6  Modern  Commercial  Wiring  Copyright  Goodheart-Willcox  Co.,  Inc.  a  parallel  circuit.  In  electrical  circuits,  the  term  parallel  does  not  necessarily  mean  physically  or  geometrically  parallel,  but  merely  signifi  es  alternate  routes  or  branches.  Parallel  circuits  are  often  called  branched  circuits.  The  following  lists  some  of  the  characteristics  of  parallel  circuits:  •  The  voltage  across  all  branches  of  the  circuit  is  the  same  as  the  total  voltage  at  the  source.  E  TOTAL  =  E  A  =  E  B  =  E  C  =  E  N  •  The  total  current  is  equal  to  the  sum  of  the  currents  fl  owing  through  each  of  the  branches.  I  TOTAL  =  I  A  +  I  B  +  I  C  +  …  +  I  N  •  The  total  resistance  in  a  parallel  circuit  is  the  reciprocal  of  the  sum  of  the  reciprocals  of  the  separate  branch  resistance.  To  express  this  in  a  much  clearer  manner,  let  us  look  at  it  mathematically:  1  ___  R  TOTAL  =  1  __  R  1  +  1  __  R  2  +  1  __  R  3  +  …  +  1  __  R  N  •  Ohm’s  law  applies  to  the  entire  circuit  and  any  of  the  branches.  •  Any  opening  or  break  of  a  branch  does  not  stop  the  current  fl  ow  through  other  branches.  For  this  reason,  parallel  circuits  have  a  distinct  advantage  over  series  circuits.  Notice  that  the  characteristics  of  a  parallel  cir-  cuit  are  in  many  ways  opposite  those  of  a  series  circuit.  The  following  example  illustrates  the  relationships  at  work  in  a  parallel  circuit.  Source  R1  R2  R3  Goodheart-Willcox  Publisher  Figure  1-6.  Current  moves  through  parallel  circuits  using  more  than  one  pathway.  SAMPLE  PROBLEM  1-2  Problem:  Using  the  following  fi  gure,  fi  nd  (a)  the  current  through  each  resistor,  (b)  the  total  current  of  the  circuit,  and  (c)  the  total  circuit  resistance.  120  V  Source  R1  =  10  Ω  R3  =  20  Ω  R2  =  30  Ω  (a)  Formulas:  E  T  =  E  1  =  E  2  =  E  3  I  1  =  E  1  __  R  1  I  2  =  E  2  __  R  2  (a)  Solution:  E  T  =  E  1  =  E  2  =  E  3  =  120  V  I  1  =  120  V  ___  10  Ω  =  12  A  I  2  =  120  V  ___  30  Ω  =  4  A  I  3  =  120  V  ___  20  Ω  =  6  A  (b)  Formula:  I  T  =  I  1  +  I  2  +  I  3  (b)  Solution:  I  T  =  12  A  +  4  A  +  6  A  =  22  A  (Continued)