For the circuit shown in FIGURE P28.62, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

We need to find the current through and the potential difference across each resistor.

To calculate the total current in circuit, both $4\Omega and12\Omega$are in parallel, and calculating their equivalent resistance

$$\begin{gathered} \frac{1}{R_{1,eq}}=\frac{1}{4\Omega }+\frac{1}{12\Omega } \\ \frac{1}{R_{1,eq}}={\left(\frac{1}{4\Omega }+\frac{1}{12\Omega }\right)}^{-1} \\ {R}_{1,eq}=3\Omega \end{gathered}$$

$5\Omega$resistor is in series with $R_{1,eq}$, In this case, the equivalent resistance is

$R_{2,eq}=R_{1,eq}+5\Omega =3\Omega +5\Omega =8\Omega$

$R_{2,eq}$combination is parallel with $24\Omega$, so, the equivalent resistance for this combination by

$$\begin{gathered} \frac{1}{R_{3,eq}}=\frac{1}{8\Omega }+\frac{1}{24\Omega } \\ \frac{1}{R_{3,eq}}={\left(\frac{1}{8\Omega }+\frac{1}{24\Omega }\right)}^{-1} \\ {R}_{3,eq}=6\Omega \end{gathered}$$

To find the current in circuit use the loop rule, If we travel clockwise, we get

$$\begin{gathered} \sum _{}V=0 \\ {\epsilon }_1=I{R}_{3,eq}-{\epsilon }_1-I\left(3\Omega \right)=0 \\ 12V-I\left(6\Omega \right)-3V-I\left(3\Omega \right)=0 \\ I=1A \end{gathered}$$

The current through the circuit is$3\Omega$

$I_3=1A$

To get voltage of $3\Omega$resistor, we use Ohm's law

$△V_3=I_3{R}_3=(1A)\left(3\Omega \right)=3V$

The voltage of $24\Omega$is calculated by

$△V_{24}=I{R}_{3,eq}=(1A)\left(6\Omega \right)=6V$

Again we have to get voltage of $24\Omega$resistor, so we have to use Ohm's law

$I_{24}=\frac{△V_{24}}{R_{24}}=\frac{6V}{24\Omega }=0.25A$

The current of $5,12,4$combination is calculated by

$I_{2,eq}=\frac{△V_{24}}{R_{24}}=\frac{6V}{8\Omega }=0.75A$

$5\Omega$is in series with $R_{1,eq}$, and both of them have $0.75A$of current

$I_5=0.75A$

To get voltage of $5\Omega$resistor, we have to use Ohm's law

$△V_5=I_5{R}_5=(0.75A)(5\Omega )=3.75V$

Parallel resistors $4\Omega and12\Omega$both have same voltage and it is calculated by

$△V_4=△V_{12}=I_5{R}_{1,eq}=(0.75A)(3\Omega )=2.25V$

Again we use Ohm's law to get the current of $4\Omega and12\Omega$by

$I_4=\frac{△V_4}{R_4}=\frac{2.25V}{4\Omega }=0.5625A$

and

$I_{12}=\frac{△V_{12}}{R_{12}}=\frac{2.25V}{12\Omega }=0.1875A$

After summerize the results in table