Chapter  1  Electrical  Fundamentals  Review  5  Copyright  Goodheart-Willcox  Co.,  Inc.  The  following  example  illustrates  the  charac-  teristics  and  relationships  of  a  series  circuit  and  how  Ohm’s  law  is  used  to  solve  circuit  problems.  Series  circuits  are  rarely  encountered  in  prac-  tical  commercial,  residential,  or  industrial  wir-  ing  applications.  These  circuits  have  several  inherent  problems:  •  Any  change  of  a  resistance  in  the  circuit  affects  the  current  throughout  the  circuit.  This,  in  turn,  changes  the  voltage  across  the  other  resistors  in  the  circuit.  •  If  any  part  of  the  path  is  broken  (opened),  the  circuit  is  dead.  If  one  element  fails,  the  entire  circuit  is  shut  off.  •  Switches  or  other  controlling  devices  cannot  be  used  for  individual  loads  within  the  circuit.  If  one  load  is  turned  off,  all  loads  lose  power.  •  Voltage  across  any  load  depends  on  the  overall  circuit  arrangement.  It  is  almost  impossible  to  design  circuit  devices  that  can  operate  at  so  many  different  voltages.  1.2.2  Parallel  Circuits  A  parallel  circuit  is  one  in  which  the  elements  are  arranged  in  such  a  manner  that  there  are  sev-  eral  paths  for  the  current.  Figure  1-6  illustrates  SAMPLE  PROBLEM  1-1  Problem:  Using  the  following  fi  gure,  calculate  the  current  fl  owing  through  the  circuit  and  the  voltage  at  each  resistor.  120  V  Source  R1  =  20  Ω  R3  =  30  Ω  R2  =  10  Ω  Formulas:  IT  =  E  T  __  R  T  R  T  =  R  1  +  R  2  +  R  3  IT  =  I1  =  I2  =  I3  E  N  =  I  N  R  N  =  I  T  R  N  Solution:  The  current  can  be  calculated  using  the  formula  IT  =  E  T  __  R  T  To  fi  nd  the  total  resistance,  add  the  individual  resistances  together:  R  T  =  R  1  +  R  2  +  R  3  =  20  Ω  +  10  Ω  +  30  Ω  =  60  Ω  The  voltage  is  120  volts,  so  the  current  is  I  T  =  120  V  ___  60  Ω  =  2  A  The  voltage  at  each  resistor  can  be  calculated  using  the  fact  that  current  is  constant  through-  out  the  circuit:  I  1  =  I  2  =  I  3  =  I  T  (Continued)  Therefore,  E  1  =  I  1  R  1  =  I  T  R  1  =  2  A  ×  20  Ω  =  40  V  E  2  =  I  2  R  2  =  I  T  R  2  =  2  A  ×  10  Ω  =  20  V  E  3  =  I  3  R  3  =  I  T  R  3  =  2  A  ×  30  Ω  =  60  V  The  current  fl  owing  through  the  circuit  is  2  amps,  with  resistor  voltages  of  40  volts  (E  1  ),  20  volts  (E  2  ),  and  60  volts  (E  3  ).