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P is the amount of power generated by the component of i(t) that's in phase with v(t). It is oscillatory but has a mean value that's equal the mean power delivered to the load. And the complex power S, the total power, is exactly the sum of these two components
In order to calculate power loss in the transmission lines, you need to know the power received by the load end and the voltage at the load end. Use these two to calculate the current drawn from the line. Here you use P=VI. Then use the this current value and the total resistance of the transmission line to calculate the power loss.
The product of the rms phasor voltage V and current I gives the complex power S: S = V ⋅I = P + jQ. where P, the real part of S, is the average power. The rms phasor voltage and current for the time domain voltage and current above are: V = Vm 2–√. I = Im 2–√. The complex power is then: S = Vm 2–√ Im 2–√ = Vm ⋅Im 2.
for finding out power supplied by the battery we use the formula, P = V*I; where V is the voltage across its terminal and I is current flowing out positive terminal. For finding out power consumed by the resistance of wire. we prefer, P = I^2*R; where I is the current flowing through the wire/resistance and R is the resistance offered by the wire.
Informally, this is referred to as the "power law" or "power formula". It seems that there is a widespread practice to refer to it, incorrectly, as "Watt's law". However, note that the first term potentially conflicts with a widely used meaning of "power law" which describes any situation in which two quantities are related via any any power ...
Power isn't "across" something. Power is the voltage across something times the current going through it. Since the small amount of current going into the base is irrelevant in power dissipation, calculate the C-E voltage and the collector current. The power dissipated by the transistor will be the product of those two.
The power is 6A * 12V = 72W; or the voltage source : you have first to calculate the current flowing in the source, which is the difference between the current in the resistor and the current of the current source. I = 4A - 6A = -2A. And the power is : -2A * 12V = -24W. The minus sign tells the voltage source will RECEIVE 24W
Power is P = IV P = I V. You can only relate it to resistance if you decided to fix one and replace the other by Ohm's law. Ohm's law states that voltage across a resistor is linearly proportional to the current flowing through it, or V = IR V = I R. This can be arranged to say that current is inversely proportional to voltage, or I = V R I = V R.
To get average power, you would only need to divide by the load. If x represents current, then the formula computes mean square current and you would multiply by the load to get average power over the interval. So the formula is related to power, even though it is not power. I think you can't compute power without knowing the load. \$\endgroup\$
The voltage drop over one piece of wire is, as calculated above, 0.049025V 0.049025 V. The current through the circuit was 0.01961A 0.01961 A. We can now calculate the power loss in one wire: Pwire = 0.049025 ⋅ 0.01961 = 0.00096138W = 0.96138mW P wire = 0.049025 ⋅ 0.01961 = 0.00096138 W = 0.96138 mW.