
Total current supplied by the battery $i = \frac{6}{{2.8 + 1.2}} = \frac{3}{2}$.
Current through $2 \,\Omega$ resistor $ = \frac{3}{2} \times \frac{3}{5} = 0.9\,A$



Reason : An ideal voltmeter draws almost no current due to very large resistance, and hence $(V)$ and $(a)$ will read zero.
