for the circuit below all values are rms with a source frequency of 60 hertz. the generator impedance is negligible (0 w). calculate the value of ib and and determine the value of vb-n. now, calculate the rms values of apparent, real and reactive power of zb. (show units for all)

Reactive power is the power in an AC circuit that is required to establish and maintain a voltage across a load. It is associated with the storage and release of energy in the form of electric and magnetic fields. Reactive power does not contribute to the actual work output of a system and is measured in Volt-amperes reactive (VARs). Power factor is a measure of reactive power relative to the total power in a system.

Source frequency = 60 Hz
Generator impedance = 0 W
Circuit:
Vg = 170 V
R1 = 10 Ω
R2 = 20 Ω
Zb = 20 + j10 Ω

The current flowing through the generator (Ig) and the branch (Ib) can be calculated from Ohm's Law:
Ig = Vg/R1 + Vg/R2 = 170/10 + 170/20 = 17 A
Ib = Ig - Vg/Zb = 17 - 170/(20 + j10) = 17 - 16.4 + j4.4 = 0.6 + j4.4 A
Since Ib is a complex number, we can find its magnitude (|Ib|) and angle (θ):
|Ib| = √(0.6² + 4.4²) = 4.46 A
θ = tan⁻¹(4.4/0.6) = 80.16°
The voltage across the branch (Vb-n) can be calculated using Ohm's Law:
Vb-n = Ib × Zb = (0.6 + j4.4) × (20 + j10) = -8.4 + j74.4 V
The apparent power of the branch (Sb) can be calculated from its voltage and current:
Sb = Vb-n × Ib* = (-8.4 + j74.4) × (0.6 - j4.4) = -45.48 + j367.04 VA
The real power of the branch (Pb) can be calculated from its voltage and current:
Pb = Vb-n × Ib = (-8.4 + j74.4) × (0.6 + j4.4) = -45.48 - j367.04 W
The reactive power of the branch (Qb) can be calculated from its voltage and current:
Qb = Vb-n × Ib* = (-8.4 + j74.4) × (0.6 - j4.4) = 676.16 VA

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