Life is too short for that too!

"http://www.physics.udel.edu/~watson/phys345/protected/exercises/answers/P4-8.html"

Last updated Oct. 1, 1998.

Copyright George Watson, Univ. of Delaware, 1998.