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(University of Nevada - Reno)
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Date Posted: 7/24/2008 12:38:52 AM  Status: Live
resistors...
Course Textbook Chapter Problem
Algebra Based Physics College Physics (7th) by Serway, Faughn, Bennett 18 10P
Question Details:
Two resistors, A and B, are connected in parallel across a 7.0 V battery. The current through B is found to be 3.0 A. When the two resistors are connected in series to the 7.0 V battery, a voltmeter connected across resistor A measures a voltage of 4.0 V. Find the resistances A and B.
RA =
RB =

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Date Posted: 7/24/2008 3:05:37 AM  Status: Live
Asker's Rating: Lifesaver   
Response:
 
 
         Given that the  voltage accross the battery is  V = 7.0 V
             when the  resistances are in  parallel the the current through the resister B is IB = 3.0 A
   when the resisters are in seires then the voltage accross the resester A is  VA = 4.0 V  
------------------------------------------------------------------------------------------
 
          if the resistances are in parallel then the 
 
 current flowing through the  resisr\ters   Ia = I*RB / (RA+RB) 
                                                       3.0 A = I*RB / (RA+RB)  ---------(1)
                      the resultant resistance is  R =  RA *RB / (RA+RB)  
 
                                              current is  I = V / R 
                                                                 =  7.0V *(RA+RB)/  RA *RB          ----------------(2)
 
 from the equations  (1) and (2) we get      
 
                                  3.0A = 7.0 V / RB 
                                        RB = 7.0 V / 3.0A = 2.33 Ω 
      if the  resisters are in seires then  voltage accross the resester A is
                                           VA  = V*RA / (RA+RB)  
                                           4.0V = 7.0V * RA / (RA+RB)   
                             RA / (RA+RB) = 4.0V/ 7.0V             ----------(3) 
 
          substitude the value of  RB  in equation (3) we get value of   RA = -------- Ω
          
 
 



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