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
There is an explanation video available below.
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