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).
This type of circuit is solved using the technique called YΔ transform (or star-delta transform or
Kennelly's Theorem), replacing the capacitors in the circuit (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 is as follows (Figure 3).
The two capacitors between points D and E, with values 8C and C in Figure 3, are
connected in series, the equivalent capacitor C4 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, with values 4C and 2C in Figure 3, are
connected in series. Applying equation (IV), the equivalent capacitor C5 between them is given by
\[
\begin{gather}
C_5=\frac{C_c\,2C}{C_c+2C} \\[5pt]
C_5=\frac{4 C\,2C}{4C+2C} \\[5pt]
C_5=\frac{\cancelto{4}8C^{\cancel{2}}}{\cancelto{3}6\cancel{C}} \\[5pt]
C_5=\frac{4}{3}C
\end{gather}
\]
The circuit can be represented as (Figure 4).
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.
\[
\begin{gather}
C_6=\frac{8}{9}C+\frac{3}{3}\times\frac{4}{3}C \\[5pt]
C_6=\frac{8}{9}C+\frac{12}{9}C \\[5pt]
C_6=\frac{20}{9}C
\end{gather}
\]
The circuit becomes two series capacitors (Figure 5).
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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