1. **State the problem:** We have a parallel circuit with four branches involving currents $i_A$, $i_B$, and $i_C$, voltage $V_x$ across an 18 \(\Omega\) resistor, and given currents and resistances. We need to find $i_A$, $i_B$, and $i_C$. 2. **Identify values and relations:** - Current source in first branch: $5.6\ A$ - Second branch: resistor $R_2=18\ \Omega$, voltage across it $V_x$, current $i_A$ - Third branch: current-controlled current source (CCCS) $i_B=0.1 V_x$ - Fourth branch: resistor $R_4=9\ \Omega$ with current $i_C$, then a $2\ A$ current source 3. **Write equations for currents in branches:** All four branches are in parallel, so voltage across each branch is the same and equal to $V_x$. - For the second branch: $i_A = \frac{V_x}{R_2} = \frac{V_x}{18}$ - For the third branch: $i_B = 0.1 V_x$ - For the fourth branch: two elements in series, current source of 2 A sets $i_C = 2\ A$ 4. **Apply Kirchhoff's Current Law (KCL) at the node:** Sum of currents into node equals sum out. Assuming currents downward are positive, total current from node = sum of branch currents: $$ 5.6 = i_A + i_B + i_C $$ Substitute known expressions: $$ 5.6 = \frac{V_x}{18} + 0.1 V_x + 2 $$ 5. **Solve for $V_x$: ** $$ 5.6 - 2 = \frac{V_x}{18} + 0.1 V_x $$ $$ 3.6 = \frac{V_x}{18} + 0.1 V_x = \frac{V_x}{18} + \frac{0.1\times 18 V_x}{18} = \frac{V_x + 1.8 V_x}{18} = \frac{2.8 V_x}{18} $$ $$ 3.6 = \frac{2.8 V_x}{18} \implies V_x = \frac{3.6 \times 18}{2.8} = \frac{64.8}{2.8} = 23.1429\ V $$ 6. **Calculate currents $i_A$, $i_B$, $i_C$ using $V_x$: ** - $i_A = \frac{V_x}{18} = \frac{23.1429}{18} = 1.2857\ A$ - $i_B = 0.1 V_x = 0.1 \times 23.1429 = 2.3143\ A$ - $i_C = 2\ A$ (given) 7. **Final answer:** $$ i_A \approx 1.29\ A $$ $$ i_B \approx 2.31\ A $$ $$ i_C = 2.00\ A $$