Figure PI-14 shows a simple single-phase ac power system with three loads. The voltage source is V = 12020° V, and the impedances of the three loads are Z - 5230° N Z - 5. 545° A Z, - 52-90° N calculate the current I,

answer is a. R=22.5 ohms, X = 30 ohms b. Vs = 747.11<34.82 Vrms     How do I solve in steps?

Add the following two impedances (a) in series and (b) in parallel. Z1 = 4 + 3i and Z2 = 12 + 5i

When a resistor is connected across the terminals of an ac generator (117 V) that has a fixed frequency, there is a current of 0.332 A in the resistor. When an inductor is connected across the terminals of the same generator, there is a current of 0.365 A in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what is (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise. (a) Number (b) Number Buda i Vrms R Irms.R Circuit 1 Units Units Vrms. Irms, L Circuit 2 > L Vrms R www Circuit 3