Question

impedance is a measure of the opposition to the flow of alternating electrical current. it consists of two parts called resistance and reactance. impedance ( z ) in ohms (( omega )) can be expressed as a complex number where the real part represents resistance and the imaginary part represents reactance. in the series circuit shown, the total impedance is the sum of the individual impedances. assuming that the light bulbs are pure resistive and the motors are pure reactive, find the total impedance in this circuit and express it in the form ( z = a + bi ).
the total impedance is (\boxed{quad} omega).
(simplify your answer. type your answer in the form ( a + bi ).)

Explanation:

Step1: Identify Resistive and Reactive Parts

Resistive (real) parts: light bulbs (80Ω, 50Ω). Reactive (imaginary) parts: motors (13Ω, 19Ω, but reactive is imaginary, so real for resistors, imaginary for reactance? Wait, pure resistive is real (a), pure reactive is imaginary (bi). So total resistance (real) = 80 + 50 = 130. Total reactance (imaginary) = 13 + 19 = 32. Wait, no: pure resistive is real (so resistance is real, reactance is imaginary). So impedance Z = (sum of resistances) + (sum of reactances)i. So resistances: 80Ω and 50Ω (light bulbs, pure resistive) → real part: 80 + 50 = 130. Reactances: 13Ω and 19Ω (motors, pure reactive) → imaginary part: 13 + 19 = 32. So Z = 130 + 32i.

Step2: Verify Calculation

Sum resistances: 80 + 50 = 130. Sum reactances: 13 + 19 = 32. So impedance is 130 + 32i ohms.

Answer:

130 + 32i