Solved Problem on Capacitors
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Find the equivalent capacitance between points A and B of the circuit represented in the figure.


Solution

Let's redesign the circuit as follows to facilitate viewing (Figure 1)

Figure 1

This type of circuit is solved using the technique called YΔ transform (or star-delta transform), replacing the capacitors in the circuit (Figure 2)

Figure 2

Ca capacitor will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {C_a=\frac{C_1 C_2+C_1 C_3+C_2 C_3}{C_3}} \end{gather} \]
\[ \begin{gather} C_a=\frac{2C\,C+2C\,2C+C\,2C}{2C}\\[5pt] C_a=\frac{2C^2+4C^2+2C^2}{2C}\\[5pt] C_a=\frac{\cancelto{4}{8}C^{\cancel 2}}{\cancel 2\cancel C}\\[5pt] C_a=4C \tag{I} \end{gather} \]
Cb capacitor will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {C_b=\frac{C_1 C_2+C_1 C_3+C_2 C_3}{C_2}} \end{gather} \]
\[ \begin{gather} C_b=\frac{2C\,C+2C\,2C+C\,2C}{C}\\[5pt] C_b=\frac{2C^2+4C^2+2C^2}{C}\\[5pt] C_b=\frac{8C^{\cancel 2}}{\cancel C}\\[5pt] C_b=8C \tag{II} \end{gather} \]
Cc capacitor will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {C_c=\frac{C_1 C_2+C_1 C_3+C_2 C_3}{C_1}} \end{gather} \]
\[ \begin{gather} C_c=\frac{2C\,C+2C\,2C+C\,2C}{2C}\\[5pt] C_c=\frac{2C^2+4C^2+2C^2}{2C}\\[5pt] C_c=\frac{\cancelto{4}{8}C^{\cancel 2}}{\cancel 2\cancel C}\\[5pt] C_c=4C \tag{III} \end{gather} \]
Then using the values (I), (II), and (III), the circuit to be solved becomes the following (Figure 3)

Figure 3

The two capacitors between points D and E, Cb and C, are in series, the C4 equivalent capacitance is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {C_{eq}=\frac{C_A C_B}{C_A+C_B}} \tag{IV} \end{gather} \]
\[ \begin{gather} C_4=\frac{C_b\,C}{C_b+C}\\[5pt] C_4=\frac{8C\,C}{8C+C}\\[5pt] C_4=\frac{8C^{\cancel 2}}{9\cancel C}\\[5pt] C_4=\frac{8}{9}C \end{gather} \]
The two capacitors between points D and F, Cc and 2C are in series, applying the expression (IV) the equivalent capacitance C5 between them will be
\[ \begin{gather} C_5=\frac{C_c\,2C}{C_c+2C}\\[5pt] C_5=\frac{4 C\,2C}{4C+2C}\\[5pt] C_5=\frac{\cancelto{4}8C^{\cancel{2}}}{\cancelto{3}6\cancel{C}}\\[5pt] C_5=\frac{4}{3}C \end{gather} \]
The circuit can be represented as (Figure 4)

Figure 4

The two capacitors obtained above are connected in parallel, the equivalent capacitance is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {C_{eq}=\sum_{i=1}^{n}C_{i}} \end{gather} \]
the equivalent capacitance C6 will be
\[ \begin{gather} C_6=\frac{8}{9}C+\frac{4}{3}C \end{gather} \]
multiplying the numerator and denominator of the second term on the right-hand side by 3, we obtain
\[ \begin{gather} C_6=\frac{8}{9}C+\frac{3}{3}\times\frac{4}{3}C\\[5pt] C_6=\frac{8}{9}C+\frac{12}{9}C\\[5pt] C_6=\frac{20}{9}C \end{gather} \]
The circuit becomes two series capacitors (Figure 5)

Figure 5

The equivalent capacitance Ceq of the circuit
\[ \begin{gather} C_{eq}=\frac{4C\times\dfrac{20}{9}C}{4C+\dfrac{20}{9}C} \end{gather} \]
In the denominator, we multiply the numerator and the denominator of the first term by 9
\[ \begin{gather} C_{eq}=\frac{4 C\times\dfrac{20}{9}C}{\dfrac{9}{9}\times 4C+\dfrac{20}{9}C}\\[5pt] C_{eq}=\frac{\dfrac{80}{9}C^2 }{\dfrac{36}{9}C+\dfrac{20}{9}C}\\[5pt] C_{eq}=\frac{\dfrac{80}{\cancel 9}C^{\cancel 2}}{\dfrac{56}{\cancel 9}\cancel C}\\[5pt] C_{eq}=\frac{\cancelto{10}{80}}{\cancelto{7}{56}}C \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {C_{eq}=\frac{10}{7}C} \end{gather} \]
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