# 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

#### 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:

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

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

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