Solved Problem on Mesh Analysis
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In the circuit below find the currents in the branches and their directions.



Problem Data:

Resistors:
  • R1 = 2 Ω;
  • R2 = 3 Ω;
  • R3 = 2 Ω;
  • R4 = 2 Ω;
  • R5 = 3 Ω;
  • R6 = 2 Ω;
  • R7 = 3 Ω;
  • R8 = 2 Ω;
emf of batteries:
  • E1 = 5 V;
  • E2 = 5 V;
  • E3 = 4 V;
Solution

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

Figure 1

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

Figure 2
\[ \begin{gather} \bbox[#99CCFF,10px] {\sum_{n} V_{n}=0} \end{gather} \]
\[ \begin{gather} R_{2}i_{1}+R_{3}(i_{1}-i_{2})+R_{1}i_{1}-E_{1}=0 \end{gather} \]
substituting the problem values
\[ \begin{gather} 3 i_{1}+2(i_{1}-i_{2})+2 i_{1}-5=0\\[5pt] 3 i_{1}+2 i_{1}-2 i_{2}+2 i_{1}=5 \\[5pt] 7 i_{1}-2 i_{2}=5 \tag{I} \end{gather} \]
Applying Kirchhoff's Second law to mesh i2 from point B in the chosen direction, forgetting meshes i1 and i3 (Figure 3)

Figure 3
\[ \begin{gather} R_{4}i_{2}-E_{2}+R_{6}(i_{2}-i_{3})+R_{5}i_{2}+R_{3}(i_{2}-i_{1})=0 \end{gather} \]
substituting the values
\[ \begin{gather} 2 i_{2}-5+2(i_{2}-i_{3})+3 i_{2}+2(i_{2}-i_{1})=0\\[5pt] 2 i_{2}+2 i_{2}-2 i_{3}+3 i_{2}+2 i_{2}-2 i_{1}=5\\[5pt] -2 i_{1}+9 i_{2}-2 i_{3}=5 \tag{II} \end{gather} \]
Applying Kirchhoff's Second Law to mesh i3 from point C in the chosen direction, forgetting meshes i1 and i2 (Figure 4)

Figure 4
\[ \begin{gather} R_{8}i_{3}-E_{3}+R_{7}i_{3}+R_{6}(i_{3}-I_{2})+E_{2}=0 \end{gather} \]
substituting the values
\[ \begin{gather} 2 i_{3}-4+3 i_{3}+2(i_{3}-i_{2})+5=0\\[5pt] 2 i_{3}+3 i_{3}+2 i_{3}-2 i_{2}+1=0\\[5pt] -2 i_{2}+7 i_{3}=-1 \tag{III} \end{gather} \]
With equations (I), (II) and (III) we have a linear system of three equations with three unknowns ( i1, i2 and i3).
\[ \left\{ \begin{array}{l} \;\;\,7 i_{1}-2 i_{2}=5\\ -2 i_{1}+9 i_{2}-2 i_{3}=5\\ -2 i_{2}+7 i_{3}=-1 \end{array} \right. \tag{IV} \]
solving the first equation for i1, and solving the third equation for i3
\[ \begin{gather} 7 i_{1}-2 i_{2}=5\\[5pt] 7 i_{1}=5+2 i_{2}\\[5pt] i_{1}=\frac{5+2 i_{2}}{7} \tag{V-a} \end{gather} \]
\[ \begin{gather} -2 i_{2}+7 i_{3}=-1\\[5pt] 7 i_{3}=-1-2 i_{2}\\[5pt] i_{3}=\frac{-1+2 i_{2}}{7} \tag{V-b} \end{gather} \]
substituting the expressions (V-a) and (V-b) in the second equation of system (IV)
\[ \begin{gather} -2\times \left(\frac{5+2 i_{2}}{7}\right)+9 i_{2}-2\times \left(\frac{-1+2 i_{2}}{7}\right)=5\\[5pt] \frac{-10-4 i_{2}}{7}+9 i_{2}+\frac{2-4 i_{2}}{7}=5 \end{gather} \]
multiplying the numerator and denominator of the second term on the left-hac side and the right-hand side of the equation by 7
\[ \begin{gather} \frac{-10-4i_{2}}{7}+\frac{7}{7} \times 9i_{2}+\frac{2-4i_{2}}{7}=5 \times \frac{7}{7}\\[5pt] \frac{-10-4i_{2}}{7}+\frac{63}{7}i_{2}+\frac{2-4i_{2}}{7}=\frac{35}{7}\\[5pt] \frac{-10-4i_{2}+63i_{2}+2-4i_{2}}{\cancel{7}}=\frac{35}{\cancel{7}} \end{gather} \]
canceling factor 7 on both sides of the equation
\[ \begin{gather} -10-4i_{2}+63i_{2}+2-4i_{2}=35\\[5pt] -8+55i_{2}=35\\[5pt] 55i_{2}=35+8\\[5pt] i_{2}=\frac{43}{55}\\[5pt] i_{2}=0.78\ \text{A} \end{gather} \]
Substituting this result in the expressions given in (V-a) and (V-b) we have the currents i1 and i3

Note: Instead of substituting the current in the decimal form, we will substitute the by the fraction to minimize rounding errors.

for the expression (V-a) we have the current i1
\[ \begin{gather} i_{1}=\frac{5+2 \times \dfrac{43}{55}}{7}\\[5pt] i_{1}=\left(5+2 \times \frac{43}{55}\;\right) \times \frac{1}{7} \end{gather} \]
multiplying the numerator and denominator of the first term in parentheses by 55
\[ \begin{gather} i_{1}=\left(5 \times \frac{55}{55}+2 \times \frac{43}{55}\;\right) \times \frac{1}{7}\\[5pt] i_{1}=\left(\frac{275}{55}+\frac{86}{55}\;\right) \times \frac{1}{7}\\[5pt] i_{1}=\frac{361}{55} \times \frac{1}{7}\\[5pt] i_{1}=\frac{361}{385}\\[5pt] i_{1}=0,94\ \text{A} \end{gather} \]
for the expression (V-b) we have the current i3
\[ \begin{gather} i_{3}=\frac{-1+2 \times \dfrac{43}{55}}{7}\\[5pt] i_{3}=\left(-1+2 \times \frac{43}{55}\right) \times \frac{1}{7} \end{gather} \]
multiplying the numerator and denominator of the first term in parentheses by 55
\[ \begin{gather} i_{3}=\left(-1 \times \frac{55}{55}+2 \times \frac{43}{55}\;\right) \times \frac{1}{7}\\[5pt] i_{3}=\left(\frac{-{55}}{55}+\frac{86}{55}\;\right) \times \frac{1}{7}\\[5pt] i_{3}=\frac{31}{55} \times \frac{1}{7}\\[5pt] i_{3}=\frac{31}{385}\\[5pt] i_{3}=0.08\ \text{A} \end{gather} \]
In the branch BG, we have the current i4 given by
\[ \begin{gather} i_{4}=i_{1}-i_{2}\\[5pt] i_{4}=\frac{361}{385}-\frac{43}{55} \end{gather} \]
multiplying the numerator and denominator of the second term on the right-hand side of the expression by 7
\[ \begin{gather} i_{4}=\frac{361}{385}-\frac{43}{55}\times\frac{7}{7}\\[5pt] i_{4}=\frac{361}{385}-\frac{301}{385}\\[5pt] i_{4}=\frac{361-301}{385}\\[5pt] i_{4}=\frac{60}{385} \end{gather} \]
dividing the numerator and denominator by 5
\[ \begin{gather} i_{4}=\frac{60:5}{385:5}\\[5pt] i_{4}=\frac{12}{77}\\[5pt] i_{4}=0.16\ \text{A} \end{gather} \]
The direction of the current i4 will be the same as the current i1 (highest value).
In the CF branch, we have the current i5 given by
\[ \begin{gather} i_{5}=i_{2}-i_{3}\\[5pt] i_{5}=\frac{43}{55}-\frac{31}{385} \end{gather} \]
multiplying the numerator and denominator of the first term on the right-hand side of the expression by 7
\[ \begin{gather} i_{5}=\frac{43}{55}\times\frac{7}{7}-\frac{31}{385}\\[5pt] i_{5}=\frac{301}{385}-\frac{31}{385}\\[5pt] i_{5}=\frac{301-31}{385}\\[5pt] i_{5}=\frac{270}{385} \end{gather} \]
dividing the numerator and denominator by 5
\[ \begin{gather} i_{5}=\frac{270:5}{385:5}\\[5pt] i_{5}=\frac{54}{77}\\[5pt] i_{5}=0.70\ \text{A} \end{gather} \]
The direction of the current i5 will be the same as the current i2 (highest value).

The currents values are i1=0.99 A, i2=0.78 A, i3=0.08 A, i4=0.16 A, and i5=0.70 A, and their directions are shown in the Figure 5.

Figure 5
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .