Question 9.9: Use the concept of source transformation to find the phasor ...

We can replace the series combination of the voltage source \left(40\angle 0^{°} \right) and the impedance of 1 + j 3 Ω with the parallel combination of a current source and the 1 + j 3 Ω impedance. The source current is

\pmb{I} =\frac{40}{1 + j 3}=\frac{40}{10}\left(1 – j 3\right) = 4 – j12 A.

Thus we can modify the circuit shown in Fig. 9.27 to the one shown in Fig. 9.28. Note that the polarity reference of the 40 V source determines the reference direction for I.

Next, we combine the two parallel branches into a single impedance,

Z =\frac{\left(1 + j 3\right)\left(9 – j 3\right)}{10} = 1.8 + j 2.4 Ω,

which is in parallel with the current source of 4 – j 12 A .Another source transformation converts this parallel combination to a series combination consisting of a voltage source in series with the impedance of 1.8 + j 2.4 Ω.The voltage of the voltage source is

\pmb{V} = \left(4 – j 12\right)\left(1.8 + j 2.4\right) = 36 – j 12 V.

Using this source transformation, we redraw the circuit as Fig. 9.29. Note the polarity of the voltage source. We added the current \pmb{I}_{0}to the circuit to expedite the solution for \pmb{V}_{0}.

Also note that we have reduced the circuit to a simple series circuit. We calculate the current \pmb{I}_{0} by dividing the voltage of the source by the total series impedance:

=\frac{39 + j27}{25} = 1.56 + j1.08 A.

We now obtain the value of V_{0} by multiplying I_{0} by the impedance : 10 – j19:

\pmb{V}_{0} = \left(1.56 + j1.08\right)\left(10 – j19\right)=36.12-j18.84 V.