Här kör vi nodanalys. $V_{A}$ till vänster om $j4\Omega$ och $V_{B}$ till höger.

Sedan fås

\bar{I}_{2} = \frac{\bar{V}_{A} - \bar{V}_{B}}{4j} \frac{\bar{V}_{A} - 10}{1} + \frac{\bar{V}_{A}}{-j2} + \frac{\bar{V}_{A} - \bar{V}_{B}}{j4} = 0 \Rightarrow 4j\bar{V}_{A} - 40j - 2\bar{V}_{A} + \bar{V}_{A} - \bar{V}_{B} = 0 \Rightarrow (1 - 4j)\bar{V}_{A} + \bar{V}_{B} = -40j \quad (1) \frac{\bar{V}_{B} - \bar{V}_{A}}{j4} + \frac{\bar{V}_{B}}{2} + \frac{\bar{V}_{B} - (1 + j)}{2 - j2} = 0 \Rightarrow \frac{\bar{V}_{B} - \bar{V}_{A}}{j4} + \frac{\bar{V}_{B}}{2} + \frac{(2 + 2j)(\bar{V}_{B} - (1 + j))}{8} = \frac{\bar{V}_{B} - \bar{V}_{A}}{j4} + \frac{\bar{V}_{B}}{2} + \frac{(1 + j)(\bar{V}_{B} - (1 + j))}{4} = 0 \Rightarrow \bar{V}_{B} - \bar{V}_{A} + 2j\bar{V}_{B} + j(1 + j)(\bar{V}_{B} - (1 + j)) = 0 \Rightarrow \bar{V}_{B} - \bar{V}_{A} + 2j\bar{V}_{B} + (-1 + j)(\bar{V}_{B} - (1 + j)) = 0 \Rightarrow \bar{V}_{B} - \bar{V}_{A} + 2j\bar{V}_{B} - \bar{V}_{B} + (1 + j) + j\bar{V}_{B} - j(1 + j) = 0 \Rightarrow - \bar{V}_{A} + (1 + 2j - 1 + j)\bar{V}_{B} + 1 + j - j + 1 = 0 \Rightarrow - \bar{V}_{A} + 3j\bar{V}_{B} + 2 = 0 \Rightarrow \bar{V}_{A} - 3j\bar{V}_{B} = 2 \quad (2) \left\{ \begin{array}{l} (1 - 4j)\bar{V}_{A} + \bar{V}_{B} = -40j \quad (1) \\ \bar{V}_{A} - 3j\bar{V}_{B} = 2 \quad (2) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} (1 - 4j)\bar{V}_{A} + \bar{V}_{B} = -40j \quad (1) \\ (1 - 4j)\bar{V}_{A} - (1 - 4j)3j\bar{V}_{B} = 2(1 - 4j) \quad (2) \end{array} \right. \Rightarrow (1) - (2) \Rightarrow \bar{V}_{B} + (1 - 4j)3j\bar{V}_{B} = -40j - 2(1 - 4j) \Rightarrow (1 + 3j + 12)\bar{V}_{B} = -40j - 2 + 8j = -2 - 32j (13 + 3j)\bar{V}_{B} = (-2 - 32j) \Rightarrow \bar{V}_{B} = - \frac{2(1 + 16j)}{(13 + 3j)} \Rightarrow \bar{V}_{A} = 2 + 3j\bar{V}_{B} = 2 - \frac{6j(1 + 16j)}{(13 + 3j)} = \frac{26 + 6j - 6j + 96}{13 + 3j} = \frac{122}{13 + 3j} \bar{I}_2 = \frac{\bar{V}_4 - \bar{V}_B}{4j} = \frac{\frac{122}{13 + 3j} - \left(-\frac{2(1 + 16j)}{(13 + 3j)}\right)}{4j} = \frac{124 + 32j}{4j(13 + 3j)} = \frac{31 + 8j}{(13 + 3j)} = \frac{\sqrt{31^2 + 8^2}}{\sqrt{13^2 + 3^2}} \angle \arctan \left(\frac{8}{31}\right) - \arctan \left(\frac{3}{13}\right) = 2.4\angle 1.5^\circ