Use Ohm's law to determine the potential difference across the 3 Ω resistor in the circuit shown in the following diagram when key is closed:

\[\begin{aligned} \mathrm{R}_{\mathrm{S}} & =\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \\ & =1+2+3=6 \Omega \end{aligned} \] \[\begin{aligned} & \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ & =\frac{2 \mathrm{~V}}{6 \Omega}=\frac{1}{3} \mathrm{~A} \end{aligned} \] \[\begin{aligned} \mathrm{V} & =\mathrm{IR} \\ & =\frac{1}{3} \mathrm{~A} \times 3(\Omega)=1 \mathrm{~V} \end{aligned} \]