Compute the equivalent resistance of the network in Fig. E26.14, and find the current in each resistor. The battery has negligible internal resistance.

In a series circuit, all the components are connected in one line head to tail so the current is the same for all the components.

In a parallel circuit, all components are connected head to head, tail to tail so the voltage drop is the same for all the components.

Here, ${\mathrm{R}}_1\mathrm{and}{\mathrm{R}}_2$are in series, so their combination $R_{12}$can be calculated as:

$$\begin{gathered} R_{12}=R_1+R_2 \\ =1.0+3.0 \\ =4.0\Omega \end{gathered}$$

The same step for $R_3$and $R_4$as they are in series,

$$\begin{gathered} R_{34}=R_3+R_4 \\ =7.0+5.0 \\ =12.0\Omega \end{gathered}$$

After reducing the circuit, the two combinations are in parallel, and equivalent resistance can be given as:

$$\begin{gathered} R_{eq}=\frac{R_{12}{R}_{34}}{R_{12}+R_{34}} \\ =\frac{\left(4.0\right)\left(12.0\right)}{4.0+12.0} \\ =3.0\Omega \end{gathered}$$

Here, as $R_{12}$and $R_{34}$are in parallel, therefore, they have the same voltages which are equal to the voltage of the battery as:

$V_{12}=V_{34}=V=48.0\hspace{0.33em}V$

Now, from Ohm’s law the current can be calculated as:

$$\begin{gathered} l_{12}=\frac{V}{R_{12}} \\ =\frac{48.0V}{4.0\Omega } \\ =12.0A \end{gathered}$$

And

$$\begin{gathered} l_{34}=\frac{48.0V}{12.0\Omega } \\ =4.0A \end{gathered}$$

The two resistors are connected in series. So, they are both equal.

Therefore, $I_1=I_2=I_{12}=12.0$and $I_3=I_4=I_{34}=4.0\hspace{0.33em}A$.

Thus, the equivalent resistance is $3.0\Omega$and the current in each resistor is $12.0\hspace{0.33em}A,12.0\hspace{0.33em}A,4.0\hspace{0.33em}A,and4.0\hspace{0.33em}A$.