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 $V_{a c}$ must be the same in both circuits, as $V_{b c}$ must be. Also, the [[../Charge/|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/|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.. $q_{1} = C_{y} V_{a c} + C_{z} V_{a b}$ From Kirchoff's second law: $V_{a b} = V_{a c} + V_{c b} = V_{a c} - V_{b c}$ $q_{1} = C_{y} V_{a c} + C_{z} (V_{a c} - V_{b c})$ Therefore we get the equation:

$q_{1} = (C_{y} + C_{z}) V_{a c} - C_{z} V_{b c}$

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

$q_{2} = - C_{z} V_{a c} + (C_{x} + C_{z}) V_{b c}$

Solving (1) and (2) simultaneously for $V_{a c}$ and $V_{b c}$ , we get: $V_{a c} = (\frac{C_{x} + C_{z}}{C_{x} C_{y} + C_{y} C_{z} + C_{z} C_{x}}) q_{1} + (\frac{C_{z}}{C_{x} C_{y} + C_{y} C_{z} + C_{z} C_{x}}) q_{2}$ $V_{b c} = (\frac{C_{z}}{C_{x} C_{y} + C_{y} C_{z} + C_{z} C_{x}}) q_{1} + (\frac{C_{y} + C_{z}}{C_{x} C_{y} + C_{y} C_{z} + C_{z} C_{x}}) q_{2}$ Keeping these in mind, we proceed to the Y network. Let us apply Kirchoff's second law to the left part: $V_{1} + V_{3} = V_{a c}$ $\frac{q_{1}}{C_{1}} + \frac{q_{3}}{C_{3}} = V_{a c}$ From conservation of charge, $q_{3} = q_{1} + q_{2}$ Simplifying the above equation yields: $V_{a c} = (\frac{1}{C_{1}} + \frac{1}{C_{3}}) q_{1} + (\frac{1}{C_{3}}) q_{2}$ Similarly for the right part: $V_{2} + V_{3} = V_{b c}$ $\frac{q_{2}}{C_{2}} + \frac{q_{3}}{C_{3}} = V_{b c}$ $V_{b c} = (\frac{1}{C_{3}}) q_{1} + (\frac{1}{C_{2}} + \frac{1}{C_{3}}) q_{2}$ The coefficients of corresponding charges in corresponding equations must be the same for both networks. i.e. we compare the equations for $V_{a c}$ and $V_{b c}$ for both networks. Immediately by comparing the coefficient of $q_{1}$ in $V_{b c}$ 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 $V_{a c}$ : $\frac{1}{C_{1}} + \frac{1}{C_{3}} = \frac{C_{x} + C_{z}}{C_{x} C_{y} + C_{y} C_{z} + C_{z} C_{x}}$ 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.

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PlanetPhysics/Capacitor Networks

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