Chapter 21, Problem 23CQ

To determine

The loop rule to loop $\text{abgefa and cbgedc}$

Answer to Problem 23CQ

  $I_1=0.53$ Amperes, $I_2=-1.1275$ amperes and $I_3=\left(I_1-I\right)=1.1275-0.53=0.5975$ amperes.

Explanation of Solution

Given information:

The given figure is

Formula used:

Kirchhoff's second rule states that in a closed loop the sum of all the voltages is equal to zero.

Let the current flowing in a closed network $\text{abgefa}$ be $\text{I}$ amperes and the current flowing in a closed network $\text{ cbgedc}$ be ${\text{I}}_1$.

Apply Kirchhoff's loop rule to loop $\text{abgefa }$

  $\begin{array}{l}-I{R}_1-\left(I-I_1\right)R_2+E_2-\left(I-I_1\right)r_2-I{R}_4+E_1-I{r}_1=0\\ \text{substitute all the values}\\ \Rightarrow -20I-6\left(I-I_1\right)+3-0.25\left(I-I_1\right)-15I+18-0.5I=0\end{array}$

  $\begin{array}{l}\Rightarrow -20I-6\left(I-I_1\right)-0.25\left(I-I_1\right)-15I-0.5I=-18-3\\ \Rightarrow -20I-6I+6I_1-0.25I+0.25I_1-15I-0.5I=-21\\ \Rightarrow -41.75I+6.25I_1=-21\mathrm{...}\text{eq}\left(1\right)\end{array}$

Apply Kirchhoff's loop rule to loop $\text{ cbgedc}$

  $\begin{array}{l}-I_1{R}_3-\left(I_1-I\right)R_2+E_2-\left(I_1-I\right)r_2+E_4-I_1{r}_4-I{r}_3-E_3=0\\ \text{substitute all the values}\\ \Rightarrow -8I_1-\left(I_1-I\right)6+3-\left(I_1-I\right)0.25+24-I_{1}0.75-0.5I-12=0\\ \Rightarrow -8I_1-6I_1+6I+3-0.25I_1+0.25I+24-I_{1}0.75-0.5I_1-12=0\\ \Rightarrow -8I_1-6I_1+6I-0.25I_1+0.25I-I_{1}0.75-0.5I_1=-24-3+12\\ \Rightarrow -15.5I_1+6.25I=-15\mathrm{...}\text{eq}\left(2\right)\end{array}$

From $\text{eq}\left(\text{1}\right)$

  $\begin{array}{l}-41.75I+6.25I_1=-21\\ \Rightarrow I_1=-21+41.75I\end{array}$

Substitute $I_1=-21+41.75I$ in $\text{eq}\left(\text{2}\right)$

  $\begin{array}{l}-15.5\left(-21+41.75I\right)+6.25I=-15\\ \Rightarrow 325.5-647.125I+6.25I=-15\\ \Rightarrow -640.875I=-340.5\\ \Rightarrow I=\frac{340.5}{640.875}\\ \Rightarrow I=0.53\end{array}$

Substitute $I=0.53$ in $\text{eq}\left(\text{1}\right)$

  $\begin{array}{l}{I}_1=-21+41.75\left(0.53\right)\\ \Rightarrow I_1=-21+22.1275\\ \Rightarrow I_1=1.1275\end{array}$

Therefore the current value of the currents indicated in the figure $I_1=0.53$ amperes, $I_2=-1.1275$ amperes and $I_3=\left(I_1-I\right)=1.1275-0.53=0.5975$ amperes.

Conclusion:

Application of the loop rule to loop $\text{abgefa }$ and $\text{ cbgedc}$ results in $I_1=0.53$ amperes, $I_2=-1.1275$ amperes and $I_3=\left(I_1-I\right)=1.1275-0.53=0.5975$ amperes.