Chapter 21, Problem 40PE

To determine

The currents flowing in the circuit.

Answer to Problem 40PE

  $I_1=0.38$ amperes, $I_2=-0.82$ amperes and $I_3=1.2$ amperes.

Explanation of Solution

Given information:

The given figure is

Concept Introduction:

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

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

Kirchhoff's second rule is also known as Kirchhoff's voltage law.

It states that all the energy is conserved.

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

  $\begin{array}{l}{E}_1-I_1{r}_1-I_1{R}_1-I_3{R}_2-I_3{r}_2+E_2-I_1{R}_4=0\\ \text{substitute all the values}\\ \Rightarrow 18-0.5I_1-I_{1}20-I_{3}6-I_{3}0.25+3-I_{1}15=0\\ \Rightarrow -6.25I_3-\text{35}\text{.5}{I}_1=-21\\ \Rightarrow I_1=\frac{-21+6.25I_3}{-\text{35}\text{.5}}\mathrm{...}\text{eq}\left(\text{1}\right)\end{array}$

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

  $\begin{array}{l}-I_2{R}_3+E_3-I_2{r}_3-I_2{r}_4-E_4+I_3{r}_2-E_2+I_3{R}_2=0\\ \text{substitute all the values }\\ \Rightarrow -I_{2}8+12-I_{2}0.5-I_{2}0.75-24+I_{3}0.25-3+I_{3}6=0\\ \Rightarrow -9.25I_2\text{+6}{\text{.25I}}_3\text{=15 }\\ \Rightarrow I_2\text{=}\frac{\text{15 }-\text{6}{\text{.25I}}_3}{-9.25}\mathrm{...}\text{eq}\left(2\right)\end{array}$

Now apply the Kirchhoff's junction rule,

  $I_1=I_2\text{ + }{I}_3\mathrm{...}\text{eq}\left(3\right)$

Equate $\text{eq}\left(\text{1}\right)$ and $\text{eq}\left(2\right)$ in $\text{eq}\left(3\right)$

  $\begin{array}{l}\frac{-21+6.25I_3}{-\text{35}\text{.5}}\text{ =}\frac{\text{15 }-\text{6}{\text{.25I}}_3}{-9.25}+I_3\\ \Rightarrow \frac{-21+6.25I_3}{-\text{35}\text{.5}}\text{ =}\frac{\text{15 }-\text{6}{\text{.25I}}_3+I_3\left(-9.25\right)}{-9.25}\\ \Rightarrow \frac{-21+6.25I_3}{-\text{35}\text{.5}}\text{ =}\frac{\text{15 }-\text{6}{\text{.25I}}_3-9.25I_3}{-9.25}\\ \Rightarrow \frac{-21+6.25I_3}{-\text{35}\text{.5}}\text{ =}\frac{\text{15 }-15.5I_3}{-9.25}\\ \Rightarrow -9.25\left(-21+6.25I_3\right)=-\text{35}\text{.5}\left(\text{15 }-15.5I_3\right)\\ \Rightarrow 194.25-57.81I_3=-532.5+550.25I_3\\ \Rightarrow -608.06I_3=-726.75\\ \Rightarrow I_3=\frac{-726.75}{-608.06}\\ \Rightarrow I_3=1.2\end{array}$

Substitute $I_3=1.2$ in $\text{eq}\left(\text{1}\right)$

  $\begin{array}{l}{I}_1=\frac{-21+6.25\left(1.2\right)}{-\text{35}\text{.5}}\\ \Rightarrow I_1=\frac{-13.5}{-\text{35}\text{.5}}\\ \Rightarrow I_1=0.38\end{array}$

Substitute $I_3=1.2$ and $I_1=0.38$ in $\text{eq}\left(3\right)$

  $\begin{array}{l}{I}_1=I_2\text{ + }{I}_3\\ \Rightarrow I_2+1.2=0.38\\ \Rightarrow I_2=-0.82\end{array}$

Conclusion:

The currents flowing in the circuit are $I_1=0.38$ amperes, $I_2=-0.82$ amperes and $I_3=1.2$ amperes.