Solved Problem on Kirchhoff's Laws

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, to each branch of the circuit, we arbitrarily choose a direction of the current. In the EFAB branch we have the current i₁ clockwise, in the branch BE the current i₃ from B to E and in the branch EDCB the current i₂ in the counterclockwise direction. Second, for each loop of the circuit, we assign a direction, also arbitrarily, to traverse the mesh. The α-mesh (ABEFA) clockwise and β-mesh (BCDEB) also clockwise. We see all these elements in Figure 1.

Figure 1

Using Kirchhoff's First Law

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

The currents i₁ and i₂ flow into node B and the current i₃ flows out

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

Using 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 β-mesh (Figure 2)

\[ \begin{gather} R i_{3}+r_{1} i_{1}-E_{1}=0 \tag{II} \end{gather} \]

Figure 2

substituting the values of the problem

\[ \begin{gather} 4700i_{3}+1i_{1}-1.5=0\\[5pt] 4700i_{3}+i_{1}=1.5 \tag{III} \end{gather} \]

For the β-mesh from point B, in the chosen direction, forgetting the α-mesh, (Figure 3)

Figure 3

\[ \begin{gather} E_{2}-r_{2}i_{2}-Ri_{3}=0 \tag{IV} \end{gather} \]

substituting the values

\[ \begin{gather} 9-2.2i_{2}-4700i_{3}=0\\[5pt] 2.2i_{2}+4700i_{3}=9 \tag{V} \end{gather} \]

Equations (I), (III) and (V) can be written as a system of linear equations with three variables (i₁, i₂, and i₃)

\[ \left\{ \begin{array}{l} \;i_{3}=i_{1}+i_{2}\\ \;4700i_{3}+i_{1}=1.5\\ \;2.2i_{2}+4700i_{3}=9 \end{array} \right. \]

solving the second equation for i₁

\[ \begin{gather} i_{1}=1.5-4700i_{3} \tag{VI} \end{gather} \]

solving the third equation for i₂

\[ \begin{gather} i_{2}=\frac{9-4700i_{3}}{2.2} \tag{VII} \end{gather} \]

substituting expressions (VI) and (VII) in the first equation

\[ \begin{gather} i_{3}=1.5-4700i_{3}+\frac{9-4700i_{3}}{2.2} \end{gather} \]

multiplying both sides of equality by 2.2

\[ \begin{gather} 2.2i_{3}=2.2 \times \left(\;1.5-4700i_{3}+\frac{9-4700i_{3}}{2.2}\;\right)\\[5pt] 2.2i_{3}=2.2 \times 1.5-2.2 \times 4700i_{3}+{\cancel{2.2}} \times \frac{9-4700i_{3}}{\cancel{2.2}}\\[5pt] 2.2i_{3}=3.3-10340i_{3}+9-4700i_{3}\\[5pt] 2.2i_{3}=12.3-15040i_{3}\\[5pt] 2.2i_{3}+15040i_{3}=12.3\\[5pt] 15042.2i_{3}=12.3\\[5pt] i_{3}=\frac{12.3}{15042.2}\\[5pt] i_{3}=8.1770 \times 10^{-4}=0.81770 \times 10^{-3}\simeq 0.82\;\text{mA} \end{gather} \]

substituting the value found above in expressions (VI) and (VII) we find i₁ and i₂ respectively

\[ \begin{gather} i_{1}=1.5-4700 \times 0.00081770\\[5pt] i_{1}=1.5-3.8432\\[5pt] i_{1}=-2.3432\;\text{A} \end{gather} \]

\[ \begin{gather} i_{2}=\frac{9-4700 \times 0.00081770}{2.2}\\[5pt] i_{2}=\frac{9-3.8432}{2.2}\\[5pt] i_{2}=\frac{5.1568}{2.2}\\[5pt] i_{2}=2.3440\;\text{A} \end{gather} \]

Since the value of current i₁ is negative, this means that its true direction is opposite to that chosen in Figure 1. The currents values are i₁=2.3432 A, i₂=2.3440 A, and i₃=0.82 mA, and its 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 .