Chapter 21, Problem 38PE

To determine

The currents flowing in the circuit and show the steps followed in the problem-solving strategies for series and parallel resistors.

Answer to Problem 38PE

  $I_1=-0.34$ amperes, $I_2=0.38$ amperes and $I_3=0.04$ amperes

Explanation of Solution

Given information:

The given figure is

  

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 is also known as Kirchhoff's voltage law.

It states that all the energy is conserved.

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

  $\begin{array}{l}{E}_2-I_2{r}_2-I_2{R}_2+I_1{R}_5+I_1{r}_1-E_1+I_1{R}_1=0\\ \text{substitute all the values}\\ \Rightarrow 48-0.50I_2-40I_2+20I_1+0.10I_1-24+5I_1=0\\ \Rightarrow -40.50I_2+25.10I_1=-24\\ \Rightarrow -20.25I_2+12.55I_1=-12\\ \Rightarrow I_1=\frac{-12+20.25I_2}{12.55}\mathrm{...}\text{eq}\left(\text{1}\right)\end{array}$

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

  $\begin{array}{l}{E}_2-I_2{r}_2-I_2{R}_2-I_3{r}_4-E_4-I_3{r}_3+E_3-I_3{R}_3=0\\ \text{substitute all the values }\\ \Rightarrow \text{48}-0.50I_2-{\text{40I}}_2-0.{\text{20I}}_3-36-\text{0}{\text{.05I}}_3\text{+6}-78{\text{I}}_3\text{=0 }\\ \Rightarrow -40.50I_2-78.25{\text{I}}_3\text{=}-\text{18 }\\ \Rightarrow 40.50I_2+78.25{\text{I}}_3\text{=18 }\\ \Rightarrow {\text{I}}_3\text{=}\frac{18-40.50I_2}{78.25}\mathrm{...}\text{eq}\left(2\right)\end{array}$

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 $\text{eq}\left(3\right)$

  $\begin{array}{l}\frac{18-40.50I_2}{78.25}=\frac{-12+20.25I_2}{12.55}+I_2\\ \Rightarrow \frac{18-40.50I_2}{78.25}=\frac{-12+20.25I_2\text{ +12}{\text{.55I}}_2}{12.55}\\ \Rightarrow \frac{18-40.50I_2}{78.25}=\frac{-12+32.8{\text{I}}_2}{12.55}\\ \Rightarrow 12.55\left(18-40.50I_2\right)=78.25\left(-12+32.8{\text{I}}_2\right)\\ \Rightarrow 225.9-508.275I_2=-939+2566.6I_2\\ \Rightarrow -3074.875I_2=-1164.9\\ \Rightarrow I_2=\frac{-1164.9}{-3074.875}\\ \Rightarrow I_2=0.38\end{array}$

Substitute $I_2=0.38$ in $\text{eq}\left(\text{1}\right)$

  $\begin{array}{l}{I}_1=\frac{-12+20.25\left(0.38\right)}{12.55}\\ \Rightarrow I_1=\frac{-4.305}{12.55}\\ \Rightarrow I_1=-0.34\end{array}$

Substitute $I_2=0.38$ and $I_1=-0.34$ in $\text{eq}\left(3\right)$

  $\begin{array}{l}{I}_1+I_2\text{ = }{I}_3\\ \Rightarrow -0.34+0.38=I_3\\ \Rightarrow I_3=0.04\end{array}$

The steps followed in the problem solving strategies for series and parallel resistors are:

At first add the series resistors and parallel resistors which will give the equivalent resistance such as Equivalent resistance of several resistors connected in series is

  ${\text{R}}_{\text{eq}}{\text{=R}}_{\text{1}}{\text{+R}}_{\text{2}}\text{+}\mathrm{............}{\text{+R}}_{\text{N}}$

For $N$ identical series resistors the equivalent resistance is

  ${\text{R}}_{\text{eq}}\text{=NR}$

Equivalent resistance of several resistors connected in parallel is

  $\frac{\text{1}}{{\text{R}}_{\text{eq}}}\text{=}\frac{\text{1}}{{\text{R}}_{\text{1}}}\text{+}\frac{\text{1}}{{\text{R}}_{\text{2}}}\text{+}\mathrm{............}\text{+}\frac{\text{1}}{{\text{R}}_{\text{N}}}$

For $\text{N}$ identical parallel resistors the equivalent resistance is ${\text{R}}_{\text{eq}}\text{=}\frac{\text{R}}{\text{N}}$

b)Now the desired term can be calculated such as current, power or voltage.

Conclusion:

The currents flowing in the circuit are $I_1=-0.34$ amperes, $I_2=0.38$ amperes and $I_3=0.04$ amperes and the steps followed in the problem solving strategies for series and parallel resistors is discussed.