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

I = 4.28 amps

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Resolving the parallel resistor branches

`(1)/(Rtotal) =(1)/(8.0ohm) +(1)/(8.0ohm) +(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=(V)/(Rtotal+R) \\I=(20.0V)/(2.67Ohm+2.0Ohm) \\\\I= 4.28A`