Solved Problem on Mesh Analysis

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

Problem Data:

Resistors:

R₁ = 1 Ω;
R₂ = 2 Ω;
R₃ = 1 Ω;
R₄ = 2 Ω;
R₅ = 1 Ω;
R₆ = 2 Ω;

Batteries:

E₁ = 10 V;
E₂ = 20 V;
E₃ = 10 V;
E₄ = 20 V;

Solution

First, to each loop of the circuit is randomly assigned a direction of current. The meshes ABGHA, BCFGB and CDEFC have, respectively, the currents i₁, i₂ and i₃ in the clockwise direction (Figure 1).

Figure 1

Applying Kirchhoff's Mesh Law to mesh i₁ from point A in the chosen direction, forgetting meshes i₂ and i₃ (Figure 2), we write

Figure 2

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

\[ \begin{gather} R_{1}i_{1}-E_{2}+R_{6}i_{1}-E_{1}=0 \end{gather} \]

substituting the values given in the problem

\[ \begin{gather} 1i_{1}-20+2i_{1}-10=0\\[5pt] 3i_{1}-30=0\\[5pt] 3i_{1}=30\\[5pt] i_{1}=\frac{30}{3}\\[5pt] i_{1}=10\;\text{A} \end{gather} \]

Applying Kirchhoff's Mesh Law to mesh i₂ from point B in the chosen direction, forgetting meshes i₁ and i₃ (Figure 3)

Figure 3

\[ \begin{gather} R_{2}i_{2}+E_{3}+R_{5}i_{2}+E_{2}=0 \end{gather} \]

substituting the values

\[ \begin{gather} 2i_{2}+10+1i_{2}+20=0\\[5pt] 3i_{2}+30=0\\[5pt] 3i_{2}=-30\\[5pt] i_{2}=\frac{-30}{3}\\[5pt] i_{2}=-10\;\text{A} \end{gather} \]

Applying Kirchhoff's Mesh Law to mesh i₃ from point C in the chosen direction, forgetting meshes i₁ and i₂ (Figure 4)

Figure 4

\[ \begin{gather} R_{3}i_{3}-E_{4}+R_{4}i_{3}-E_{3}=0 \end{gather} \]

substituting the values

\[ \begin{gather} 1i_{3}-20+2i_{3}-10=0\\[5pt] i_{3}+2i_{3}-30=0\\[5pt] 3i_{3}=30\\[5pt] i_{3}=\frac{30}{3}\\[5pt] i_{3}=10\;\text{A} \end{gather} \]

In branch BG will circulate the current i₄ given by

\[ \begin{gather} i_{4}=i_{1}-i_{2}\\[5pt] i_{4}=10-(\;-10\;)\\[5pt] i_{4}=10+10\\[5pt] i_{4}=20\;\text{A} \end{gather} \]

The direction of the current i₄ will be the same as current i₁.
In the CF branch will circulate the current i₅ given by

\[ \begin{gather} i_{5}=i_{3}-i_{2}\\[5pt] i_{5}=10-(\;-10\;)\\[5pt] i_{5}=10+10\\[5pt] i_{5}=20\;\text{A} \end{gather} \]

The direction of current i₅ will be the same as current i₃.
Since the current i₂ is negative, its direction is contrary to the one chosen in Figure 1. The current values are i₁=10 A, i₂=10 A, i₃=10 A, i₄=20 A, and i₅=20 A, and its directions are shown in Figure 5.

Figure 5

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