The current I in the given circuit will be

The circuit can be redrawn as :

We can see that the circuit is a balanced Wheatstone bridge

$\Rightarrow R_{\mathrm{AB}}=4 \mathrm{\Omega }$

Current through cell $I=\frac{V}{R_{\mathrm{AB}}}=\frac{40}{4} \mathrm{A}=10 \mathrm{A}$