Question

Calculate the three currents I₁, I₂, and I₃ indicated in the circuit diagram shown in Fig. P26.59.

Short answer

The three currents are$I_1=0.848\hspace{0.33em}A,\hspace{0.33em}{I}_2=2.14\hspace{0.33em}A,\hspace{0.33em}and\hspace{0.33em}{I}_3=0.171\hspace{0.33em}A$

Step-by-step solution

Concept Introduction

Given data

$$\begin{gathered} E_1=12.0\hspace{0.33em}V \\ {E}_2=9.0\hspace{0.33em}V \\ {R}_1=1.00\hspace{0.33em}\Omega \\ {R}_2=5.00\hspace{0.33em}\Omega \end{gathered}$$

Apply Kirchoff’s loop law

Use loop rule to loop (1),

$-E_1+I_2{R}_1+(I_2-I_3)R_2=0$

$-E_1+I_2(R_1+R_2)-I_3{R}_2=0$ (1)

For loop 2,

${\mathrm{E}}_2-({\mathrm{I}}_1+{\mathrm{I}}_3){\mathrm{R}}_3-{\mathrm{I}}_1{\mathrm{R}}_1=0$

${\mathrm{E}}_2-{\mathrm{I}}_3{\mathrm{R}}_3-{\mathrm{I}}_1({\mathrm{R}}_1+{\mathrm{R}}_3)=0$ (2)

Loop 3,

$-E_1+I_2{R}_1-I_1{R}_1+E_2+I_3{R}_4=0$ (3)

Use the values in the above expressions

Put ${\mathrm{R}}_2$= 5 ${\mathrm{R}}_1$, ${\mathrm{R}}_3$= 8 ${\mathrm{R}}_1$, ${\mathrm{R}}_4$= 10 in equations 1,2, and 3, and we get,

localid="1668335255112" $$\begin{gathered} -{\mathrm{E}}_1+6{\mathrm{I}}_2{\mathrm{R}}_1-5{\mathrm{I}}_3{\mathrm{R}}_1=0 \\ {\mathrm{E}}_2-8{\mathrm{I}}_3{\mathrm{R}}_1-9{\mathrm{I}}_1{\mathrm{R}}_1=0 \\ -{\mathrm{E}}_1+{\mathrm{I}}_2{\mathrm{R}}_1-{\mathrm{I}}_1{\mathrm{R}}_1+{\mathrm{E}}_2+10{\mathrm{I}}_3{\mathrm{R}}_1=0 \end{gathered}$$

Divide the above equation by $R_1$,

$$\begin{gathered} -\frac{E_1}{R_1}+6I_2-5I_3=0 \\ 6I_2-5I_3=\frac{E_1}{R_1}\dots \dots \dots \dots \left(4\right) \end{gathered}$$

localid="1668330342001" $$\begin{gathered} \frac{E_2}{R_1}-8I_3-9I_1=0 \\ 8I_3+9I_1=\frac{E_2}{R_1}\dots \dots \dots \dots \dots \left(5\right) \end{gathered}$$

$$\begin{gathered} -\frac{E_1}{R_1}+I_2-I_1+\frac{E_2}{R_1}+10I_3=0 \\ {I}_2-I_1+10I_3=\frac{E_1}{R_1}-\frac{E_2}{R_1}\dots \dots \left(6\right) \end{gathered}$$

Use the values from equations (4), and (6) and we get,

$\frac{5I_3}{6}+\frac{E_1}{6R_1}-I_1+10I_3=\frac{E_1}{R_1}-\frac{E_2}{R_1}$

Multiply by factor 6, and we get,

$\begin{array}{r}5l_3-6l_1+60I_3=\frac{5E_1}{R_1}-\frac{6E_2}{R_1}\\ -6l_1+65l_3=\frac{5E_1}{R_1}-\frac{6E_2}{R_1}\end{array}$ (7)

Multiply the equation (5) by factor 2/3,

$\frac{16l_3}{3}+6l_1=\frac{2E_2}{3R_1}$

Add the above expression from equation (7), and we get

$$\begin{gathered} \frac{16{\mathrm{I}}_3}{3}+65{\mathrm{I}}_3=\frac{2{\mathrm{E}}_2}{3{\mathrm{R}}_1}+\frac{5{\mathrm{E}}_1}{{\mathrm{R}}_1}-\frac{6{\mathrm{E}}_2}{{\mathrm{R}}_1} \\ \frac{211{\mathrm{I}}_3}{3}=\frac{15{\mathrm{E}}_1-16{\mathrm{E}}_2}{3{\mathrm{R}}_1} \\ {\mathrm{I}}_3=\frac{15{\mathrm{E}}_1-16{\mathrm{E}}_2}{211{\mathrm{R}}_1} \end{gathered}$$

Calculate the value of currents 

Substituting all the values we get,

$$\begin{gathered} I_3=\frac{15{\mathrm{E}}_1-16{\mathrm{E}}_2}{211{\mathrm{R}}_1} \\ =\frac{15\left(12.0\mathrm{V}\right)-16\left(9.00\mathrm{V}\right)}{211\left(1.00\mathrm{\Omega }\right)} \\ =0.17\mathrm{A} \end{gathered}$$

from this equation substitute the second one in (5), and we get,

$$\begin{gathered} {\mathrm{I}}_1=\frac{{\mathrm{E}}_2}{9{\mathrm{R}}_1}-\frac{8{\mathrm{I}}_3}{9} \\ =\frac{9.00\mathrm{V}}{9\left(1.00\mathrm{\Omega }\right)}-\frac{8\left(0.171\mathrm{A}\right)}{9} \\ =0.848\mathrm{A} \end{gathered}$$

Substitute to the first equation in (4),

$$\begin{gathered} {\mathrm{I}}_2=\frac{5\left(0.171\mathrm{A}\right)}{6}-\frac{\left(12.0\mathrm{V}\right)}{6\left(1.00\mathrm{\Omega }\right)} \\ =2.14\mathrm{A} \end{gathered}$$

Thus, $I_1=0.848\hspace{0.33em}A,\hspace{0.33em}{I}_2=2.14\hspace{0.33em}A,\hspace{0.33em}and\hspace{0.33em}{I}_3=0.171\hspace{0.33em}A$.