Solved Problem on Kirchhoff's Laws
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In the circuit below find the currents in the branches and their 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 branch of the circuit, we arbitrarily choose a direction of the current. In the branch GHAB we have the current i1 clockwise, in the branch BC the current i2 going from B to C, in the branch CDEF the current i3 clockwise, in the branch CF the current i4 going from C to F, in the branch FG the current i5 going from F to G, and in the branch BG the current i6 going from B to G. Second for each loop of the circuit we assign a direction, also random, to traverse the loop. α-mesh (GHABG), β-mesh (BCFGB), and γ-mesh (CDEFC) all clockwise (Figure 1).

Figure 1
  • Applying Kirchhoff's First Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\sum_{n} i_{n}=0} \end{gather} \]
The current i1 flows into node B, and the currents i2 and i6 flow out
\[ \begin{gather} i_{1}=i_{2}+i_{6} \tag{I} \end{gather} \]
The current i2 reaches node C, and the currents i3 and i4 flow out
\[ \begin{gather} i_{2}=i_{3}+i_{4} \tag{II} \end{gather} \]
The currents i3 and i4 reach node F, and the current i5 flows out
\[ \begin{gather} i_{5}=i_{3}+i_{4} \tag{III} \end{gather} \]
  • Applying Kirchhoff's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\sum_{n} V_{n}=0} \end{gather} \]
For the α-mesh from point A in the chosen direction, forgetting the meshes β and γ (Figure 2)

Figure 2
\[ \begin{gather} R_{1}i_{1}-E_{2}+R_{6}i_{1}-E_{1}=0 \end{gather} \]
substituting the problem values
\[ \begin{gather} 1i_{1}-20+2i_{1}-10=0\\[5pt] 3i_{1}-30=0\\3i_{1}=30\\[5pt] i_{1}=\frac{30}{3}\\[5pt] i_{1}=10\;\text{A} \end{gather} \]
For the β-mesh from point B in the chosen direction, forgetting the meshes α and γ (Figure 3)

Figure 3
\[ \begin{gather} R_{2}i_{2}+E_{3}+R_{5}i_{5}+E_{2}=0 \end{gather} \]
substituting the values
\[ \begin{gather} 2i_{2}+10+1i_{5}+20=0\\[5pt] 2i_{2}+i_{5}+30=0 \\[5pt] 2i_{2}+i_{5}=-30 \tag{IV} \end{gather} \]
For the γ-mesh from point C in the chosen direction, forgetting the meshes α and β (Figure 4)

Figure 4
\[ \begin{gather} R_{3}i_{3}-E_{4}+R_{4}i_{3}-E_{3}=0 \end{gather} \]
substituting the values
\[ \begin{gather} 1i_{3}-20+2i_{3}-10=0\\[5pt] i_{3}+2i_{3}-30=0\\[5pt] 3i_{3}=30\\[5pt] i_{3}=\frac{30}{3}\\[5pt] i_{3}=10\;\text{A} \end{gather} \]
Substituting the values of i1 and i3 in (I), (II) and (III), we have the equations (I), (II), (III) and (IV) as a system of four equations with four unknowns (i2, i4, i5, and i6)
\[ \left\{ \begin{array}{l} \;i_{2}+i_{6}=10\\ \;i_{2}-i_{4}=10\\ \;i_{5}-i_{4}=10\\ \;2i_{2}+i_{5}=-30 \end{array} \right. \]
factoring i4 in the second equation
\[ \begin{gather} i_{4}=i_{2}-10 \tag{V} \end{gather} \]
substituting the expression (V) in the third equation
\[ \begin{gather} i_{5}-(i_{2}-10)=10\\[5pt] i_{5}-i_{2}+10=10 \\[5pt] i_{5}-i_{2}=10-10 \\[5pt] i_{5}-i_{2}=0 \\[5pt] i_{5}=i_{2} \tag{VI} \end{gather} \]
substituting the expression (VI) in the fourth equation
\[ \begin{gather} 2i_{2}+i_{2}=-30\\[5pt] 3i_{2}=-30\\[5pt] i_{2}=\frac{-{30}}{3}\\[5pt] i_{2}=-10\ \text{A} \end{gather} \]
The expression (VI) gives
\[ \begin{gather} i_{5}=-10\ \text{A} \end{gather} \]
Substituting the value of i2 in expression (V)
\[ \begin{gather} i_{4}=-10-10\\[5pt] i_{4}=-20\ \text{A} \end{gather} \]
Substituting the value of i2 in the first equation
\[ \begin{gather} -10+i_{6}=10\\[5pt] i_{6}=10+10\\[5pt] i_{6}=20\ \text{A} \end{gather} \]
Since the values of the currents i2, i4 and i5 are negative, their real directions are contrary to those chosen in Figure 1. The values of currents are i1=10 A, i2=10 A, i3=10 A, i4=20 A, i5=10 A, and i6=20 A and their directions are shown in Figure 5.

Figure 5
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