Page 312 - Schaum's Outline of Theory and Problems of Applied Physics
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CHAP. 25]                        DIRECT-CURRENTCIRCUITS                               297



        IMPEDANCE MATCHING
        When a source of electric energy is connected to a load, the power transfer is a maximum when both source and
        load have the same resistance. This is an example of impedance matching: When energy is being transferred
        from one system to another (here from an emf source to a load), the efficiency is greatest when both systems have
        the same impedance, which is a general term for resistance to the flow of energy. Another example of impedance
        matching occurs when a moving body strikes a stationary one. When both bodies have the same mass (which
        is a measure of the resistance of a body to a change in its state of motion or of rest), the energy transfer to the
        stationary body is a maximum.
            Figure 25-7 shows an emf source of internal resistance r connected to a load resistance R. If the current that
        flows is I,

                                                     V emf
                                                I =
                                                    R + r
        The power in the load is

                                                       V 2  R
                                                 2      emf
                                            P = I R =
                                                      (R + r) 2
        The maximum value of P corresponds to R = r, when

                                                        2
                                                      V emf
                                                P max =
                                                       4R
        Figure 25-8 shows how P/P max varies with the resistance ratio R/r.




                                              1.00
                                           Power dissipated in R relative  to maximum  0.75




                                              0.50

                                              0.25

                                                0
                                                 0       1        2       3        4       5
                                                                Resistance ratio, R/r
                                          Fig. 25-8.  (From Modern Technical Physics, 6th Ed., Arthur Beiser,
                  Fig. 25-7                c  1992. Reprinted by permission of Pearson Education, Inc.)



        SOLVED PROBLEM 25.20
              Verify that P max occurs when R = r by calculating the power delivered by a battery of emf 10 V and
              internal resistance 0.5   when it is connected to (a) 0.25- ,(b) 0.5- , and (c)1-  resistors.
                                                 V e       10 V
              (a)                          I 1 =     =             = 13.3A
                                               R 1 + r  0.25   + 0.5

                                                           2
                                                2
                                           P 1 = I R 1 = (13.3A) (0.25  ) = 44 W
                                                1
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