[tex]\begin{gathered} V=IR \\ Series \\ V=I(R1+R2+R3)=I(15) \\ \frac{12}{15}=I=0.8A \end{gathered}[/tex][tex]\begin{gathered} Parallel \\ V=I(\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3})^{-1} \\ 12V=I(\frac{5}{3}) \\ 12\cdot\frac{3}{5}=7.2A=I \end{gathered}[/tex]
The maximum current is in a parallel circuit. On the other hand, the minimum current is in a series circuit. An arrangement with parallel and series would be between these two currents.
The third level would be the combination.
[tex]\begin{gathered} Combination \\ V=I(\frac{1}{R1}+\frac{1}{R2+R3})^{-1};\text{ }R2,R3\text{ in series} \\ 12=I(\frac{3}{10})^{-1} \\ 12\cdot\frac{3}{10}=I=3.6A \end{gathered}[/tex]
