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.

Mar 12, 2017

Please see the explanation.

Explanation:

Given:

${R}_{\text{series}} = 2.0 \textcolor{w h i t e}{.} \Omega$
${R}_{1} = {R}_{2} = {R}_{3} = 8 \textcolor{w h i t e}{.} \Omega$
${V}_{\text{battery"= 20color(white)(.)"Volts}}$

Let ${R}_{\text{equivalent}}$ 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)

${R}_{\text{equivalent}} \approx 2.7 \textcolor{w h i t e}{.} \Omega$

Let ${R}_{\text{total}}$ be the total resistance as seen by the battery:

${R}_{\text{total" = R_"series" + R_"equivalent}}$

${R}_{\text{total}} = 2.0 \textcolor{w h i t e}{.} \Omega + 2.7 \textcolor{w h i t e}{.} \Omega$

${R}_{\text{total}} = 4.7 \textcolor{w h i t e}{.} \Omega$

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

$I = {V}_{\text{battery"/R_"total}}$

$I = \frac{20 \textcolor{w h i t e}{.} \text{Volts}}{4.7 \textcolor{w h i t e}{.} \Omega}$

$I \approx 4.3 \textcolor{w h i t e}{.} \text{Amperes}$