Applying Ohm's law, current I is 12/1,000 ampere, or 12 ma. The power delivered to the load is found by using the

formula: P=I2RL, where I is in milliamperes, RL is in kilohms, and P is in millowatts. The power delivered to the 400-

ohm (.4 kilohm) load is 57.6 mw.

(2) B shows the same circuit with the load resistance changed to 900 ohms. The current is now 12/1,500 ampere,

or 8 ma. The power delivered to the load is therefore (8)2 times .9, or 57.6 mw. Note that this is the same load power as

that produced for the conditions given in A.

(3) In C, the load is changed to 600 ohms, the same value as RG. The current becomes 12/1,200 ampere, or 10

ma, and the load power is now (10) times .6, or 60 mw. This is larger than the load power obtained under the conditions

shown in A and B.

(4) If the load resistance is changed to values above and below 600 ohms, the corresponding values of power

delivered to the load can be calculated by the method used in the previous examples. The results of such calculations are

shown in the table below. From the data, a power transfer curve, such as that shown in figure 58, may be plotted. It can

be seen from the tabulation of load power versus load resistance that maximum power is transferred from a generator to a

load when the resistance of the load equals the internal resistance of the generator. This relationship is called the

maximum power transfer theorem for resistive networks.

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