b. Application of maximum power transfer theorem to transmission line. Figure 59 illustrates the application of
the maximum power transfer theorem to transmission lines. For simplicity, the transmission line under consideration is
assumed to be replaced by a tee section composed of pure resistances. The characteristic impedance of such a section has
been shown to be a pure resistance of 600 ohms (fig. 55).
(1) In A, figure 59, the line is shown to be terminated in a resistance of 66.7 ohms. By using the methods
previously discussed for combining series-parallel resistances at the input terminals of the line, ZI, is found to be 400
ohms. From the table in a(4) above, the current and power in a 400-ohm resistance connected to a 12-volt generator in
series with an internal resistance of 600 ohms are 12 ma and 57.6 mw, respectively. However, these are values for current
and power at the input terminals of the circuit. By applying the laws of division of current in parallel circuits, the line
current of 12 ma divides so that 3 ma flow through the shunt 800-ohm resistance of the tee section, and 9 ma flow through
the actual load of 66.7 ohms. The power delivered to this load is therefore 5.4 mw.
(2) In B, the load resistance is changed to 5,400 ohms. This value makes the resistance at the input terminals of
the line 900 ohms. Therefore, the line current is 8 ma, and the power delivered to the input terminals of the line is 57.6
mw. Again, applying the laws of division of current in parallel circuits, the current through the shunt 800-ohms resistance
is found to be 7 ma, so that the load current is 1 ma. The load power again is shown to be 5.4 mw.
(3) Finally, in C, the line is terminated in 600 ohms, the characteristic resistance of the line. Therefore, the input
resistance of the line is 600 ohms (by the definition of characteristic resistance). Referring to C, figure 59, the line current
and the power delivered to the input terminals of the line are now 10 ma and 60 mw, respectively. Since the two parallel
branches are equal in resistance (800 ohms), the load current is one-half of the line current, or 5 ma. This makes the load
power (5)2 times .6, or 15 mw. This is only one-fourth of the input power because of the attenuation of the line, but it is
considerably greater than the load power obtained for the values of load impedance shown in A and B.