1. **Problem Statement:** Find the current $i_x$ in the given AC circuit using nodal analysis. 2. **Given:** Voltage source $v_s = 20\cos 4t$ V, resistor $R = 10\ \Omega$, capacitor $C = 0.1$ F, inductors $L_1 = 1$ H and $L_2 = 0.5$ H, and a dependent current source $2i_x$. 3. **Step 1: Convert to phasor domain.** The source voltage phasor is $V_s = 20\angle 0^\circ$ V. Angular frequency $\omega = 4$ rad/s. 4. **Step 2: Calculate impedances.** - Resistor impedance: $Z_R = 10\ \Omega$ - Capacitor impedance: $Z_C = \frac{1}{j\omega C} = \frac{1}{j4 \times 0.1} = -j2.5\ \Omega$ - Inductor $L_1$ impedance: $Z_{L1} = j\omega L_1 = j4 \times 1 = j4\ \Omega$ - Inductor $L_2$ impedance: $Z_{L2} = j\omega L_2 = j4 \times 0.5 = j2\ \Omega$ 5. **Step 3: Define node voltages.** Let the node at the junction after the resistor be $V_1$ (with respect to ground). 6. **Step 4: Write nodal equation at $V_1$.** Currents leaving node $V_1$ are: - Through resistor: $\frac{V_1 - V_s}{Z_R} = \frac{V_1 - 20}{10}$ - Through capacitor: $\frac{V_1}{Z_C} = \frac{V_1}{-j2.5} = j0.4 V_1$ - Through inductor $L_1$: $\frac{V_1}{Z_{L1}} = \frac{V_1}{j4} = -j0.25 V_1$ - Through dependent current source and $L_2$: The dependent current source is $2i_x$ upward, where $i_x$ is the current through the capacitor (downward). Since $i_x = \frac{V_1}{Z_C} = j0.4 V_1$, the dependent current source is $2i_x = 2 \times j0.4 V_1 = j0.8 V_1$ upward. The inductor $L_2$ is connected to ground, so the voltage at ground is zero. 7. **Step 5: Apply KCL at node $V_1$: sum of currents leaving node = 0** $$ \frac{V_1 - 20}{10} + j0.4 V_1 - j0.25 V_1 + i_x - 2i_x = 0 $$ Note: $i_x$ is downward current through capacitor, so it leaves node $V_1$ downward, counted positive leaving. Substitute $i_x = j0.4 V_1$: $$ \frac{V_1 - 20}{10} + j0.4 V_1 - j0.25 V_1 + j0.4 V_1 - 2 \times j0.4 V_1 = 0 $$ Simplify imaginary terms: $j0.4 V_1 - j0.25 V_1 + j0.4 V_1 - j0.8 V_1 = (0.4 - 0.25 + 0.4 - 0.8)j V_1 = (-0.25)j V_1$ So equation becomes: $$ \frac{V_1 - 20}{10} - j0.25 V_1 = 0 $$ Multiply both sides by 10: $$ V_1 - 20 - j2.5 V_1 = 0 $$ Group terms: $$ V_1 (1 - j2.5) = 20 $$ 8. **Step 6: Solve for $V_1$:** $$ V_1 = \frac{20}{1 - j2.5} $$ Multiply numerator and denominator by conjugate $1 + j2.5$: $$ V_1 = \frac{20 (1 + j2.5)}{1^2 + (2.5)^2} = \frac{20 (1 + j2.5)}{1 + 6.25} = \frac{20 (1 + j2.5)}{7.25} $$ $$ V_1 = 2.7586 + j6.8966 $$ 9. **Step 7: Calculate $i_x$:** Recall $i_x = \frac{V_1}{Z_C} = \frac{V_1}{-j2.5} = -\frac{V_1}{j2.5} = j \frac{V_1}{2.5}$ $$ i_x = j \frac{2.7586 + j6.8966}{2.5} = j (1.1034 + j2.7586) = j 1.1034 + j^2 2.7586 = j 1.1034 - 2.7586 $$ $$ i_x = -2.7586 + j1.1034 $$ 10. **Step 8: Express $i_x$ in magnitude and phase:** $$ |i_x| = \sqrt{(-2.7586)^2 + (1.1034)^2} = \sqrt{7.61 + 1.217} = \sqrt{8.827} = 2.971 $$ $$ \theta = \tan^{-1} \left( \frac{1.1034}{-2.7586} \right) = \tan^{-1}(-0.4) = -21.8^\circ $$ Since real part is negative and imaginary part positive, angle is in second quadrant: $$ \theta = 180^\circ - 21.8^\circ = 158.2^\circ $$ 11. **Final answer:** $$ i_x(t) = 2.971 \cos(4t + 158.2^\circ)\ \text{A} $$