91Ó°ÊÓ

Kirchoff’s law state that at any junction the algebraic sum of all currents should be always equal to zero.

The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero

Consider the following circuit, where

We need to find the current through each one of the resistors and in the battery(by using the junction rule we will find the current). Apply the loop rule to the left loop(clockwise) we get

$$\begin{gathered} \mathrm{Î\mu }-2I{R}_1+3I_2{R}_1=0 \\ or \\ \mathrm{Î\mu }-I_1(R_1+R_2)+I_2{R}_2=0 \end{gathered}$$

Apply the loop rule to the right upper loop(clockwise), and we get,

$$\begin{gathered} I_1{R}_1-(I-I_1)R_2+I_2{R}_2=0 \\ {I}_1(R_1+R_2)-I{R}_2+I_2{R}_1=0 \end{gathered}$$

And to the right bottom loop( also clockwise), we get,

$$\begin{gathered} -I_2{R}_1-(I-I_1+I_2)R_1+(I_1-I_2)R_2=0 \\ {I}_1(R_1+R_2)-I{R}_1+I_2(2R_1+R_2)=0 \end{gathered}$$

To simplify the problem, we will use the substitution $R_2=2R_1$, and we get

$\left\{\begin{array}{l}\mathrm{Î\mu }−3I_1{R}_1+I_2{R}_1=0\\ 3l_1{R}_1−2âˆ\pounds R_1+I_2{R}_1=0\\ 3l_1{R}_1−âˆ\pounds R_1+4I_2{R}_2=0\end{array}\right.$

We solve this equation for $\mathrm{I}, {\mathrm{I}}_2, \mathrm{and} {\mathrm{I}}_1$the first we subtract the third equation in (4) from the second equation to eliminate

$$\begin{gathered} -I{R}_1+5I_2{R}_1=0 \\ I=5I_2 \end{gathered}$$

And the second equation in (4) to the first one, so we get,

$\mathrm{Î\mu }-2I{R}_1+3I_2{R}_1=0$

Substitute from (5)

$\mathrm{Î\mu }-10I_2{R}_1+3I_2{R}_1=0$

Now, we will have,

$l_2=\frac{Â\pounds }{7R_1}$

Substitution of the given data, we get,

$$\begin{gathered} {\mathrm{I}}_2=\frac{\mathrm{Î\mu }}{7{\mathrm{R}}_1} \\ =\frac{14.0\mathrm{V}}{7×1.0\mathrm{Î\copyright }} \\ =2.0\mathrm{A} \end{gathered}$$

Substitute into (5) we get,

$$\begin{gathered} \mathrm{I}=5{\mathrm{l}}_2 \\ =5×\left(2.00\mathrm{A}\right) \\ =10.0\mathrm{A} \end{gathered}$$

Which is the current through the battery, substitute into the first equation into (4), so we get,

$$\begin{gathered} \mathrm{Î\mu }−3{\mathrm{I}}_1{\mathrm{R}}_1+{\mathrm{I}}_2{\mathrm{R}}_1=0 \\ {\mathrm{I}}_1=\frac{14.0\mathrm{V}}{3×1.0\mathrm{Î\copyright }}+\frac{2.0\mathrm{A}}{3}{ \\ }_1=\frac{\mathrm{Î\mu }}{3{\mathrm{R}}_1}+\frac{{\mathrm{I}}_2}{3} \\ {\mathrm{I}}_1=5.34\mathrm{A} \end{gathered}$$

Thus,

$$\begin{gathered} I=10.0 A \\ {I}_2=2.0 A \\ {I}_1=5.34 A \end{gathered}$$

using these currents you get any current in the circuit as shown in the figure.

(b) The equivalent resistance equals the potential difference across the resistor network, divided by the current through the network which is I so,

$$\begin{gathered} {\mathrm{R}}_{\text{eq}}=\frac{\mathrm{Î\mu }}{1} \\ =\frac{14.0\mathrm{V}}{10.0\mathrm{A}} \\ =1.40\mathrm{Î\copyright } \end{gathered}$$

Hence the equivalent resistance is $1.40 \mathrm{Î\copyright }$