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Therefore the power put out by the current source is \$ 6V x 2A = 12W \$, and so it absorbs \$-12W\$. … The power absorbed by the resistor is \$10\Omega \times (2A)^2 = 40W\$, whereas the power put out by the voltage source is \$2A \times 14V = 28W\$ hence it absorbs \$-28W\$. …

Using phasor (Steinmetz') notation with peak values (not RMS): \$ V = 250 \angle{45°}\$ \$ I = 4 \angle{-30°}\$ Complex power is \$ S=\dfrac{1}{2}\; V\cdot I^{*}\$ where the asterisk means "complex conjugate … Without entering in further details, the average power is also called active power, whereas the reactive power is defined as: \begin{align*} Q = \dfrac{V_m \, I_m}{2} \sin \phi \end{align*} On the …

You can use a lower voltage, and the motor will have lower speed, but also lower maximum power. …

You correctly calculated the total power dissipation of your BJT. … Then calculate the power that that equivalent circuit provides to the base of the BJT and ask where does that power go, if not into the BJT? Conservation of energy is a basic law of nature. …

For example, if you are computing the maximum power that a particular power line can deliver to a load, you can say that a load is particularly power-hungry by saying "it consumes a lots of power", or … "it consumes 1000W (of power)"; talking about how much energy it consumes will sound strange in this case. …

Even if the MCU is just used as a PWM generator and not in a full closed-loop control system, it will allow you to set the power in a very simple way. … If DC conversion is not an option, you could use a TRIAC-based power controller using phase-control. …

In all this landscape high power signals enter the picture only in some very limited areas, typically power supply systems, final amplifier stages (BF or RF) or final drivers for power control systems … For example switching power supplies, before the final DC conversion stage, may be viewed as power oscillators, since they produce an high power oscillating signal that is then converted to the needed …

As hinted by others in comments, that question is formulated so that you might reflect on the benefits of high-voltage power transmission lines. … Power is not transmitted over long distances at such a low voltage as 115Vac (230-240Vac in EU). …

Your power supply voltage is too low for the opamp to be able to drive that MOSFET. …

The charger must convert AC mains voltage to the smallish ~5V required to power modern phones, therefore any sensible designer should have designed it to have reasonable isolation between the parts connected …

Since the power (i.e. the "wattage") is, by definition, the energy per unit time, that means that the power exchanged in the process is: $$ P = \frac {E}{\Delta t} = \frac{V_{AB} \cdot Q} {\Delta t … If you divide that by the time in which the process happens, you get the (average) power of the process. …

TL;DR You need either a specialized BMS (Battery Management System) that can block charging at low temperatures or highly specialized cells that are very expensive. Charging Li-ION batteries below ~5° …

These two windings are wound on the same ferromagnetic core in a way that exploits the different path that power line current and CMN currents take in the circuit. … Another interesting document is this application note about line filters in switching power supplies. …

Of course, since any current provided to the load must come from the power supply, you must add that to the quiescent current whenever you have a load that draws significantly more current than the quiescent …

Hence the I0 current will be non-zero (this is called shoot-through) because the power rail will be "shorted" to ground with a relatively small resistance for a (hopefully) small amount of time. …