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Number of turns on a transformer matters greatly. The turns ratio is one of many considerations in designing a transformer. The following are high-level considerations when designing a transformer.
As Tobalt stated, magnetizing inductance is important. Too few turns will consume excess current, even without a load.
You need a minimum of turns to prevent ...
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Efficiency of light-to-electric converters (aka "solar cells") is in the 15-25% range, and efficiency of the emitter is below 50%. Nothing stops you from trying, but with this low end-to-end efficiency it isn't going to compete with wires.
If you had a requirement to supply a low power circuit with extra high voltage isolation, or you need a ...
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Any given core material, cross section and frequency has a maximum volts per turn. If you want to use the winding at a certain voltage, then you need enough turns to support that voltage.
For instance, low frequency mains transformer steel will only run up to a peak field of 1.7 T or so before it saturates. If you had a core that was 10 mm x 20 mm, and ...
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Obviously, B has some advantages indeed. Otherwise all transformers would use a tiny number of turns. More turns generate more magnetizing inductance \$L_M \propto N^2\$.
Imagine a transformer with the secondary open, which is essentially a large inductor. If you connect an AC voltage with frequency \$f\$ to the primary, it will see an impedance of \$2 \pi f ...
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Loss in a fiber cable can be as low as 0,2dB per kilometer if the wavelength is optimal. That means 15km cable would dissipate half of the power. I guess 50...60Hz AC in copper or aluminium cables get attenuated substantially less if the voltage is high enough. This article http://insideenergy.org/2015/11/06/lost-in-transmission-how-much-electricity-...
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That is an inverter. There are multiple ways how it can convert 12V DC to 220V AC.
It can first step up the voltage and then output it as AC with an H-bridge, or it can first use the 12V to generate 12V AC with H-bridge and then step up the voltage with a transformer.
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Yes, once you have calculated the impedance matrix for the line, \$Z_{abc}\$, you just need to pre-multiply it with \$A^{-1}\$ and post-multiply it with \$A\$.
$$Z_{012}=\begin{pmatrix}Z_{00} & Z_{01} & Z_{02} \\\ Z_{10} & Z_{11} & Z_{12} \\\ Z_{20} & Z_{21} & Z_{22} \end{pmatrix} =A^{-1}Z_{abc}A$$
Turan Gönen called this ...
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If in a circuit there is real power positive and reactive power negative, what does mean in terms of power flow?
If you have positive real power it means that some component is consuming it and converting it into work, be it mechanical work or heat or both.
If you have negative reactive power it means that some component is changing the phase of the voltage ...
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Real power flow is controlled primarily by angular differences, flowing from leading angles toward lagging. Reactive power is controlled by voltage magnitudes, flowing from higher voltage toward lower. This assumes a system where all bus voltages are reasonably near 1 p.u. and relatively small angles. This is the basis of a decoupled load flow by the way.
...
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The "charging" speed.
It will be the current, as the charging is current regulated and voltage is pretty much set by the chemistry (it does vary during charging).
Most chemistry, besides NIMH will use the voltage to determine when to stop charging.
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Transformer design has to optimize a lot of different variables. And I am not an expert. But the transformer primary needs to have enough turns so that it acts like a big inductor and prevents excessive current from flowing. If you take a particular transformer core, you can either wind the primary with many turns of fine wire (high voltage primary) or wind ...
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That appears to be a typo in the help for the graph. From the ERCOT glossary:
HSL:
High Sustained Limit.
High Sustained Limit (HSL) for a Generation Resource:
The limit established by the QSE, continuously updated in Real-Time, that describes the maximum sustained energy production capability of the Resource.
High Sustained Limit (HSL) for a Load Resource:
...
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I will take a SWAG and assume it does not work because the 7805 is oscillating at a high frequency. I would suggest reading the data sheet for the 7805 you have and follow the recommendations it gives for the input and output. Not all 7805s are the same. You can also post a schematic of your circuit, pictures and frizzy things do not count. You probably will ...
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Figure 1. A simplified inverter schematic. Image source: ElecCircuit.
How it works:
Q1 is turned on. Current flows from +12V through CT (centre-tap) to the top of the transformer.
Q1 is turned off.
Q2 is turned on. Current flows from +12V through CT (centre-tap) to the bottom of the transformer.
Q2 is turned off.
The result is pulses of current running in ...
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I do not have experience with fibre optic cables to draw upon. I can, however, link a similar thread: here ...that deals with attenuation of power in long versus short fibre optic cables. While the question posed in the other thread is poorly supported (and the measurements given flimsy at best, as many of the answers agree), there is quite a bit of good ...
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