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10

The loadspeaker is an inductor; so you have to create a good notion about its behavior. Here is a simple explanation. Both capacitors and inductors are accumulating elements. They accumulate energy; so they can be considered as sources... rechargeable sources. The capacitor can be thought of as a voltage source containing potential energy (as a tensioned ...


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I'm aware of that, in a circuit, inductive loads need to be protected with a flyback diode because if the circuit gets open the magnetic field will collapse forcing a current flow in the opposite direction. And the reason why it doesn't make sense is because it is wrong; current continues to flow in the inductive load in the same direction until all the ...


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As an interview question the point might be just to get you to say that it isn't an "either/or" answer — of course the two are interrelated, and you can't have a current without a voltage! However, consider the way an LED works: every time (well, hopefully most of the time) when an electron and a hole meet at the junction and recombine, a photon is ...


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From your search results I just picked this one to go off in my explanation. https://www.diodes.com/assets/Datasheets/ds30018.pdf (B170/B) Its all about how the system is designed. If your system is designed to work at 50V and there are limiters in place to keep max voltage under 60V then a 60V diode would be fine. Therefore, you should pick a diode with a ...


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The meter you are using is a SGM30C2. The question, as I see it, is how to meter the aggregate current of multiple houses using this meter. The maximum total current that the houses may use is, at the moment, unknown. Since the metering is intended for the purposes of technical monitoring, rather than billing customers, there is greater legal flexibility for ...


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Take a look at the curves they published, What you are getting sounds somewhat close. I know you are losing voltage in the current measurement, that is the way they work. The max with one volt in is 150mA at about 60% efficiency. I would assume you are below 1 volt which takes you completely off the curve. Consider putting two batteries in series, your ...


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From this wiki article: - A light-emitting diode (LED) is a semiconductor light source that emits light when current flows through it. But, you're right in that to get current to flow you need to apply a voltage however, it is the flow of current that causes the effect: - Electrons in the semiconductor recombine with electron holes, releasing energy in ...


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It seems like your understanding is pretty much correct. There is just one clarification. The Voc and Vmpp are only slightly sensitive to irradiance. See attached graphs which are real DC data from one string in my grid-tie solar array. The grid-tie inverter logs data at 5 minute intervals. As you can see, Vmpp does not vary much during the day. And its ...


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The .08A is most likely just a guide to power consumption, not a spec. The fan is designed to work on 12VDC, and that's what you should give it. The current draw will depend on the load on the fan which usually won't change much. You could try upping the voltage a bit but it's really pot luck how far and at some point you will damage it if you go too far. (...


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You could do it like the first diagram. The input current generates a voltage across R1, which is buffered and amplified by the amplifier. Unfortunately, you run into a very nasty compromise if you want speed from this arrangement. The voltage on the input of the amplifier has to move, which means it has to charge Cstray. Cstray and R1 form a lowpass time ...


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