How do I find the current in this battery?

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 battery? Show your work.

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Answer 1 · 2017-03-12T16:20:34

Please see the 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:

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

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

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