A limited amount of current can be drawn from a single cell or battery
There are situations where single cell fails to meet the current requirement in a circuits
To overcome the problem cells can be grouped in series and in parallel combinations or mixed grouping of cells is done in order to obtain a large value
of electric current
(A) Series combination
Figure below shows the two cells of emfs E_{1} and E_{2} and internal resistance r_{1} and r_{2} respectively connected
in series combination through external resistance
Points A and B in the circuit acts as two terminals of the combination
Applying Kirchhoff's loop rule to above closed circuit
-Ir_{2}-Ir_{1}-IR+E_{1}+E_{2}=0
or
I=E_{1}+E_{2}/R+(r_{1}+r_{2})
Where I is the current flowing through the external resistance R
Let total internal resistance of the combination by r=r_{1}+r_{2} and also let E=E_{1}+E_{2} is the total EMF of the two cells
Thus this combination of two cells acts as a cell of emf E=E_{1}+E_{2} having total internal resistance r=r_{1}+r_{2} as shown above in the figure
(B) Parallel combinations of cells
Figure below shows the two cells of emf E_{1} and E_{2} and internal resistance r_{1} and r_{2} respectively connected in parallel combination through external resistance
Applying Kirchhoff's loop rule in loop containing E_{1} ,r_{1} and R,we find
E_{1}-IR-I_{1}r_{1}=0 ------------------------(1)
Similarly applying Kirchhoff's loop rule in loop containing E_{2} ,r_{2} and R,we find
E_{2}-IR-(I-I_{1})r_{2}=0 ------------------------(2)
Now we have to solve equation 1 and 2 for the value of I,So multiplying 1 by r_{2} and 2 by r_{1} and then adding these equations
results in following equation
IR(r_{1}+r_{2})+r_{2}r_{1}I-E_{1}r_{2}-E_{2}r_{1}=0
which gives
We can rewrite this as
E is the resulting EMF due to parallel combination of cells and r is resulting internal resistance.