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$$
\begin{array}{l}{\text { Use the concept of source transformation to find the }} \\ {\text { phasor voltage } \mathrm{V}_{0} \text { in the circuit shown }}\end{array}
$$

$$
Z_{e q}=\frac{Z_{1} Z_{2}}{Z_{1}+Z_{2}}=\frac{(1+3J)(9-3J}{1+3J+9-3J}=1.8+J2.4
$$

$$
\overline{V}=\overline{I} \overline{z}
$$

$$
\overline{V}=(4-J12)(1.8+J2.4)
$$

$$
\overline{V}=36-J12
$$

$$
V_{0}=\frac{V Z_{1}}{Z_{1}+Z_{2}}
$$

$$
V_{0}=\frac{(36-J{12}(10-J19)}{1.8+J{2.4}+0.2+J0.6+10-J19}
$$

$$
V_{0}=36.12-J{18.84}
$$                    (V)

 

 

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