CapacitorNetworks 2009-06-06 09:44:43 2009-06-06 09:46:03 Topic % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 {\em 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} \begin{center} \includegraphics{circuit1.eps} \caption{The delta network} \end{center} \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 {\em Y network}. The name "Y network" also refers to the shape of the network. \begin{figure} \begin{center} \includegraphics{circuit2.eps} \caption{The Y network} \end{center} \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 $$C_1 = (C_xC_y + C_yC_z + C_zC_x)/C_x$$ $$C_2 = (C_xC_y + C_yC_z + C_zC_x)/C_y$$ $$C_3 = (C_xC_y + C_yC_z + C_zC_x)/C_z$$ 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.. $$q_1 = C_yV_{ac} + C_zV_{ab}$$ From Kirchoff's second law: $V_{ab} = V_{ac} + V_{cb} = V_{ac} - V_{bc}$ $$q_1 = C_yV_{ac} + C_z(V_{ac} - V_{bc})$$ Therefore we get the equation: \begin{equation} q_1 = (C_y + C_z)V_{ac} - C_zV_{bc} \end{equation} Similarly, we apply the rule to the right part of the circuit: $$q_2 = q_x - q_z$$ $$q_2 = C_xV_{bc} - C_z(V_{ac} - V_{bc})$$ We then get the second equation \begin{equation} q_2 = -C_zV_{ac} + (C_x + C_z)V_{bc} \end{equation} Solving (1) and (2) simultaneously for $V_{ac}$ and $V_{bc}$, we get: $$V_{ac} = \left( \frac{C_x + C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_1 + \left( \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_2$$ $$V_{bc} = \left( \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_1 + \left( \frac{C_y + C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_2$$ 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: $$\frac{1}{C_3} = \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}$$ $$C_3 = (C_xC_y + C_yC_z + C_zC_x)/C_z$$ Now compare the coefficient of $q_2$: $$\frac{1}{C_2} + \frac{1}{C_3} = \frac{C_y + C_z}{C_xC_y + C_yC_z + C_zC_x}$$ Substitute the expression we got for $C_3$, and solve for $C_2$ to get: $$C_2 = (C_xC_y + C_yC_z + C_zC_x)/C_y$$ Now we look at the coeffcient of $q_1$ in the equation for $V_{ac}$: $$\frac{1}{C_1} + \frac{1}{C_3} = \frac{C_x + C_z}{C_xC_y + C_yC_z + C_zC_x}$$ Again substituting the expression for $C_3$ and solving for $C_1$ we get: $$C_1 = (C_xC_y + C_yC_z + C_zC_x)/C_x$$ We have derived the required transformation equations mentioned at the top.