For the circuit shown in FIGURE $28.59$, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

We have given the circuit.

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

When the resistors are connected at both the ends of the point then it is said that the resistors are connected in parallel. This is obvious when the resistors are aligned side by side. The potential difference across the resistors is the same in parallel connection. Equation $(28.24)$shows the equivalent resistance for the parallel connection in the form

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_N}$

The two resistors $6\Omega$and $12\Omega$are in parallel, so we use the equation $\left(1\right)$to get their combination by

$$\begin{gathered} \frac{1}{R_{1,eq}}=\frac{1}{6\Omega }+\frac{1}{12\Omega } \\ {R}_{1,eq}=\frac{1}{6\Omega }+{\frac{1}{12\Omega }}^{-1} \\ {R}_{1,eq}=4\Omega \end{gathered}$$

The resistor $4\Omega$is in series with a combination $R_{1,eq}$,so they have the same current. We apply the loop rule to get the current through resistors $4\Omega$and the combination $R_{1,eq}$. Traveling clockwise in the left loop and get the next

$$\begin{gathered} V=0 \\ {}_{\in -}I\left(2\Omega \right)-I_4\left(4\Omega \right)-I_4{R}_{1,eq}=0 \\ 24-2I-4I_4-4I_4=0 \\ 24-2I-8I_4=0 \end{gathered}$$ $\left(2\right)$

Appling the loop rule on the right loop and travel clockwise to get

$$\begin{gathered} V=0 \\ -I_8\left(8\Omega \right)+I_4{R}_{1,eq}+I_4\left(4\Omega \right)=0 \\ -8I_8+4I_4+4I_4=0 \\ {I}_4=I_8 \end{gathered}$$

The current $I$from the junction rule equals $I=I_4+I_8=2I_4=2I_8$.Use this into equation $\left(2\right)$to get $I_4$by

$$\begin{gathered} 24-2I-8I_4=0 \\ 24-2\left(2I_4\right)-8I_4=0 \\ {I}_4=2A \end{gathered}$$

Hence, the current through $8\Omega$is $I_8=2A$

Using Ohm's law to get the voltage across both resistors by

$∆V_4=I_4{R}_4=\left(2A\right)\left(4\Omega \right)=8V$

and

$∆V_8=I_8{R}_8=\left(2A\right)\left(8\Omega \right)=16V$

The current $I$is the same for resistors $2\Omega$, so the current through it is calculated by

$I_2=I_4+I_8=2A+2A=4A$

Use Ohm's law to get the voltage across this resistor by

$∆V_2=I_2{R}_2=\left(4A\right)\left(2\Omega \right)=8V$

Both resistors $6\Omega$and $12\Omega$have the same voltage and their combination current is the same for $I_4$, so using Ohm's law, the voltage across both of them is

$∆V_6=∆V_{12}=I_4{R}_{1,eq}=\left(2A\right)\left(4\Omega \right)=8V$

Again using Ohm's law, the current through $6\Omega$and $12\Omega$will be

$I_6=\frac{∆V_6}{R_6}=\frac{8V}{6\Omega }=1.33A$

and

$I_{12}=\frac{∆V_{12}}{R_{12}}=\frac{8V}{12\Omega }=0.66A$

The results are summarised in below's table