Solved Problem on Mesh Analysis

Two batteries whose emf and internal resistances are respectively E₁=1.5 V, E₂=9 V, and r₁=1 Ω, r₂=2.2 Ω are connected by wires of negligible resistance to a resistor R=4.7 kΩ. Find the currents in the branches of the circuit.

Problem data:

Batteries internal resistances:

r₁ = 1 Ω;
r₂ = 2.2 Ω;

External resistance:

R = 4.7 kΩ = 4700 Ω;

emf of the batteries:

E₁ = 1.5 V;
E₂ = 9 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 counterclockwise (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)

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

\[ \begin{gather} R(i_{1}+i_{2})+r_{1}i_{1}-E_{1}=0 \end{gather} \]

Figure 2

substituting the values of the problem

\[ \begin{gather} 4700(i_{1}+i_{2})+1i_{1}-1.5=0\\[5pt] 4700i_{1}+4700i_{2}+i_{1}=1.5 \\[5pt] {4701i_{1}+4700i_{2}=1.5 \tag{I}} \end{gather} \]

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

\[ \begin{gather} R(i_{1}+i_{2})+r_{2}i_{2}-E_{2}=0 \end{gather} \]

Figure 3

substituting the values of the problem

\[ \begin{gather} 4700(i_{1}+i_{2})+2.2i_{2}-9=0\\[5pt] 4700i_{1}+4700i_{2}+2.2i_{2}=9\\[5pt] {4700i_{1}+4702.2i_{2}=9 \tag{II}} \end{gather} \]

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

\[ \left\{ \begin{array}{l} \;4701i_{1}+4700i_{2}=1.5\\ \;4700i_{1}+4702.2i_{2}=9 \end{array} \right. \]

solving the first equation for i₂

\[ \begin{gather} i_{2}=\frac{1.5-4701i_{1}}{4700} \tag{III} \end{gather} \]

substituting this value in the second equation

\[ \begin{gather} 4700i_{1}+4702.2 \times \left(\;\frac{1.5-4701i_{1}}{4700}\;\right)=9 \end{gather} \]

multiplying both sides of the equation by 4700

\[ \begin{gather} 4700 \times 4700i_{1}+4700 \times 4702.2 \times \left(\;\frac{1.5-4701i_{1}}{4700}\;\right)=4700 \times 9\\[5pt] 22090000i_{1}+\cancel{{4700}} \times 4702.2 \times \left(\;\frac{1.5-4701i_{1}}{\cancel{{4700}}}\;\right)=4700 \times 9\\[5pt] 22090000i_{1}+4702.2 \times 1.5-4702.2 \times 4701i_{1}=42300\\[5pt] 22090000i_{1}+7053.3-2210542.2i_{1}=42300\\[5pt] -15042.2i_{1}=42300-7053.3\\[5pt] -15042.2i_{1}=35246.7\\[5pt] i_{1}=\frac{35246.7}{-15042.2}\\[5pt] i_{1}=-2.34319\;\text{A} \end{gather} \]

Substituting this value into expression (III)

\[ \begin{gather} i_{2}=\frac{1.5-4701 \times (\;-2.34319\;)}{4700}\\[5pt] i_{2}=\frac{1.5+11015.33619}{4700}\\[5pt] i_{2}=\frac{11016.83619}{4700}\\[5pt] i_{2}=2.34401\;\text{A} \end{gather} \]

In the branch BE the current i₃ is

\[ \begin{gather} i_{3}=i_{1}+i_{2}\\[5pt] i_{3}=-2.34319+2.34401\\[5pt] i_{3}=0.00082=0.82.10^{-3}=0.82\;\text{mA} \end{gather} \]

The direction of the current i₃ will be the same as the current i₂ (the highest value).
Since the value of current i₂ is negative, this means that its real direction is opposite to that chosen in Figure 1. The current values are i₁=2.3432 A, i₂=2.3440 A, and i₃=0.82 mA, 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 .