V = IR from ohms law I = \(\frac{V}{R}\) Since the resistances are in parallel \(\frac{1}{Reff}\) = \(\frac{1}{18}\) + \(\frac{1}{12}\) + \(\frac{1}{6}\) = \(\frac{2 + 3 + 6}{36}\) = \(\frac{11}{36}\) R\(_{eff}\) = \(\frac{36}{11}\) = 3.27 I = I\(_1\) + I\(_2\) + I\(_3\) = \(\frac{V}{Reff}\)= \(\frac{12}{3.27}\) = 3.669A I = 3.7A