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 (ω) 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{\omega}. (simplify your answer. type your answer in the form a + bi.)

Explanation:

Step1: Identify Resistive and Reactive Parts

Light bulbs (resistive): \(80\,\Omega\) and \(40\,\Omega\) (real parts, \(a\) components).
Motors (reactive): \(14\,\Omega\) and \(19\,\Omega\) (imaginary parts, \(b\) components, but wait—wait, no: Wait, the problem says light bulbs are pure resistive (so their impedance is real: \(80 + 0i\), \(40 + 0i\)) and motors are pure reactive (so their impedance is imaginary: \(0 + 14i\), \(0 + 19i\))? Wait, no—wait, the diagram: Let's re-express. Wait, the total impedance is sum of individual impedances. So resistive (light bulbs) have real parts, reactive (motors) have imaginary parts. So first, sum the resistances (real parts): \(80 + 40\). Then sum the reactances (imaginary parts): \(14 + 19\). Wait, no—wait, the problem says "light bulbs are pure resistive" (so their impedance is \(R + 0i\)) and "motors are pure reactive" (so their impedance is \(0 + Xi\), where \(X\) is reactance). So let's check the diagram: the two light bulbs are \(80\,\Omega\) and \(40\,\Omega\) (resistive, real), and two motors are \(14\,\Omega\) and \(19\,\Omega\) (reactive, imaginary). So total impedance \(Z = (80 + 40) + (14 + 19)i\).

Step2: Calculate Real and Imaginary Parts

Real part (resistance sum): \(80 + 40 = 120\).
Imaginary part (reactance sum): \(14 + 19 = 33\).

So \(Z = 120 + 33i\).

Answer:

\(120 + 33i\)