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The impedance of the headphones isn't the only factor. You also need to think about efficiency, which tells you how much power you need to produce a certain loudness. … The typical efficiency for over-the-ear high-impedance headphones is 100 dB/mW[1], So the few volts coming out of your laptop is more than enough. [1] https://blog.son-video.com/en/2016/08/understanding-the-impedance-and-sensitivity-of-audio-headphones …

That's how you get to complex impedance in the first place: $$v(t) = A\mathrm e^{\mathrm{j} (\omega t + \theta)}$$ $$i(t) = B \mathrm e^{\mathrm j (\omega t + \phi)}$$ $$\frac {v(t)} {i(t)} = Z = \frac … So where does impedance come in? Well, think about the difference between DC and the sinusoidal steady state. At DC, node voltages are constant values with different magnitudes. …

The solutions are just the definition of impedance: $$\hat V = \hat IZ$$ $$Z = \frac {\hat V} {\hat I}$$ Your voltage is 230 V. … So your phasors are: $$\hat V = 230\ \mathrm V$$ $$\hat I = 7.5\angle{-30^\circ}\ \mathrm A = 7.5\angle{-\frac{\pi}{6}} \ \mathrm A = 7.5e^{-j\frac{\pi}{6}}\ \mathrm A$$ And the impedance is the voltage …

Find the voltage across the terminals, then divide by 1 amp to get the Thevenin impedance: $$R_{th} = \frac{v_{in}}{1\ A}$$ The Thevenin voltage is found in the usual way, by determining the open-circuit … The Thevenin equivalent is just an impedance. …

EDIT: Yes, the primary impedance (also known as the reflected impedance) depends on the turns ratio due to mutual inductance. … The actual load impedance on the secondary does not -- it's the physical impedance connected across the wires, what you're calling "appliances". …

No, what you see is what you'd expect. As jonk points out in the comments, putting a 100k resistor in parallel with a 1k resistor gives you something very close to a 1k resistor. Let's look at this i …

The impedance of a capacitor is given by the formula: $$Z_C = \frac 1 {j \omega C} = \frac 1 {j 2 \pi f C}$$ where \$j = \sqrt{-1}\$. … negative sign: $$\frac 1 j = \frac j j \cdot \frac 1 j = \frac j {j^2} = \frac j {-1} = -j$$ $$Z_C = \frac 1 j \cdot \frac 1 {\omega C} = \frac {-j} {\omega C}$$ The reactance is the imaginary part of the impedance …

Impedance matching would allow high frequency noise to enter the IC through its supply pins. …