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posted by  stanbee on 5/28/2008 1:12:17 AM  |  status: Live  

PLEASEEEEEEEEE HELP !!! TEST SOON!!! WILL GIVE LIFE

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Question Details:
A 3Ω resistor is connected in parallel with a 6Ω resistor.  This combination is connected in series with a 4Ω resistor.  The resistors are connected to a 12 volt battery.  How much power is dissipated in the 3Ω resistor?

the answer is 5.3 W but explain how

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posted by Varmonds on 5/28/2008 3:45:53 AM  |  status: Live
Asker's Rating: Lifesaver   
stanbee's comment:
"THANK YOU SO MUCH"
Response Details:
given
  the eqvivalent resistance of the parallel combination is
               R' = 1/3 +1/6
              or R'  = 2Ω
hence this is in series with 4Ω & hence the equivalent resistance is R" = 2+4 = 6Ω
      hence the current is I = V/R" =  12V/ 6Ω = 2A
now the power dissipation  in the equivalent resistance R' circuit is
     P = I2R'
     P = 22*2
     P = 8W
in the parallel circuit of resistors we have generally
       Pα1/R
     so P1 : P2 = 1/R1 :1/R2
    or P1 : P2 = 1/3 : 1/6
     or  P1 : P2 = 2:1
hence the power dissipated in 3Ω resistor is
           P3Ω = 8*2/3
                   = 5.33W
 
 
 
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