four cells in a battery , each of internal resistance 0.8 ohm and emf 1.4V are connected (a) in series , (b) in parallel . the terminals of the battery are joined to to a lamp of resistance 10 ohm. find the current through thelamp and the cell in each case.

For series combination,

total emf of cells=1.4 x 4=5.6 V

Total resistance=0.8 x 4+10=13.2 ohm

Current through the lamp=5.6/13.2=0.424 A

Now

 for parallel combination,

resistance of lamp=10 ohm

Let the internal resistance of each cell=r

total internal resistance r'=r/4,So total resistance for the circuit=10+r/4=10+0.8/4=10.2 ohm

total emf=1.4 V

So current through the lamp=1.4/10.2=0.137 A

Again ,

current through the cell in each of the above case will be same.you can calculate it by using the individual resistance and emf of the cell.