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 $$C_1 = (C_xC_y + C_yC_z + C_zC_x)/C_x$$ $$C_2 = (C_xC_y + C_yC_z + C_zC_x)/C_y$$ $$C_3 = (C_xC_y + C_yC_z + C_zC_x)/C_z$$

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: $$ q_1 = (C_y + C_z)V_{ac} - C_zV_{bc} $$ 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 $$ q_2 = -C_zV_{ac} + (C_x + C_z)V_{bc} $$ Solving (1) and (2) simultaneously for $$V_{ac}$$ and $$V_{bc}$$, we get: $$V_{ac} = \left( \frac{C_x + C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_1 + \left( \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_2$$ $$V_{bc} = \left( \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_1 + \left( \frac{C_y + C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_2$$ 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.