Power

Part 1: Fundamental Concepts: Power

Earlier in this chapter we discussed, in an indirect way, the reason a resistor (or lightbulb) limits the current in the circuit is that work is being done.  In the case of the resistor, the work being done results in the resistor becoming warm (it is dissipating power as heat) and, in the case of the light bulb, some of the power being dissipated appears as light and some as heat.  This occurs because it takes energy to move a charge through a change in voltage.

Power is the amount of energy per unit of time.  Current is the number of charges per unit of time.  Putting these pieces together, the amount of power, due to a change in voltage across a resistance, for a given amount of current flowing through it is:

In this equation, P is the power in Watts (abbreviated as W which is equivalent to a joule/sec), and I and V are the current passing through the resistance and the voltage that appears across it.  The power dissipated is due to the change in energy level (the voltage difference across the resistor) and the amount of charge (the current) to which that change of energy level occurs. 

The power being delivered from a source is also found by the same relationship.  In the case of a resistor, the charge is moving from a higher potential to a lower one.  In the case of a power source, such as a battery, there is an increase in potential.  Suppose you had a battery with a resistor connected across it.  The current and voltage are the same for both the battery and the resistor.  The power being delivered by the battery is equal to the power being dissipated by the resistor.

Given that both I and V are common in the equations V=IR and P=IV,  there are a number of permutations that can be formed from these two very simple, and useful, equations.  The figure below lists those permutations.  To use it, select a unit that you would like to solve for from the inner circle then, from the quadrant that unit is in, select one of the three equations that are shown.

Figure 1.  Equation wheel for Ohm’s and power laws.

Suppose you were interested in the power dissipated by a light bulb.  To see what else is needed to solve the problem, find the three equations that are in the quadrant where “P=” is in the inner circle.  From those equations you see that you either need to know V and I, I and R, or V and R.  If you have the measurements for the voltage across the light bulb and the current going through it you select “V * I” to solve for the power (in other words P = I * V).

The term resistance has been used so far implies that these equations apply just to resistors, or things that act like them, such as lightbulbs.  This is not true.  In any component, the power dissipated by it or converted to another form, as is the case with a light bulb or motor for example, can be found by the voltage drop across it and the current that goes through it.  Note that this applies only to components that are not specifically designed for energy transfer or energy storage.  Components, which we have not discussed yet such as transformers and capacitors, are excluded because they either transform the energy and deliver it to another load or they store it to be returned to the circuit later.

Maximum Power Transfer

The voltage source used to investigate how a circuit operates is not something that exists in practice.  It has been treated as a source of voltage with no internal resistance with the ability to source an unlimited amount of current.  Everything has resistance, including power supplies and voltage references which are the practical equivalents of a voltage source.  Even if it were possible to produce a perfect voltage source, the act of connecting it to a load would add resistance due to the interconnecting wire. The following figure depicts a voltage source with an internal resistance driving a load.  The dashed lines indicate the boundaries of a power source which includes its internal resistance.

Figure 2.  A practical power source.

A voltage sources has no internal resistance.  It can supply an unlimited amount of power.  The lower the load resistance, the greater the current, and the more power delivered to the load.  With the addition of an internal resistance, the situation is quite a bit different.

The internal resistance, Rint, and the external load, Rload, form a voltage divider.  As the resistance of Rload increases, more voltage appears across it.  At the same time, the current decreases, which in turn reduces the amount of power dissipated.  When Rload becomes smaller, the current increases, but less voltage appears across Rload which again decreases the power dissipated by it.  The addition of Rint complicates matters.

The power dissipated by Rload is given in the following equation.  The equation is derived by applying Ohm’s Law and the Power Law to a voltage divider.

Equation 2.  Power dissipated by Rload.

Although it is a bit messy to look at, the power dissipated by Rload is at its maximum when Rload = Rint.  This means that the power dissipated internally is the same as the power dissipated externally.  You can never get more than one half of the power from a source due to its internal resistance and the maximum power transfer occurs when the source resistance is equal to the load resistance.  

In circuits where power transfer is important, the resistance of the source should match the resistance of the load. If this seems hard to accept, let’s take a look at what happens using some values for Rin, Rload, and V.  For simplicity, V is assigned a value of 1V and Rint is kept at 2Ω while Rload is varied.

To begin, take the two extreme cases when Rload is open and a short.  When Rload is open, no current flows and there is no power delivered by the source.  When Rload is a short (0Ω) the current flow is at its maximum of 0.5A.  Using P=I^2*R, the power dissipated by Rint is 0.5W and the power dissipated by Rload is 0W.

Looking at three more resistances, where Rload is 1Ω, 2Ω, and 3Ω aids is showing where the maximum power transfer occurs.  When Rload is 1Ω, it dissipates 0.111W.  When it is 2Ω, it dissipates 0.25W and when it is 3Ω, it dissipates 0.222W.  The maximum power transfer occurs with a 2Ω load where it, and the internal 2Ω resistance, both dissipate 0.25W.

When Rload was a short, all of the power (in this case 0.5W) was dissipated internally.  When Rload and Rint are equal, half of the power (0.25W) is dissipated internally and half externally in Rload.

 Even though maximum power transfer occurs when the source resistance matches the load resistance, this is not how most power sources are used.  Typically the power source is made with as low and internal resistance as possible because the goal is to maintain a constant output voltage with varying loads.  As is the case with power dissipation in general, this rule only applies to components that dissipate power.  Items such as transformer, which will be discussed later, are treated differently.

In a previous section it was stated that power supplies and voltage references are the real world equivalents of ideal voltage sources.  To maintain a constant output voltage, voltage references and power supplies often use a mechanism that adjusts their output voltage to compensate for Rint.  By adjusting their internal voltage, the voltage across the load is constant despite the fact they have internal resistance.  This does not get around the fact that Rint is dissipating power.  It simply hides the fact that it is happening and makes the power supply or voltage reference appear as if they have no internal resistance.  There are limits to this however.  This is why power supplies and voltage reference only operate correctly over a limited range of output current.