Power
We have already defined power, and we will represent it by P or p. If one joule of energy is expended in transferring one coulomb of charge through the device in one second, then the rate of energy transfer is one watt. The absorbed power must be proportional both to the number of coulombs transferred per second (current) and to the energy needed to transfer one coulomb through the element (voltage). Thus,
p = vi ............[4].
Dimensionally, the right side of this equation is the product of joules per coulomb and coulombs per second, which produces the expected dimension of joules per second, or watts. The conventions for current, voltage, and power are shown in Fig. 2.12. We now have an expression for the power being absorbed by a circuit element in terms of a voltage across it and current through it. Voltage was defined in terms of an energy expenditure, and power is the rate at which energy is expended. However, no statement can be made concerning energy transfer in any of the four cases shown in Fig. 2.9, for example, until the direction of the current is specified. Let us imagine that a current arrow is placed alongside each upper lead, directed to the right, and labeled “+2 A.” First, consider the case shown in Fig. 2.9c. Terminal A is 5 V positive with respect to terminal B, which means that 5 J of energy is required to move each coulomb of positive charge into terminal A, through the object, and out of terminal B. Since we are injecting +2 A (a current of 2 coulombs of positive charge per second) into terminal A, we are doing (5 J/C) × (2 C/s) = 10 J of work per second on the object. In other words, the object is absorbing 10 W of power from whatever is injecting the current.
We know from an earlier discussion that there is no difference between Fig. 2.9c and d, so we expect the object depicted in Fig. 2.9d to also be absorbing 10 W. We can check this easily enough: we are injecting +2 A into terminal A of the object, so +2 A flows out of terminal B. Another way of saying this is that we are injecting −2 A of current into terminal B. It takes −5 J/C to move charge from terminal B to terminal A, so the object is absorbing (−5 J/C) × (−2 C/s) = +10 W as expected. The only difficulty in describing this particular case is keeping the minus signs straight, but with a bit of care, we see the correct answer can be obtained regardless of our choice of positive reference terminal (terminal A in Fig. 2.9c, and terminal B in Fig. 2.9d)
Now let’s look at the situation depicted in Fig. 2.9a, again with +2 A injected into terminal A. Since it takes −5 J/C to move charge from terminal A to terminal B, the object is absorbing (−5 J/C) × (2 C/s) = −10 W. What does this mean? How can anything absorb negative power? If we think about this in terms of energy transfer, −10 J is transferred to the object each second through the 2 A current flowing into terminal A. The object is actually losing energy—at a rate of 10 J/s. In other words, it is supplying 10 J/s (i.e., 10 W) to some other object not shown in the figure. Negative absorbed power, then, is equivalent to positive supplied power. Let’s recap. Figure 2.12 shows that if one terminal of the element is v volts positive with respect to the other terminal, and if a current i is entering the element through that terminal, then a power p = vi is being absorbed by the element; it is also correct to say that a power p = vi is being delivered to the element. When the current arrow is directed into the element at the plus-marked terminal, we satisfy the passive sign convention. This convention should be studied carefully, understood, and memorized. In other words, it says that if the current arrow and the voltage polarity signs are placed such that the current enters the terminal on the element marked with the positive sign, then the power absorbed by the element can be expressed by the product of the specified current and voltage variables. If the numerical value of the product is negative, then we say that the element is absorbing negative power, or that it is actually generating power and delivering it to some external element. For example, in Fig. 2.12 with v = 5 V and i = −4 A, the element may be described as either absorbing −20 W or generating 20 W. Conventions are only required when there is more than one way to do something, and confusion may result when two different groups try to communicate. For example, it is rather arbitrary to always place “North” at the top of a map; compass needles don’t point “up,” anyway. Still, if we were talking to people who had (unknown to us) chosen the opposite convention of placing “South” at the top of their maps, imagine the confusion that could result! In the same fashion, there is a general convention that always draws the current arrows pointing into the positive voltage terminal, regardless of whether the element supplies or absorbs power. This convention is not incorrect, but sometimes results in counterintuitive currents labeled on circuit schematics. The reason for this is that it simply seems more natural to refer to positive current flowing out of a voltage or current source that is supplying positive power to one or more circuit elements.
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