From the diagram above calculate the total curr... - JAMB Physics 2017 Question

V = IR from ohms law

I = \(\frac{V}{R}\)

Since the resistance are in parallel

\(\frac{1}{Reff}\) = \(\frac{1}{8}\) + \(\frac{1}{12}\) + \(\frac{1}{6}\)

= \(\frac{2 + 3 + 6}{36}\) = \(\frac{11}{36}\)

Reff = \(\frac{36}{11}\) = 3.27

I = I₁ + I₂ + I₃

= \(\frac{V}{Reff}\)= \(\frac{12}{3.27}\)

= 3.669A

I = 3.7A