Mixed combination of cells

# Mixed combination: 

In this combination n-cell are connected in series and a row is formed and then m-rows of  cell are connected in parallel to each other with external resistance.

In above figure shows the mixed combination of cells.  Each cell having EMF E and internal resistance r.  The current drawn from the battery so formed in the external resistance R.

EMF of each cell =  E volt
Thus, total EMF of a cell are joined in series of  a row = nE volt

Effective EMF of m-rows are joined in parallel combination = nE volt

and internal resistance of each cell = r

Let internal resistance of n cell are joined in series be r' ohm
Hence,
r' = r+r+r+............up to n terms
r' = nr

Let effective internal resistance of all cell in m-rows are joined in parallel combination be r"
Hence,
1/r" = 1/nr + 1/nr +.........up to m terms
1/r" = m/nr
Hence, {r" = nr/m}
( Equivalent resistance of a battery)

Total resistance of the circuit = (R+nr/m)

And current drawn from battery
I = (Total emf) / (Total resistance)

I = nE / ( R + nr/m )
I = mnE / (mR+nr)  ........(1)

It is clear from eqn(1) that to obtain maximum current in external circuit the condition is required that (mR+nr) is to be minimum.

We know that
[ mR+nr = (√mR - √nr)² + 2 √mR √nr ]

It is clear from above relation
If √mR - √nr = 0 , then (mR+nr) become minimum.

Hence,   √mR - √nr = 0

√mR = √nR

Squaring both sides

mR = nr

R = nr/m
[ External resistance = Effective internal 
                                     resistance of battery]

On Putting  mR = nr in enq(1)
then I = I (Max)

I(Max) = mnE / (nr+nr)

I(Max) = mnE/2nr

{ I(Max) = mE/2r }