PlanetPhysics/Capacitor Networks

Capacitors in networks cannot always be grouped into simple series or parallel combinations. As an example, the figure shows three capacitors $C_{x}$, $C_{y}$, and $C_{z}$ in a delta network , so called because of its triangular shape. This network has three terminals $a$, $b$, and $c$ and hence cannot be transformed into a sinle equivalent capacitor. \begin{figure}

\includegraphics{circuit1.eps} \caption{The delta network}

\end{figure} It can be shown that as far as any effect on the external circuit is concerned, a delta network is equivalent to what is called a Y network . The name "Y network" also refers to the shape of the network. \begin{figure}

\includegraphics{circuit2.eps} \caption{The Y network}

\end{figure} I am going to show that the transformation equations that give $C_{1}$, $C_{2}$, and $C_{3}$ in terms of $C_{x}$, $C_{y}$, and $C_{z}$ are $C_{1}=(C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x})/C_{x}$ $C_{2}=(C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x})/C_{y}$ $C_{3}=(C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x})/C_{z}$

The potential difference $V_{ac}$ must be the same in both circuits, as $V_{bc}$ must be. Also, the charge $q_{1}$ that flows from point $a$ along the wire as indicated must be the same in both circuits, as must $q_{2}$. Now, let us first work with the delta circuit. Suppose the charge flowing through $C_{z}$ is $q_{z}$ and to the right. According to Kirchoff's first rule: $q_{1}=q_{y}+q_{z}$ Lets play with the equation a little bit.. $q_{1}=C_{y}V_{ac}+C_{z}V_{ab}$ From Kirchoff's second law: $V_{ab}=V_{ac}+V_{cb}=V_{ac}-V_{bc}$ $q_{1}=C_{y}V_{ac}+C_{z}(V_{ac}-V_{bc})$ Therefore we get the equation:

$q_{1}=(C_{y}+C_{z})V_{ac}-C_{z}V_{bc}$

Similarly, we apply the rule to the right part of the circuit: $q_{2}=q_{x}-q_{z}$ $q_{2}=C_{x}V_{bc}-C_{z}(V_{ac}-V_{bc})$ We then get the second equation

$q_{2}=-C_{z}V_{ac}+(C_{x}+C_{z})V_{bc}$

Solving (1) and (2) simultaneously for $V_{ac}$ and $V_{bc}$, we get: $V_{ac}=\left({\frac {C_{x}+C_{z}}{C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x}}}\right)q_{1}+\left({\frac {C_{z}}{C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x}}}\right)q_{2}$ $V_{bc}=\left({\frac {C_{z}}{C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x}}}\right)q_{1}+\left({\frac {C_{y}+C_{z}}{C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x}}}\right)q_{2}$ Keeping these in mind, we proceed to the Y network. Let us apply Kirchoff's second law to the left part: $V_{1}+V_{3}=V_{ac}$ ${\frac {q_{1}}{C_{1}}}+{\frac {q_{3}}{C_{3}}}=V_{ac}$ From conservation of charge, $q_{3}=q_{1}+q_{2}$ Simplifying the above equation yields: $V_{ac}=\left({\frac {1}{C_{1}}}+{\frac {1}{C_{3}}}\right)q_{1}+\left({\frac {1}{C_{3}}}\right)q_{2}$ Similarly for the right part: $V_{2}+V_{3}=V_{bc}$ ${\frac {q_{2}}{C_{2}}}+{\frac {q_{3}}{C_{3}}}=V_{bc}$ $V_{bc}=\left({\frac {1}{C_{3}}}\right)q_{1}+\left({\frac {1}{C_{2}}}+{\frac {1}{C_{3}}}\right)q_{2}$ The coefficients of corresponding charges in corresponding equations must be the same for both networks. i.e. we compare the equations for $V_{ac}$ and $V_{bc}$ for both networks. Immediately by comparing the coefficient of $q_{1}$ in $V_{bc}$ we get: ${\frac {1}{C_{3}}}={\frac {C_{z}}{C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x}}}$ $C_{3}=(C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x})/C_{z}$ Now compare the coefficient of $q_{2}$: ${\frac {1}{C_{2}}}+{\frac {1}{C_{3}}}={\frac {C_{y}+C_{z}}{C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x}}}$ Substitute the expression we got for $C_{3}$, and solve for $C_{2}$ to get: $C_{2}=(C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x})/C_{y}$ Now we look at the coeffcient of $q_{1}$ in the equation for $V_{ac}$: ${\frac {1}{C_{1}}}+{\frac {1}{C_{3}}}={\frac {C_{x}+C_{z}}{C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x}}}$ Again substituting the expression for $C_{3}$ and solving for $C_{1}$ we get: $C_{1}=(C_{x}C_{y}+C_{y}C_{z}+C_{z}C_{x})/C_{x}$ We have derived the required transformation equations mentioned at the top.