University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2020
Views:

255

Pages:

5
Author:

Arielle Armstrong

In fig 1 below, the 3-phase source is balanced 1000 watt, reads upscale 4411 watt, while P2 reads downscale -1000 watt, & VAB = 440∠0° V rms. Determine: (2 marks) The value of PF for the load (is it leading or lagging?) (4 marks) I_AN and Z_P (2 marks) V_AN (2 marks) The PF at the source (is it leading or lagging?) & the total complex power of the source. [Diagram of a 3-phase circuit with a balanced source and load] Calculations: tanθ = √3(P_R - P_I) / (P_R + P_I) = √3(-1000 - 4411) / (-1000 + 4411) = -2.742 ⇒ θ = tan⁻¹(-2.742) = -70° (leading) ⇒ PF_L = cos(-70°) = 0.342 (leading) P_1 = |V_AB||I_AN|cos(30° + θ) ⇒ 4411 / 440 x √3 x |I_AN| = 13.0867 ⇒ |I_AN| = 13.0867 / √3 = 7.556 ⇒ Z_P = 440 / 7.556 = 58.2348∠-70° V_AN = V_AB / √3 = 440 / √3 = 254∠-30° I_An = 13.0867 ∠-70° S_source = 3 × V_AN × I_AN* = 3 × 254∠-30° × 7.556∠70° = 5732.35∠40° P_R = 4411 - 1000 = 3411 W ⇒ P_L = 3411 W P_L = |V_AB||I_AL|PF_L ⇒ |I_AL| = 440 x 0.342 / √3 = 7.555 A ⇒ I_AN = 7.555 ∠-70° A ⇒ I_AA = 13.0867 ∠-70° A V_AN = (8-j) I_AA + V_AB = (8-j)7.555 + 440∠0° = 316.39 ∠-12.73° V OR V_AB = (8-j) I_AA + V_BA - (8-j) I_BA ⇒ I_BA = 7.0367 ∠-80° A V_BN = (8-j) I_BA + V_AB = 316.39 ∠-12.73° V S_R = 3V_AN * I_AN* = 3 × 254∠-30° × 7.556∠70° = 5732.35∠40° VA PF_s = cos(-12.73°) = 0.95055 (leading) Q(2)(a) (6 marks) For the circuit shown in Fig.2, find V2 and Ix. circuit diagram with a voltage source, resistors, and a current source Mesh Analysis: Mesh 1: -30∠0° + 20I1 + V1 = 0 Mesh 2: V1 - 20I2 + V2 = 0 Solving these equations, we get: I1 = 0.4 ∠-20° A I2 = -0.4 ∠-20° A V1 = 8 ∠-20° V V2 = 8.9 ∠-17.8° V Nodal Analysis: 30∠0° - V2 + V2 = 20I1 50I1 + V1 + V2 = 20 Solving these equations, we get: Ix = 0.76 ∠-4.6° A V2 = 8.9 ∠-17.8° V

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