First of all, let's calculate the equivalent resistance of the three resistors in parallel:
[tex] \frac{1}{R_p}= \frac{1}{8 \Omega}+ \frac{1}{8 \Omega}+ \frac{1}{8 \Omega}= \frac{3}{8 \Omega} [/tex]
[tex]R_p= \frac{8}{3} \Omega=2.67 \Omega[/tex]
The 2.0-ohm resistance is in series with these resistors, so the equivalent resistance of the circuit is
[tex]R_{eq}=2.0 \Omega + R_p = 2.0 \Omega + 2.67 \Omega = 4.67 \Omega[/tex]
And the current in the circuit can now be calculated by using Ohm's law:
[tex]I= \frac{V}{R_{eq}}= \frac{20.0 V}{4.67 \Omega}=4.28 A [/tex]
