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11

If you want to call it PoE, you need to stick to IEEE 802.3af-2003 or IEEE 802.3at-2009. These standards call for isolation. This answer could end here: no, you cannot build a PoE device without isolation. This is also assuming that the data lines are already isolated, so the only potentials without isolation is the power lines. You also will have a ...


6

Your friend is right, and so are you. Power is the derivative of Energy over time. Since the field energy in your capacitor changes over time, there's obviously power needed to effect that. However, when you look at what happens over a full period (or multiples thereof), the capacitor stores as much energy as it releases; the average (over an arbitrarily ...


4

It sounds like the regulator is getting damaged because of the Power Loss, if I'm understanding this correctly. The power consumed by a Linear Regulator is P = (Vin-Vout) *Iout. In your case, if your Vout is 12V, and your load is drawing 10A, then your power consumption on the regulator is (15V - 12V) × 10A = 30W. This amount of power is waaay too ...


4

The imaginary part is energy that is bouncing back and forth between reactive components without being dissipated. Essentially, you can think of it as energy that gets stuck in the wiring due to the existence of capacitance and inductance. Now to your actual question: I am trying to understand how the mathematical expression Imaginary(s) is what you just ...


3

However if I connect the two terminals of a battery together with a simple wire then voltage will flow until the battery is dead. So if there is voltage flowing through the circuit why is there no current? simulate this circuit – Schematic created using CircuitLab Figure 1. A battery model with its internal resistance and a short circuit. We don't ...


3

However if I connect the two terminals of a battery together with a simple wire then voltage will flow until the battery is dead. So if there is voltage flowing through the circuit why is there no current? Current will flow in this scenario. Who told you it won't? Voltage doesn't flow in this circuit or any other. As an analogy, say we have a river whose ...


3

You need to provide current to the sum of the maximum of all components. Your powersupply must be able to handle it if all of your components try to pull their maximum current at the same time. The only way to guarantee that is to have a powersupply that can deliver the full total of the maximum of all components. You don't need to regulate the current to ...


3

Motors want lots of current to start, they want lots, given the opportunity they will take lots, and this will cause your 2A powersuply to quit. I was wondering if I used a 5V, 2A power adapter (that I already have), will it only supply 2A even if the motor tries to draw more than that and will this damage the adapter by any chance? Typically they go ...


3

I quess some images can be useful. The wanted formula is at the end. V1 is 100V peak sinusoidal voltage f=50Hz. There's 100 Ohm resistor R1 and 0.5H inductor L1 as load. The load current is about 530mA peak and its phase lags from V1 as it should be with inductive load. In the next image the red curve is the momentary power from the source to the load i.e. ...


3

I don't have the spec in front of me but I've done a number of PoE designs. As I recall the PoE isolation spec is to meet safety requirements and avoid ground loops. If your device takes the 48 Vdc and uses it within a fully enclosed, non-conductive enclosure, with no other cables, then the 1500 Vac power isolation (typically done with a flyback circuit) isn'...


2

If I’m given an electronmagnetic wave with an energy of 1 eV, how do I calculate its intensity ? You don't have enough information to calculate the intensity. In addition to the energy received, you also need to know over how wide an area the wave front was spread, and how long it took to receive the given amount of energy. Practically, when we want to ...


2

I suspect that you are measuring noise or induce voltage on the cable while using a digital multimeter with a very high (1 to 10 MΩ) input impedance. Figure 1. A Fluke 117 meter with a LO-Z measurement range. Interestingly, Fluke have a range of meters with a Lo-Z range to avoid this problem on DC and AC circuits. The input impedance is about 38kΩ (...


2

Regarding the usage circuit of the LNK586, you find it on page 4 of the datasheet and I copied it here: The blue part is a rectifier (AC to DC), the orange part is the IC which forms the control part of a DC/DC convertor (green part) Regarding your second question: You can find many ICs that make the same thing (control a DC/DC convertor), however, to ...


2

The excitation current supplied to a wound-field synchronous generator (WFSG)(alternator) determines the leading or lagging reactive volt-amperes that the generator can supply while operating at a given voltage. Here are three possible operating conditions: For a system with just a generator and a load, the excitation is used to maintain the generator ...


2

Take a tumbler filled with tap water and add a tablespoon of table salt and mix. Put in a couple forks, separated by a couple inches with a sponge or dishrag. Put that makeshift resistor in series with your LED and it won’t likely burn out. Don’t worry about getting it backwards, most LEDs can withstand quite a bit of reverse voltage.


2

A voltmeter will well you which lead is what. Otherwise, sometimes the cord will have a small 'ridge' on the + side, but that's not for certain. An LED with a 1K or so load resistor in series can also work. But be warned that LEDs don't much like reverse polarity, you can protect it with a regular diode in parallel, wired in opposite polarity to the LED. A ...


2

25 feet is about 8 meters, which is about 8 microHenries. For a single wire. But power cords have the return wire about 5mm away, so assume 10:1 reduction in inductance, or less than 1uH. How much effect on your power? Impedance (like resistance, but for changing currents) is computed as Zl = 2 pi * frequency * inductance Zl = 6.3 * 60 * 1e-6 = 377 e-6 ...


2

Its not a question of „who should be next to who“. If you want a design that is robust concerning EMI (I assume that‘s what you‘re after), then you have to defeat 3 enemies: 1) inductive coupling (loop to loop). Current only flow in loops and loops carrying current affect each other. So you must figure out all loops and find a way not to interfere. ...


2

simulate this circuit – Schematic created using CircuitLab EDIT: You get oscillations, which is the correct scenario. Look this video, the turbine valve has to move very slowly, additional deflector is used for fast rpm control, not the valve , it would tear off the entire mountain.


2

Reactive power is when X amount of REAL power flows to the load, does no work, and then X amount of REAL power flows back to the source. The reactive power in that case is X. That means that in order to view reactive power you have to observe over some time span, since if you observe a sufficiently short time span all you will see is real power flowing one ...


2

If you have a simple circuit with an AC voltage source, a resistor, and a capacitor in series then the 'apparent' power will be composed of a 'true' part and a 'reactive' part. The true power is consumed by the resistor and dissipated by heating it up. The reactive power is alternately absorbed by the capacitor to charge it up, and then contributed back ...


1

Because the losses are due to the current squared times resistance. So as the voltage increase, so the current reduces for the same resistance.


1

The higher the frequency, the more the power dissipation will be. All else equal, a processor that consumes 10 W at 25 MHz will need 20 W at 50 MHz*. The power density (I'm not familiar with this name for it, but I'll assume you have it correct) is roughly constant for a given processor, so if you have a power density of 0.4 W/MHz, you can predict how much ...


1

A critical piece of information is missing here: battery voltage. Though it is probably 12 V lead-acid, you should state it clearly just to be sure. Assuming this is correct, then you have approximately 173 Wh of capacity. Your router presents a 6 W load and your monitor a 24.7 W load, so a total of 30.7 W. Dividing one by the other, you would get about 5.6 ...


1

Not sure about your "the USB-C wall adapter", but if it is compliant to USB Type-C specifications, you will NOT have any power. Per Type-C specifications a Type-C port will NOT output the VBUS until a cable-device connection is made. If there is no Rd on consumer side, the port (charger) will not sense the connection and will not power the link. More, not ...


1

As far as wire specs go, you can use any "magnet wire" catalog as a reference. I like https://prod.essexwire.com/sites/essexwire.com/files/2017-08/Essex-Wire-Engineering-Data-Handbook-EN.pdf for general wire physical properties, but you will need to check each of the many insulation materials to determine the breakdown voltage for a specific insulation ...


1

If you have an impedance with a negative real part, it delivers power, yes. But remember: no actual passive component has a negative value! To build that negative resistance, you need active circuitry that do consume dc power. The very fact you mentioned (infinite power loop) is the concept behind designing oscillators. They consume dc power to be able to ...


1

I'm worried that line inductance would deform my input voltage and lead to a non ideal power factor. Power factor irregularities relate to the load and not the feed wire (no matter how inductive it might be). A maximum inductance of 1 uH per metre for an individual wire is impossible to get on any feeder cable so at 25 feet (7.62 metres) you won't see ...


1

To determine if an element is absorbing or providing power you need to apply the passive sign convention, which says that current enters the node of an element that is assumed to be at the higher voltage. If the product of current and voltage is positive, meaning that both current and voltage are positive or both are negative, then the power is positive. ...


1

5MW is the mechanical power the motor shaft puts out. 6kV is the source line voltage in to the motor. What the motor needs to run at rated operation. To get 5MW mechanical power out of the motor at an efficiency of 90%, means that 5MW + 10% electrical power must be put in (5.56MW). $$ \eta = \frac {P_{out}}{P_{in}} \times 100%$$ Finally, the power ...


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