Consider the circuit shown in Fig. E26.16. The current through the$\mathbf{6}\mathbf{\Omega }$resistor is 4A, in the direction shown. What are the currents through thelocalid="1664186296772" $\mathbf{25}\mathbf{\Omega }$and $\mathbf{20}\mathbf{\Omega }$resistors?

${\mathrm{R}}_1=6.09,{\mathrm{R}}_2=8.09{\mathrm{R}}_3=25.09{\mathrm{R}}_4=20.09,$are in parallel where in parallel the two resistors have the same voltage and equals to ${\mathrm{R}}_{11}=4.0\mathrm{A}$

the voltage of their combination

${\mathrm{V}}_1={\mathrm{V}}_2={\mathrm{V}}_{12}$

The combination of both resistors is Req as R1 and R2 in parallel and we could get it by equation 26.3 in the form

$$\begin{gathered} {\mathrm{V}}_1={\mathrm{V}}_2={\mathrm{V}}_{12} \\ {\mathrm{R}}_{\mathrm{eq}}{\mathrm{R}}_1{\mathrm{andR}}_2 \\ {\mathrm{R}}_{12}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2} \\ =\frac{\left(6.0\mathrm{\Omega }\right)\left(8.0\mathrm{\Omega }\right)}{6+8} \\ =3.42 \end{gathered}$$

Now let us get the voltage across R1 by using Ohm's law

$\begin{array}{l}{\mathrm{v}}_1={\mathrm{l}}_1{\mathrm{R}}_1\\ =\left(4.0\mathrm{A}\right)\left(6.0\right)\\ =24.0\mathrm{V}\end{array}$

As We discussed here, Vl = Vz so We can use Ohm's law again to get I2 as next

$$\begin{gathered} {\mathrm{I}}_2={\mathrm{V}}_2/{\mathrm{R}}_2 \\ =\left(24.0\mathrm{V}\right)/\left(8.052\right) \\ =3.0\mathrm{A} \end{gathered}$$

The combination R12 is in series with R3 so, the current I12 is the same I3 which equal I123 and as R1 and R2 are in parallel,

ThecombinationR12isinserieswithR3so,thecurrentI12isthesameI3whichequalI123andasR1andR2areinparalu

therefore the current of their combination I12 will equal the sum of the individual current

$$\begin{gathered} {\mathrm{I}}_{123}={\mathrm{I}}_3={\mathrm{I}}_{12}= \\ {\mathrm{I}}_1+{\mathrm{I}}_2=4.0\mathrm{A}+3.0\mathrm{A} \\ =7.0\mathrm{A} \end{gathered}$$

The current flows in 25 ohms is 7A

ThecombinationR12isinserieswithR3sotheircombinationR123couldbecalculatedbyusingequation26]intheform
$$\begin{gathered} {\mathrm{R}}_{123}={\mathrm{R}}_{12}+{\mathrm{R}}_3 \\ =3.429+25.09 \\ =28.42 \end{gathered}$$
Again, the combination R123 and R4 are in parallel, therefore, they have the same voltage V123 = V4- Where the voltage
V123 could be calculated by using Ohm's law
${\mathrm{R}}_{123}{\mathrm{andR}}_4$

$$\begin{gathered} {\mathrm{V}}_{123}={\mathrm{I}}_{123}{\mathrm{R}}_{123} \\ =(7.0\mathrm{A})\left(28.429\right) \\ =199\mathrm{V} \end{gathered}$$