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.

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.



Contributors to this entry (in most recent order):

As of this snapshot date, this entry was owned by curious.