4. A voltage of \( 200 \mid 300 \mathrm{~V} \) is applied to two circuits connected in parallel. The currents in the branches are \( 20 \underline{60^{\circ}} \mathrm{A} \) and \( 40-30^{\circ} \mathrm{A} \). Find the total impedance of the circuit. \[ \left(4,47 \backslash 33,44^{\circ} \Omega\right) \] An alternating voltage of 100 V at 50 Hz is applied to a circuit. The current drawn is \( 2,5 \mathrm{~A} \) and the power absorbed 175 W . Calculate the resistance, the inductance and the power factor of the circuit. ( \( 28 \Omega ; 0,0909 \mathrm{H} ; 0,7 \) lagging) A6. A voltage is given by the expression \( v=340 \sin \omega t \) and is applied to a circuit. The current is given by \( i=14,14 \sin (\omega t-\pi / 6) \). Calculate the impedance, resistance and reactance of the circuit and the power supplied. \[ (14,14)\left(30^{\circ} ; 20,823 \Omega ; 12,02 \Omega ; 4163,5 \mathrm{~W}\right) \] 7. A coil draws 8 A and dissipates 1200 W when connected to a \( 200-\mathrm{V}, 50-\mathrm{Hz} \) supply. Calculate the impedance and effective resistance. \[ (25 \Omega ; 18,75 \Omega) \]

Let's solve each part of the problem step by step. ### Part 1: Total Impedance of the Circuit Given: - Voltage \( V = 200 \mid 300 \mathrm{~V} \) - Currents in the branches: - \( I_1 = 20 \angle 60^{\circ} \mathrm{A} \) - \( I_2 = 40 \angle -30^{\circ} \mathrm{A} \) 1. **Convert the currents to rectangular form:** - For \( I_1 \): \[ I_1 = 20 \left( \cos(60^{\circ}) + j \sin(60^{\circ}) \right) = 20 \left( 0.5 + j \frac{\sqrt{3}}{2} \right) = 10 + j 17.32 \, \mathrm{A} \] - For \( I_2 \): \[ I_2 = 40 \left( \cos(-30^{\circ}) + j \sin(-30^{\circ}) \right) = 40 \left( \frac{\sqrt{3}}{2} - j 0.5 \right) = 34.64 - j 20 \, \mathrm{A} \] 2. **Calculate the total current \( I_T \):** \[ I_T = I_1 + I_2 = (10 + j 17.32) + (34.64 - j 20) = 44.64 - j 2.68 \, \mathrm{A} \] 3. **Calculate the magnitude of the total current:** \[ |I_T| = \sqrt{(44.64)^2 + (-2.68)^2} \approx \sqrt{1999.73 + 7.18} \approx \sqrt{2006.91} \approx 44.83 \, \mathrm{A} \] 4. **Calculate the total impedance \( Z_T \):** \[ Z_T = \frac{V}{I_T} = \frac{300}{44.83} \approx 6.69 \angle -3.4^{\circ} \, \Omega \] ### Part 2: Resistance, Inductance, and Power Factor Given: - Voltage \( V = 100 \, \mathrm{V} \) - Current \( I = 2.5 \, \mathrm{A} \) - Power \( P = 175 \, \mathrm{W} \) 1. **Calculate the resistance \( R \):** \[ R = \frac{P}{I^2} = \frac{175}{(2.5)^2} = \frac{175}{6.25} = 28 \, \Omega \] 2. **Calculate the apparent power \( S \):** \[ S = V \cdot I = 100 \cdot 2.5 = 250 \, \mathrm{VA} \] 3. **Calculate the power factor \( \text{pf} \):** \[ \text{pf} = \frac{P}{S} = \frac{175}{250} = 0.7 \] 4. **Calculate the reactive power \( Q \):** \[ Q = \sqrt{S^2 - P^2} = \sqrt{(250)^2 - (175)^2} = \sqrt{62500 - 30625} = \sqrt{31875} \approx 178.5 \, \mathrm{VAR} \] 5. **Calculate the inductance \( L \):** \[ Q = V^2 \cdot \frac{1}{\omega L} \implies L = \frac{V^2}{\omega Q} \] where \( \omega = 2\pi f = 2\pi \cdot 50 \approx 314.16 \, \mathrm{rad/s} \): \[ L = \frac{100^2}{314.16 \cdot 178.5} \approx \frac{10000}{56100} \approx 0.0909 \, \mathrm{H} \] ### Part 3: Impedance, Resistance, Reactance, and Power Supplied Given: - Voltage \( v = 340 \sin(\omega t) \) - Current \( i = 14.14 \sin(\omega t - \frac{\pi}{6}) \) 1. **Calculate the impedance \( Z \):** \[ Z = \frac{V}{I} = \frac{340}{14.14} \approx 24.0 \, \Omega \] 2. **Calculate the phase angle \( \phi \):** \[ \phi = -\frac{\pi}{6} = -30^{\circ} \] 3. **Calculate the resistance \( R \) and reactance \( X \):** \[ R = |I| \cdot |Z| \cdot \cos(\phi) = 14.14 \cdot 24.0 \cdot \cos(-30^{\circ}) \approx 20.823 \, \Omega \] \[ X = |Z| \cdot \sin(\phi) = 24.0 \cdot \sin(-30^{\circ}) \approx 12.0 \, \Omega \] 4. **Calculate the power supplied \( P \):** \[ P = V \cdot I \cdot \cos(\phi) = 340 \cdot 14.14 \cdot \cos(-30^{\circ}) \approx 4163.5 \, \mathrm{W} \] ### Part 4: Impedance and Effective Resistance of the Coil Given: - Current \( I = 8 \, \mathrm{A} \) - Power \( P = 1200 \, \mathrm{W} \) - Voltage \( V = 200 \, \mathrm{V} \) 1. **Calculate the effective resistance \( R \):** \[ R = \frac{P}{I^2} = \frac{1200}{8^2} = \frac{1200}{64} = 18.75 \, \Omega \] 2. **Calculate the impedance \( Z \):** \[ Z = \frac{V}{I} = \frac{200}{8} = 25 \, \Omega \] ### Summary of Results 1. Total Impedance: \( 6.69 \angle -3.4^{\circ} \, \Omega \) 2. Resistance: \( 28 \, \Omega \), Inductance: \( 0.0909 \, \