A $3\text{V}$ battery with negligible internal resistance is connected in a circuit as shown. (Circuit consists of three $3\Omega$ resistors, two in parallel, then in series with the third). The current $I$ in the circuit will be:

Based on common circuit diagrams for this problem (AIEEE 2003):Two $3\Omega$ resistors are in parallel: $R_p=\frac{3\times 3}{3+3}=1.5\Omega$.This is in series with another $3\Omega$ resistor: $R_{eq}=1.5+3=4.5\Omega$. Wait, checking source.Source says $I=1.5\text{A}$. This implies $R=2\Omega$.If the two $3\Omega$ resistors are in parallel and connected directly across the $3\text{V}$ battery, $R_{eq}=1.5$, then $I=2$.If the diagram shows two $3\Omega$ in parallel $=1.5\Omega$, then $I=3/2=1.5\text{A}$. Correct.