Solved Problem on Mesh Analysis

In the circuit below find the currents in the branches and their true directions.

Problem data:

Resistors

R₁ = 0.5 Ω;
R₂ = 0.5 Ω;
R₃ = 1 Ω;
R₄ = 0.5 Ω;
R₅ = 0.5 Ω;
R₆ = 3 Ω;
R₇ = 1 Ω.

Batteries

E₁ = 20 V;
E₂ = 20 V;
E₃ = 6 V;

Solution

First, at each circuit loop, we arbitrarily assign a direction of the current. In the ABEFA loop, we have the current i₁ clockwise, and in the BCDEB loop we have the current i₂ clockwise (Figure 1)

Figure 1

Using the Mesh Analysis to the mesh i₁ from point A in the chosen direction, forgetting the mesh i₂ (Figure 2)

Figure 2

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum_{n} V_{n}=0} \end{gather} \]

\[ \begin{gather} R_{2}i_{1}+R_{4}(\;i_{1}-i_{2}\;)+E_{2}+R_{5}(\;i_{1}-i_{2}\;)+R_{3}i_{1}+R_{1}i_{1}-E_{1}=0 \end{gather} \]

substituting the values of the problem

\[ \begin{gather} 0.5i_{1}+0.5(\;i_{1}-i_{2}\;)+20+0.5(\;i_{1}-i_{2}\;)+1i_{1}+0.5i_{1}-20=0\\[5pt] 0.5i_{1}+0.5(\;i_{1}-i_{2}\;)+0.5(\;i_{1}-i_{2}\;)+1i_{1}+0.5i_{1}=0\\[5pt] 0.5i_{1}+0.5i_{1}-0.5i_{2}+0.5i_{1}-0.5i_{2}+1i_{1}+0.5i_{1}=0\\[5pt] 3i_{1}-i_{2}=0 \tag{I} \end{gather} \]

Forgetting the mesh i₁ and using the Mesh Analysis to the mesh i₂, as was done above, we have for Figure 3, from point B

Figure 3

\[ \begin{gather} R_{6}i_{2}+E_{3}+R_{7}i_{2}+R_{5}(\;i_{2}-i_{1}\;)-E_{2}+R_{4}(\;i_{2}-i_{1}\;)=0 \end{gather} \]

substituting the values of the problem

\[ \begin{gather} 3i_{2}+6+1i_{2}+0.5(\;i_{2}-i_{1}\;)-20+0.5(\;i_{2}-i_{1}\;)=0\\[5pt] 3i_{2}+i_{2}+0.5i_{2}-0.5i_{1}-14+0.5i_{2}-0.5i_{1}=0\\[5pt] -i_{1}+5i_{2}=14 \tag{II} \end{gather} \]

Equations (I) and (II) can be written as a system of linear equations with two unknowns (i₁ and i₂)

\[ \left\{ \begin{array}{l} \;3i_{1}-i_{2}=0\\ -i_{1}+5i_{2}=14 \end{array} \right. \]

solving the first equation for i₂

\[ \begin{gather} i_{2}=3i_{1} \tag{III} \end{gather} \]

substituting this value in the second equation

\[ \begin{gather} -i_{1}+5 \times 3i_{1}=14\\[5pt] -i_{1}+15i_{1}=14\\[5pt] 14i_{1}=14\\[5pt] i_{1}=\frac{14}{14}\\[5pt] i_{1}=1\;\text{A} \end{gather} \]

Substituting this value into expression (III)

\[ \begin{gather} i_{2}=3 \times 1\\[5pt] i_{2}=3\;\text{A} \end{gather} \]

In the branch BE a current i₃ is given by

\[ \begin{gather} i_{3}=i_{2}-i_{1}\\[5pt] i_{3}=3-1\\[5pt] i_{3}=2\;\text{A} \end{gather} \]

The direction of the current i₃ will be the same as the current i₂ (the highest value).
As the values of the currents are all positive, this indicates that the directions chosen in Figure 1 are correct. The values of the currents are i₁=1 A, i₂=2 A, and i₃=3 A, and their directions are shown in Figure 4.

Figure 4

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .