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Homework Help: PHYSICS 2

Posted by PLEASE HELP on Friday, January 25, 2013 at 6:39pm.

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.99 A. When the resistors are connected in parallel to the battery, the total current from the battery is 10.7 A. Determine the two resistances.

• PHYSICS 2 - drwls, Friday, January 25, 2013 at 7:25pm

  Let the smaller resistance be R1 and the larger R2.
  
  1.99 = 12.0/(R1 + R2)
  
  10.7 = 12/R1 + 12/R2
  
  Solve the simultaneous equations.
  
  R1 + R2 = 6.03
  R2 = 6.03 - R1
  10.7 = 12/R1 + 12/(6.03 - R1)
  10.7*R1*(6.03-R1) = 12(6.03-R1) +12R1
  64.51 R1 -10.7 R1^2 = 72.36
  

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