capacitor networks | Physics Library

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.

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.

I am going to show that the transformation equations that give C₁, C₂, and C₃ in terms of C_x, C_y, and C_z are

C = (C C + C C + C C )∕C 1 x y y z z x x

C2 = (CxCy + CyCz + CzCx )∕Cy

C3 = (CxCy + CyCz + CzCx )∕Cz

The potential difference V _ac must be the same in both circuits, as V _bc must be. Also, the charge q₁ that flows from point a along the wire as indicated must be the same in both circuits, as must q₂. 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:

q1 = qy + qz

Lets play with the equation a little bit..

q1 = CyVac + CzVab

From Kirchoff’s second law: V _ab = V _ac + V _cb = V _ac − V _bc

q1 = CyVac + Cz(Vac − Vbc)

Therefore we get the equation:

q1 = (Cy + Cz)Vac − CzVbc

(1)

Similarly, we apply the rule to the right part of the circuit:

q2 = qx − qz

q2 = CxVbc − Cz(Vac − Vbc)

We then get the second equation

q2 = − CzVac + (Cx + Cz )Vbc

(2)

Solving (1) and (2) simultaneously for V _ac and V _bc, we get:

( ) ( ) Vac = -------Cx-+-Cz------- q1 + ---------Cz---------- q2 CxCy + CyCz + CzCx CxCy + CyCz + CzCx

( ) ( ) V = ---------Cz---------- q + ------Cy-+--Cz------- q bc CxCy + CyCz + CzCx 1 CxCy + CyCz + CzCx 2

Keeping these in mind, we proceed to the Y network. Let us apply Kirchoff’s second law to the left part:

V + V = V 1 3 ac

q1-+ -q3 = V C1 C3 ac

From conservation of charge, q₃ = q₁ + q₂ Simplifying the above equation yields:

( ) ( ) V = 1--+ -1- q + 1-- q ac C1 C3 1 C3 2

Similarly for the right part:

V + V = V 2 3 bc

q2-+ -q3 = V C2 C3 bc

( ) ( ) 1-- -1- 1-- Vbc = C3 q1 + C2 + C3 q2

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 q₁ in V _bc we get:

-1- ---------Cz---------- C = C C + C C + C C 3 x y y z z x

C3 = (CxCy + CyCz + CzCx )∕Cz

Now compare the coefficient of q₂:

1 1 Cy + Cz --- + ---= --------------------- C2 C3 CxCy + CyCz + CzCx

Substitute the expression we got for C₃, and solve for C₂ to get:

C = (C C + C C + C C )∕C 2 x y y z z x y

Now we look at the coeffcient of q₁ in the equation for V _ac:

1 1 Cx + Cz --- + ---= --------------------- C1 C3 CxCy + CyCz + CzCx

Again substituting the expression for C₃ and solving for C₁ we get:

C1 = (CxCy + CyCz + CzCx )∕Cx

We have derived the required transformation equations mentioned at the top.

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