91Ó°ÊÓ

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

To calculate the total current in circuit, both $4\mathrm{Î\copyright }and12\mathrm{Î\copyright }$are in parallel, and calculating their equivalent resistance

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

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

$R_{2,eq}=R_{1,eq}+5\mathrm{Î\copyright }=3\mathrm{Î\copyright }+5\mathrm{Î\copyright }=8\mathrm{Î\copyright }$

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

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

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

$$\begin{gathered} \underset{}{∑}V=0 \\ {\mathrm{Î\mu }}_1=I{R}_{3,eq}-{\mathrm{Î\mu }}_1-I\left(3\mathrm{Î\copyright }\right)=0 \\ 12V-I\left(6\mathrm{Î\copyright }\right)-3V-I\left(3\mathrm{Î\copyright }\right)=0 \\ I=1A \end{gathered}$$

The current through the circuit is$3\mathrm{Î\copyright }$

$I_3=1A$

To get voltage of $3\mathrm{Î\copyright }$resistor, we use Ohm's law

$â–³V_3=I_3{R}_3=(1A)\left(3\mathrm{Î\copyright }\right)=3V$

The voltage of $24\mathrm{Î\copyright }$is calculated by

$â–³V_{24}=I{R}_{3,eq}=(1A)\left(6\mathrm{Î\copyright }\right)=6V$

Again we have to get voltage of $24\mathrm{Î\copyright }$resistor, so we have to use Ohm's law

$I_{24}=\frac{â–³V_{24}}{R_{24}}=\frac{6V}{24\mathrm{Î\copyright }}=0.25A$

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

$I_{2,eq}=\frac{â–³V_{24}}{R_{24}}=\frac{6V}{8\mathrm{Î\copyright }}=0.75A$

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

$I_5=0.75A$

To get voltage of $5\mathrm{Î\copyright }$resistor, we have to use Ohm's law

$â–³V_5=I_5{R}_5=(0.75A)(5\mathrm{Î\copyright })=3.75V$

Parallel resistors $4\mathrm{Î\copyright }and12\mathrm{Î\copyright }$both have same voltage and it is calculated by

$â–³V_4=â–³V_{12}=I_5{R}_{1,eq}=(0.75A)(3\mathrm{Î\copyright })=2.25V$

Again we use Ohm's law to get the current of $4\mathrm{Î\copyright }and12\mathrm{Î\copyright }$by

$I_4=\frac{â–³V_4}{R_4}=\frac{2.25V}{4\mathrm{Î\copyright }}=0.5625A$

and

$I_{12}=\frac{â–³V_{12}}{R_{12}}=\frac{2.25V}{12\mathrm{Î\copyright }}=0.1875A$

After summerize the results in table