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.

1 Answer
Mar 12, 2017

Answer:

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"#