This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Haya, twende na swali la 2.1. Data uliyopewa: Resistor, R = 50 Inductor, L = 0.3 H Capacitor, C = 60 = 60 × 10^-6 F Voltage ya supply, V = 200 V (RMS) Frequency ya supply, f = 50 Hz Hatua ya 1: Kokotoa frequency ya angular na reactances. Frequency ya angular: = 2 f = 2 (50 Hz) = 100 rad/s ≈ 314.159 rad/s Inductive reactance: X_L = L = (100 rad/s)(0.3 H) = 30 ≈ 94.248 Capacitive reactance: X_C = (1)/( C) = (1)/((100 rad/s))(60 × 10^-6 F) = (1)/(0.006) = (500)/(3) ≈ 53.052 2.1.1 The current flowing through each branch Current kupitia resistor (I_R): I_R = (V)/(R) = 200 V50 = 4.00 A Current kupitia inductor (I_L): I_L = (V)/(X_L) = 200 V30 = (20)/(3) A ≈ 2.12 A Current kupitia capacitor (I_C): I_C = (V)/(X_C) = 200 V(500)/(3) = (6)/(5) A ≈ 3.77 A 2.1.2 The total current Katika mzunguko wa parallel RLC, jumla ya current inapatikana kwa kutumia formula: I_total = sqrt(I_R^2 + (I_C - I_L)^2) I_total = sqrt((4.00 A))^2 + (3.77 A - 2.12 A)^2 I_total = sqrt(16 + (1.65)^2) I_total = sqrt(16 + 2.7225) I_total = sqrt(18.7225) ≈ 4.33 A 2.1.3 The total impedance Jumla ya impedance (Z_total) inaweza kukokotolewa kwa kutumia sheria ya Ohm: Z_total = (V)/(I_total) = 200 V4.33 A ≈ 46.19 2.1.4 The phase angle of the circuit Phase angle () inapatikana kwa kutumia formula: = ((I_C - I_L)/(I_R)) = (3.77 A - 2.12 A4.00 A) = ((1.65)/(4.00)) = (0.4125) ≈ 22.42^ Kwa kuwa I_C > I_L, mzunguko ni wa capacitive, na jumla ya current inaongoza voltage. 2.1.5 Draw the phasor diagram. Chora voltage V kama reference kwenye positive x-axis. Chora current ya resistor I_R ikiwa in-phase na V (kwenye positive x-axis). Chora current ya inductor I_L ikiwa inalaga V kwa 90^ (kwenye negative y-axis). Chora current ya capacitor I_C ikiwa inaongoza V kwa 90^ (kwenye positive y-axis). Chora net reactive current I_C - I_L (kwenye positive y-axis, kwa sababu I_C > I_L). Chora jumla ya current I_total kama vector sum ya I_R na (I_C - I_L). Itakuwa katika quadrant ya kwanza, ikiongoza voltage kwa angle ya 22.42^. (Kumbuka: Siwezi kuchora mchoro hapa, lakini maelezo hapo juu yanaeleza jinsi ya kuuchora.) 3 done, 2 left today. You're making progress.