# See image below. What is the current through the 8 Ω resistor?

Jul 25, 2018

0.387A

#### Explanation:

Resistors in series: $R = {R}_{1} + {R}_{2} + {R}_{3} + \ldots . .$

Resistors in parallel: $\frac{1}{R} = \frac{1}{R} _ 1 + \frac{1}{R} _ 2 + \frac{1}{R} _ 3 + \ldots . .$

Start by combining the resistances so that we can work out the current flowing in the various paths.

The $8 \Omega$ resistor is in parallel with $14 \Omega$ ($3 + 5 + 6$) so the combination (let's call it ${R}_{a}$) is

$\frac{1}{R} = \left(\frac{1}{8} + \frac{1}{14}\right) = \frac{11}{28}$

${R}_{a} = \frac{28}{11} \text{ } \left(= 2.5454 \Omega\right)$

${R}_{a}$ is in series with $4 \Omega$ and the combination is in parallel with $10 \Omega$, so

$\frac{1}{R} _ b = \left(\frac{1}{10} + \frac{1}{4 + \frac{28}{11}}\right) = 0.1 + \frac{1}{\frac{72}{11}} = 0.1 + \frac{11}{72} = 0.2528$

${R}_{b} = 3.9560 \Omega$

${R}_{b}$ is in series with $2 \Omega$ so

${R}_{T o t a l} = 2 + 3.9560 = 5.9560 \Omega$

$I = \frac{V}{R} = \frac{12}{5.9560} = 2.0148 A$ (total current flowing from battery)

This current flowing through the $2 \Omega$ resistor divides into 2 paths, the $10 \Omega$ resistor and ${R}_{b}$

It is possible to proportion the currents through ${R}_{b}$ and then ${R}_{a}$, but easier to subtract the voltage drops across the $2 \Omega$ and $4 \Omega$ resistors.

${V}_{{R}_{a}} = 12 - \left(2 \cdot 2.0148\right) = 7.9705 V$

so current through the $4 \Omega$ resistor

$= \left(\frac{7.9705}{4 + {R}_{a}}\right) = \frac{7.9705}{4 + \left(\frac{28}{11}\right)} = 1.2177 A$

so ${V}_{4 \Omega} = 4 \cdot 1.2177 = 4.8708 V$

Voltage across $8 \Omega$ is $7.9705 - 4.8708 = 3.0997 V$

So current through the $8 \Omega$ resistor is $\frac{3.0997}{8} = 0.387 A$