# Tag Info

1

The problem arises from the 1mohm that's in series with the inductor: When I calculated the current through R1 with the extra 1mohm added my calculations line up perfectly with the simulation results. Alternatively, you can force the series resistance to be 0. Doing this also gave me equivalent results for both circuits from LTSPICE.

6

I see you have some accrurate but probably difficult to understand answers. I'll try to give you a better intuitive feel. Consider what happens when you first apply a voltage to the end of a long cable. The cable has some capacitance, so it will draw some current. If that were all there was to it, you get a big current spike, then nothing. However, it ...

5

In theory, if the cable in your example is infinitely long, then you will measure a 50Ω impedance between the two leads. If your cable is shorter than infinite, but longer than approximately 10% of the signal's wavelength* $\lambda = \dfrac{c}{f}$ (where $c\approx 3\cdot 10^8 \text{[m/s]}$), then you enter the area of transmission lines. So for a ...

5

The characteristic impedance of a cable is nothing to do with its physical length. It's quite complex to visualize but if you consider a long length of cable with a 100 ohm load at one end and a 10 volt battery at the other end and ask yourself how much current will flow down the cable when the 10 volt battery is connected. Eventually 100 mA will flow but, ...

11

When we talk about a 50-Ohm cable, we are talking about characteristic impedance which is not quite the same as a lumped impedance. When there is a signal propagating in the cable, there will be a voltage waveform and a current waveform associated with that signal. Because of the balance between capacitive and inductive characteristics of the cable, the ...

2

I would use the nominal impedance of the headphones, and pad generously. You will need a large electrolytic capacitor, well in the hundreds of microfarads or more. If you are after fidelity, then the filter calculations you have in mind might as well be tossed out of the window, because if you're after fidelity, speaker coupling capacitors must be seriously ...

4

Relying on the headphone impedance for filtering is probably a bad idea. Different kinds of headphones will be significantly different. Even different instances of the same model of headphones can be different. The headphone impedance isn't just resistive but also reactive, and it will take some work to characterize it. Someone might plug your thing into a ...

1

It will have a varying impedance with different frequencies but this is largely governed by the acoustic properties of the headphones when wearing them. At anything but the really low frequencies the capacitor you choose will be a small impedance and any variation from 200Hz upwards will be laregely unnoticeable. If your headphones are 30 ohm impedance then ...

2

Then, how can I find the voltage across the 30mH winding For coupled inductors we have the following: $$M = k\sqrt{L_1 L_2}$$ $$V_1 = L_1 \dot I_1 + M \dot I_2$$ $$V_2 = M \dot I_1 + L_2 \dot I_2$$ Now, if $I_2$ is constant, we have: $$V_1 = L_1 \dot I_1$$ $$V_2 = M \dot I_1$$ Thus: V_2 = M \dfrac{V_1}{L_1} =V_1 \cdot k\sqrt{\frac{L_2}{L_1}} ...

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