Solved Problem on Kirchhoff's Laws

The circuit below is under steady-state conditions, find:
a) The current in the circuit;
b) The potential difference in the capacitor;
c) The electric charge stored in the capacitor;
d) The electric potential energy in the capacitor.

Problem data:

Resistors

R₁ = 1 Ω;
R₂ = 3 Ω;
R₃ = 5 Ω;

Capacitor

C = 2 µ C;

Batteries

E₁ = 12 V;
E₂ = 2 V;
E₃ = 4 V;

Solution

a) Under the steady-state conditions, the current in the branch BE is zero (i_BE=0 - Figure 1).

Figure 1

We can disregard the BE branch, we get a single loop circuit (ACDFA).
First, we arbitrarily assign a clockwise direction to the current. Second, we assign a clockwise direction to traverse the mesh (Figure 2).

Figure 2

Applying Kirchhoff's Second Law

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

From point A in the chosen direction

\[ \begin{gather} R_{2}i+E_{3}+R_{3}i-E_{1}+R_{1}i=0\\[5pt] R_{2}i+R_{3}i+R_{1}i=E_{1}-E_{3} \end{gather} \]

factoring the current i on the left-hand side of the equation

\[ \begin{gather} i(\;R_{2}+R_{3}+R_{1}\;)=E_{1}-E_{3}\\[5pt] i=\frac{E_{1}-E_{3}}{R_{2}+R_{3}+R_{1}} \end{gather} \]

substituting the values, we have

\[ \begin{gather} i=\frac{12-4}{3+5+8}\\[5pt] i=\frac{8}{16} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {i=0.5\;\text{A}} \end{gather} \]

Since the current is positive, the direction of the current chosen in Figure 2 is correct.

b) To find the potential difference in the capacitor, we use the complete circuit again (Figure 3).

Figure 3

We use ABEFA mesh, the current with the direction determined in item (a), and traversing the mesh in the same direction as the current.

Applying Kirchhoff's Second Law

From point A, we have

\[ \begin{gather} E_{2}+\Delta V_{\text{c}}-E_{1}+R_{1}i=0\\[5pt] \Delta V_{\text{c}}=-E_{2}+E_{1}-R_{1}i \end{gather} \]

substituting the values and the current found in the previous item

\[ \begin{gather} \Delta V_{\text{c}}=-2+12-8 \times 0.5\\[5pt] \Delta V_{\text{c}}=10-4 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta V_{\text{c}}=6\;\text{V}} \end{gather} \]

Note: We could have done the calculation using the BCDEB mesh, but the ABEFA mesh was used because it had fewer elements for the calculation.
If we used the BCDEB mesh, we would have from point B

\[ \begin{gather} R_{2}i+E_{3}+R_{3}i+\Delta V_{\text{c}}-E_{2}=0\\[5pt] \Delta V_{\text{c}}=-R_{2}i-E_{3}-R_{3}i+E_{2}\\[5pt] \Delta V_{\text{c}}=-3 \times 0.5-4-5 \times 0.5+2\\[5pt] \Delta V_{\text{c}}=-1.5-4-2.5+2\\[5pt] \Delta V_{\text{c}}=-1.5-4-2.5+2\\[5pt] \Delta V_{\text{c}}=-6\text{V} \end{gather} \]

the negative sign signifies that we traversed the capacitor in the opposite direction of the voltage drop.

c) The charge stored in the capacitor is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {Q=C \Delta V} \end{gather} \]

substituting the value of the capacitor given in the problem, and the potencial difference ΔV_c found in item (b)

\[ \begin{gather} Q=2 \times 10^{-6} \times 6\\[5pt] Q=12 \times 10^{-6} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {Q=12\;\mu \text{C}} \end{gather} \]

d) The potential electric energy stored in a capacitor is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {U=\frac{Q \Delta V}{2}} \end{gather} \]

substituting the value of the potential difference of item (b) and the charge found in item (c)

\[ \begin{gather} U=\frac{12 \times 10^{-6} \times 6}{2}\\[5pt] U=\frac{72 \times 10^{-6}}{2}\\[5pt] U=36 \times 10^{-6} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {U=36\;\mu \text{J}} \end{gather} \]

Note: Other formulas could be used to calculate potential electric energy

\[ \begin{gather} U=\frac{Q \Delta V}{2}=\frac{C(\Delta V)^{2}}{2}=\frac{Q^{2}}{2C} \end{gather} \]

all quantities that appear in these formulas are known in the problem

\[ \begin{gather} Q=12 \times 10^{-6}\;\text{C};\ \ C=2 \times 10^{-6}\;\text{F};\ \ \Delta V=U=6\;\text{V} \end{gather} \]

applying any of these formulas would lead to the same solution.

\[ \begin{gather} U=\frac{12 \times 10^{-6} \times 6}{2}=\frac{2 \times 10^{-6} \times 6^{2}}{2}=\frac{(\;12 \times 10^{-6}\;)^{2}}{2 \times 2 \times 10^{-6}}\\[5pt] U=\frac{72 \times 10^{-6}}{2}=\frac{2 \times 10^{-6} \times 36}{2}=\frac{144 \times 10^{-12}}{4 \times 10^{-6}}\\[5pt] U=36 \times 10^{-6}\;\text{J} \end{gather} \]

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