Question

For three cell of emf e1 ,e2 and e3 with internal resistance r1,r2,r3 connected in parallel.obtain an expression for the net internal resistance and effective current.what would be the maximum current possible if the emf of each cell is e and internal resistance is r each..?

Answer 1

Let the potential across the load (the junction of the terminals of the batteries) be V.  Let the total current that flows through load from the batteries be I.  Let the net EMF be E and the effective internal resistance be R.
I = i1 + i2 + i3

V = E1 - i1 R1   =>   i1 = (E1 - V) / R1
V = E2 - i2 R2   =>   i2 = (E2 - V) / R2
V = E3 - i3 R3   =>   i3 = (E3 – V) / R3
 
  i1 + i2 + i3 = (E1 R2R3 + E2 R3R1+E3 R1R2)/(R1 R2 R3)
                              - V (R1R2+ R2R3+R3R1)/(R1R2R3)

   V = (E1 R2 R3+ E2 R3R1 +E3 R1R2) / (R1R2+R2R3+R3R1)
                - (i1+i2+i3) (R1 R2 R3) / (R1R2+R2R3 +R3 R1)       --- (1)

we also have   V = E - (i1+i2+i3) R = E – I R    --- (2)

These two equations are true for all i1, i2..  

Hence,      E = (E1 R2 R3 + E2 R1 R3 + E3 R1 R2) / (R1R2 +R2 R3 + R3 R1)

           R = (R1 R2 R3) / (R1 R2+ R2 R3 + R3 R1)

Or,      E = (E1 / R1 + E2 / R2 + E3 / R3) R