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capacitor networks (Topic)

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.

Figure: The delta network
\includegraphics{circuit1.eps}
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.
Figure: The Y network
\includegraphics{circuit2.eps}
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

$\displaystyle C_1 = (C_xC_y + C_yC_z + C_zC_x)/C_x$

$\displaystyle C_2 = (C_xC_y + C_yC_z + C_zC_x)/C_y$

$\displaystyle C_3 = (C_xC_y + C_yC_z + C_zC_x)/C_z$

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:

$\displaystyle q_1 = q_y + q_z$
Lets play with the equation a little bit..

$\displaystyle q_1 = C_yV_{ac} + C_zV_{ab}$
From Kirchoff's second law: $V_{ab} = V_{ac} + V_{cb} = V_{ac} - V_{bc}$

$\displaystyle q_1 = C_yV_{ac} + C_z(V_{ac} - V_{bc})$
Therefore we get the equation:
$\displaystyle q_1 = (C_y + C_z)V_{ac} - C_zV_{bc}$ (1)
Similarly, we apply the rule to the right part of the circuit:

$\displaystyle q_2 = q_x - q_z$

$\displaystyle q_2 = C_xV_{bc} - C_z(V_{ac} - V_{bc})$
We then get the second equation
$\displaystyle q_2 = -C_zV_{ac} + (C_x + C_z)V_{bc}$ (2)
Solving (1) and (2) simultaneously for $V_{ac}$ and $V_{bc}$, we get:

$\displaystyle V_{ac} = \left( \frac{C_x + C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_1 + \left( \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_2$

$\displaystyle V_{bc} = \left( \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_1 + \left( \frac{C_y + C_z}{C_xC_y + C_yC_z + C_zC_x}\right)q_2$
Keeping these in mind, we proceed to the Y network. Let us apply Kirchoff's second law to the left part:

$\displaystyle V_1 + V_3 = V_{ac}$

$\displaystyle \frac{q_1}{C_1} + \frac{q_3}{C_3} = V_{ac}$
From conservation of charge, $q_3 = q_1 + q_2$ Simplifying the above equation yields:

$\displaystyle V_{ac} = \left( \frac{1}{C_1} + \frac{1}{C_3}\right)q_1 + \left(\frac{1}{C_3}\right)q_2$
Similarly for the right part:

$\displaystyle V_2 + V_3 = V_{bc}$

$\displaystyle \frac{q_2}{C_2} + \frac{q_3}{C_3} = V_{bc}$

$\displaystyle V_{bc} = \left( \frac{1}{C_3}\right)q_1 + \left(\frac{1}{C_2} + \frac{1}{C_3}\right)q_2$
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:

$\displaystyle \frac{1}{C_3} = \frac{C_z}{C_xC_y + C_yC_z + C_zC_x}$

$\displaystyle C_3 = (C_xC_y + C_yC_z + C_zC_x)/C_z$
Now compare the coefficient of $q_2$:

$\displaystyle \frac{1}{C_2} + \frac{1}{C_3} = \frac{C_y + C_z}{C_xC_y + C_yC_z + C_zC_x}$
Substitute the expression we got for $C_3$, and solve for $C_2$ to get:

$\displaystyle C_2 = (C_xC_y + C_yC_z + C_zC_x)/C_y$
Now we look at the coeffcient of $q_1$ in the equation for $V_{ac}$:

$\displaystyle \frac{1}{C_1} + \frac{1}{C_3} = \frac{C_x + C_z}{C_xC_y + C_yC_z + C_zC_x}$
Again substituting the expression for $C_3$ and solving for $C_1$ we get:

$\displaystyle C_1 = (C_xC_y + C_yC_z + C_zC_x)/C_x$
We have derived the required transformation equations mentioned at the top.



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Cross-references: work, charge

This is version 4 of capacitor networks, born on 2009-06-06, modified 2009-06-09.
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Physics Classification: 07.50.Ek (Circuits and circuit components )
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