esercizi

Let's go through the exercises step by step, starting with Exercise 14 as an example and then providing a general approach for solving these types of problems. ### Exercise 14 Given: - \( R_1 = 5 \Omega \) - \( C_1 = 200 \mu F \) - \( R_2 = 10 \Omega \) - \( L_2 = 10 mH \) - \( C_3 = 200 \mu F \) - \( v_{G1}(t) = 4\cos(2\pi t + 10^\circ) V \) - \( v_{G2}(t) = 4\cos(2\pi t - 10^\circ) V \) We need to find the currents and powers for each generator. #### Step-by-Step Solution: 1. **Convert the generators' voltages to phasors:** - \( v_{G1} = 4 \angle 10^\circ \) - \( v_{G2} = 4 \angle -10^\circ \) 2. **Calculate the impedance of each component:** - Capacitor \( C_1 \): \( Z_{C1} = \frac{1}{j\omega C_1} = \frac{1}{j2\pi \cdot 500 \times 200 \times 10^{-6}} = -j4 \Omega \) - Inductor \( L_2 \): \( Z_{L2} = j\omega L_2 = j2\pi \cdot 10^3 \times 10 \times 10^{-3} = 20\pi j \approx 62.83j \Omega \) - Capacitor \( C_3 \): \( Z_{C3} = \frac{1}{j\omega C_3} = \frac{1}{j2\pi \cdot 500 \times 200 \times 10^{-6}} = -j4 \Omega \) 3. **Find the total impedance for each branch:** - Branch 1 (G1 to i1): \( Z_{total,1} = R_1 + j\omega C_1 = 5 - j4 \) - Branch 2 (G2 to i2): \( Z_{total,2} = R_2 + j\omega L_2 = 10 + 62.83j \) 4. **Calculate the currents:** - For branch 1: \( I_1 = \frac{v_{G1}}{Z_{total,1}} = \frac{4e^{j10^\circ}}{5 - j4} \) Simplify using complex division: \( I_1 = \frac{4e^{j10^\circ}(5 + j4)}{(5 - j4)(5 + j4)} = \frac{20e^{j10^\circ} + 16je^{j10^\circ}}{25 + 16} = \frac{36e^{j10^\circ} + 16je^{j10^\circ}}{41} \) \( I_1 = \left( \frac{36}{41} + j\frac{16}{41} \right) e^{j10^\circ} \approx (0.878 + 0.390j)e^{j10^\circ} \) \( I_1 = 0.878e^{j10^\circ} + j0.390e^{j10^\circ} \) - For branch 2: \( I_2 = \frac{v_{G2}}{Z_{total,2}} = \frac{4e^{-j10^\circ}}{10 + 62.83j} \) Simplify using complex division: \( I_2 = \frac{4e^{-j10^\circ}(10 -