a battery made of 5 cells each of 2v having internal resistance 0.1,0.2,0.3,0.4,0.5ohm are connected across 10 resistance. draw a circuit diagram and calculate current flowing through 10ohm.

Circuit diagram for the said arrangement will look like as,
Here, emf of cells are, ${\epsilon }_1=2V, {\epsilon }_2=2V, {\epsilon }_3=2V, {\epsilon }_4=2V, {\epsilon }_5=2V$
Internal resistances, $r_1=0.1\Omega , r_2=0.2 \Omega , r_3=0.3 \Omega , r_4=0.4 \Omega , r_5=0.5 \Omega$
External resistance, $R= 10 \Omega$

Since the cells are joined in series, the effective emf of all the cells is

                                                                               $$\begin{gathered} \epsilon ={\epsilon }_1+{\epsilon }_2+{\epsilon }_3+{\epsilon }_4+{\epsilon }_5 \\ =2+2+2+2+2 \\ =10 V \end{gathered}$$

Total internal resistance of all the cells is

                                                           $$\begin{gathered} r=r_1+r_2+r_3+r_4+r_5 \\ =0.1+0.2+0.3+0.4+0.5 \\ =1.5 \Omega \end{gathered}$$

Thus, current in the circuit, $I=\frac{\epsilon }{R+r}=\frac{10}{10+1.5}=0.869 A$