A 2.0-ohm resistor is connected in a series with a 20.0 -V battery and a three-branch parallel network with branches whose resistance are 8.0 ohms each. Ignoring the battery’s internal resistance, what is the current in the batter? Show your work.
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]
