Question 9.12: Use the mesh-current method to find the voltages V1, V2,and ...

The circuit has two meshes and a dependent voltage source, so we must write two mesh-current equations and a constraint equation. The reference direction for the mesh currents \pmb{I}_{1} and \pmb{I}_{2}is clockwise, as shown in Fig. 9.37. Once we know \pmb{I}_{1} and\pmb{I}_{2} we can easily find the unknown voltages. Summing the voltages around mesh 1 gives

150 = \left(1 + j 2\right)\pmb{I}_{1} + \left(12 – j16\right)\left(\pmb{I}_{1} – \pmb{I}_{2}\right),

or

150 = \left(13 -j 14\right)\pmb{I}_{1} -\left(12 – j16\right)\pmb{I}_{2}.

Summing the voltages around mesh 2 generates the equation

0 = \left(12 – j16\right)\left(\pmb{I}_{2} – \pmb{I}_{1}\right) + \left(1 + j 3\right)\pmb{I}_{2} + 39\pmb{I}_{x}.

Figure 9.37 reveals that the controlling current \pmb{I}_{x}is
the difference between \pmb{I}_{1} and\pmb{I}_{2} that is, the constraint is

\pmb{I}_{x}=\pmb{I}_{1} – \pmb{I}_{2}.

Substituting this constraint into the mesh 2 equation and simplifying the resulting expression gives

0 = \left(27 + j16\right)\pmb{I}_{1} – \left(26 + j13\right)\pmb{I}_{2}.

Solving for \pmb{I}_{1} and\pmb{I}_{2}yields

\pmb{I}_{1} = -26 – j52 A,

\pmb{I}_{x}= -2 + j6 A.

The three voltages are

\pmb{V}_{1} = \left(1 + j 2\right)\pmb{I}_{1} = 78 – j104 V,

\pmb{V}_{2} =\left (12 – j16\right)\pmb{I}_{x} = 72 + j104 V,

\pmb{V}_{3} = \left(1 + j3\right)\pmb{I}_{2} = 150 – j130 V.

Also

39\pmb{I}_{x} = -78 + j 234 V.

We check these calculations by summing the voltages around closed paths:

-150 + \pmb{V}_{1} + \pmb{V}_{2} = -150 + 78 – j104 + 72+ j104 = 0,