For the circuit shown in FIGURE P28.60, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

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

The circuit can be simplified as shown in figure. in case if resistors are connected at both ends of two points then it is called resistors 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 shows the equivalent resistance for the parallel connection in the form

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+....+\frac{1}{R_N}\left(1\right)$

$4\Omega ,48\Omega and16\Omega$, these three resistors are in parallel, so use equation $\left(1\right)$to get their combination by

$$\begin{gathered} \frac{1}{R_{1,eq}}=\frac{1}{4\Omega }+\frac{1}{16\Omega }+\frac{1}{48\Omega } \\ \frac{1}{R_{1,eq}}={\left(\frac{1}{4\Omega }+\frac{1}{16\Omega }+\frac{1}{48\Omega }\right)}^{-1} \\ {R}_{1,eq}=3\Omega \end{gathered}$$

The above combination is in the series with resistor $3\Omega$, so the equivalent resistance of the circuit is

$R_{eq}=R_{1,eq}+3\Omega =3\Omega +3\Omega =6\Omega$

Also, the current is same for resistors in series, so the current through both resistors and the combination is calculated by using Ohm's law

$I_3=\frac{\epsilon }{R_{eq}}=\frac{12V}{6\Omega }=2A$

After using Ohm's law to get the voltage across resisors 3$\Omega$by

localid="1649086636375" $△V=I_3{R}_3=(2A)\left(3\Omega \right)=6V$

localid="1649086640807" $4\Omega ,48\Omega and16\Omega$resistors have the same voltage and their combination current is the same for $I_3$, Using Ohm's law to get the voltage across each of the by

localid="1649086645233" $△V_4=△V_{48}=△V_{16}=I_3{R}_{eq}=(2A)\left(3\Omega \right)=6V$

Also we have to use Ohm's law to get current through $4\Omega ,48\Omega and16\Omega$by

$I_4=\frac{△V_4}{R_4}=\frac{6V}{16\Omega }=1.5A$

and

$I_{16}=\frac{△V_{16}}{R_{16}}=\frac{6V}{16\Omega }=0.375A$

and

$I_{16}=\frac{△V_{48}}{R_{48}}=\frac{6V}{16\Omega }=0.125A$

The results are in the next table