Friday 15 March 2013

MILLMAN’S THEOREM



In Millman's Theorem, the entire circuit can be drawn as a parallel network of branches, each branch containing a resistor in series with battery combination.











Rules
1)Convert all voltage sources in series with resister to current source in parallel with resistance by using source transformation technique









 
2) Combine all current sources and resistances
  IT = I1+I2+I3
      GT = G1+G2+G3







  

3)Convert the resulting current source to a voltage source   












       EEQ = IT/GT
         = ( I1+I2+I3 ) / ( G1+G2+G3 )
   EEQ = ( E1G1 + E2G2+ E3G3 ) / G1+G2+G3
    REQ= 1 / (G1+G2+G3 )





















Example  
Find voltage across RL
 Solution 
IEQ = I1 + I2 + I3
I1 = 10/5 = 2A
I2 = -16/4 = -4A
I3 = 8 / 2 = 4A
IEQ = 2 -4 +4 = 2A
REQ = parallel combinarion of 5Ω,4Ω and 2Ω
        = 1.053Ω
EEQ = REQ*IEQ= 2 * 1.053 = 2.106V

By voltage division rule we get VL
VL = 2.106 * 3 /( 1.053 + 3 )
     = 1.558 V
IL = 2.106 /( 1.053 + 3 )
    = 0.519A

                  VL = 1.558 V   IL = 0.519A


 

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