Page 366 - Schaum's Outline of Theory and Problems of Applied Physics
P. 366

CHAP. 29]                     ALTERNATING-CURRENT CIRCUITS                            351



                                         2
        What is needed is the average value of I R over a complete cycle, which is
                                                2
                                        2
                                                              2
                                                       2
                                      [I R] av = I R = I max  R[sin ωt] av
                                                eff
        The average value of I over a complete cycle is zero because I is positive over half the cycle and negative over
                             2
        the other half. However, I is always positive, and its average is therefore a positive quantity. As it happens, the
                                                1
                        2
        average value of sin ωt over a complete cycle is . Hence,
                                                2
                                                             I max
                               2
                                     1 2
                              I R = I R       and so   I eff = √
                               eff   2 eff                       = 0.707 I max
                                                              2
        Similarly, the effective voltage in an ac circuit is
                                                V max
                                          V eff = √  = 0.707 V max
                                                  2
        Currents and voltages in an ac circuit are usually expressed in terms of their effective values. The values of I
        and V at any particular moment are called the instantaneous values of these quantities.
            The effective values of I and V are sometimes called root-mean-square or rms values because of the way
        they are derived.
        SOLVED PROBLEM 29.1

              Find the maximum voltage across a 120-V ac power line.

                                                 ±V eff  ±120 V
                                         ±V max =    =        =±170 V
                                                 0.707  0.707
              The voltage across the power line varies between +170 V and −170 V.


        SOLVED PROBLEM 29.2
              The dielectric used in a certain capacitor breaks down at a potential difference of 300 V. Find the maxi-
              mum effective ac potential difference that can be applied to it.

                                       V eff = 0.707V max = (0.707)(300 V) = 212 V


        SOLVED PROBLEM 29.3
              Alternating current with a maximum value of 10 A is passed through a 20-  resistor. At what rate does
              the resistor dissipate energy?
                  The effective current is

                                        I eff = 0.707I max = (0.707)(10 A) = 7.07 A
              and so the power dissipated is
                                                         2
                                              2
                                         P = I R = (7.07 A) (20  ) = 1000 W
                                             eff


        REACTANCE
        The inductive reactance of an inductor is a measure of its effectiveness in resisting the flow of an alternating
        current by virtue of the self-induced back emf that the changing current causes in it. Unlike the case of a resistor,
        there is no power dissipated in a pure inductor. The inductive reactance X L of an inductor whose inductance is
   361   362   363   364   365   366   367   368   369   370   371