Consider the circuit shown in the figure, the current supplied by the ac voltage source  is 

$$\begin{gathered} Z_1=R_1+x_{L}i=8+6i \\ {Z}_2=R_2-x_{c}i=3-4i \\ Now,\frac{1}{Z^{\text{'}}}=\frac{1}{Z_1}+\frac{1}{Z_2} \\ \frac{1}{Z^{\text{'}}}=\frac{1}{8+6i}+\frac{1}{3-4i} =\frac{8-6i}{100}+\frac{3+4i}{100}=\frac{20+10i}{100} \\ \frac{1}{Z^{\text{'}}}=\frac{(2+i)}{10}\Rightarrow \frac{10}{(2+i)} \\ {Z}^{\text{'}}=2(2-i) \\ {Z}^{\text{'}}=(4-2i) \to \left(1\right) \\ {Z}_3=R_3+X_L^{\text{'}}i=(2+10i) \to \left(2\right) \\ Z=6+8i \\ \left|Z\right|=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}=10\Omega \\ I=\frac{80}{\left|Z\right|}=\frac{80}{10}=8amp \end{gathered}$$