Question

How do I find the current in this battery?

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 battery? Show your work.

Answer 1

Please see the explanation.

Explanation:

Given:

`R_"series"= 2.0color(white)(.)Omega`
`R_1=R_2=R_3=8color(white)(.)Omega`
`V_"battery"= 20color(white)(.)"Volts"`

Let `R_"equivalent"` be the equivalent resistance of the 3 resistors in parallel:

`R_"equivalent" = 1/(1/R_1+1/R_2+1/R_3`

`R_"equivalent" = 1/(1/(8color(white)(.)Omega)+1/(8color(white)(.)Omega)+1/(8color(white)(.)Omega)`

`R_"equivalent" ~~ 2.7color(white)(.)Omega`

Let `R_"total"` be the total resistance as seen by the battery:

`R_"total" = R_"series" + R_"equivalent"`

`R_"total" = 2.0color(white)(.)Omega + 2.7color(white)(.)Omega`

`R_"total" = 4.7color(white)(.)Omega`

Let `I` be the current supplied by the battery:

`I = V_"battery"/R_"total"`

`I = (20color(white)(.)"Volts")/(4.7color(white)(.)Omega)`

`I ~~ 4.3color(white)(.)"Amperes"`