Question

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.

Answer 1

Answer:

Explanation:

Resolving the parallel resistor branches

$\frac{1}{Rtotal} =\frac{1}{8.0ohm} +\frac{1}{8.0ohm} +\frac{1}{8.0ohm} =2.67Ohm$

The equivalent resistor is now in series with the 2.0 Ohm resistor

so, by using Ohm's Law

$I=\frac{V}{Rtotal+R} \\I=\frac{20.0V}{2.67Ohm+2.0Ohm} \\\\I= 4.28A$