# AC Bridge Networks

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The basic structure of an AC bridge network is similar to that of a DC Wheatstone bridge: four impedances (arms), a power source, and a balance detector. The power source provides the bridge with alternating current at the required frequency and amplitude.
The basic bridge configuration was discussed in some detail in Bridge Networks for dc networks. We now continue to examine bridge networks by considering those that have reactive components and a sinusoidal ac voltage or current applied.
We will first analyze various familiar forms of the bridge network using mesh analysis and nodal analysis (the format approach). The balance conditions will be investigated throughout the section.
Fig. 1: Maxwell bridge.
The network is redrawn in [Fig. 2],
Fig. 2: Assigning the mesh currents and subscripted impedances for the network of Fig. 1.
Apply mesh analysis to the network of [Fig. 1].
$${ 1 \over Z_1 } = {1 \over R_1} + { 1\over -j X_C}$$
or
$$Y_1 = G_1 + j B_C$$
where
$$Z_1 = {1 \over Y_1} = { 1 \over G_1 + j B_C}$$
$$Z_2 = R_2 \,\,\, Z_3 = R_3$$
$$Z_4 = R_4 + j X_L \,\,\, Z_5 = R_5$$
Applying the format approach:
$$\begin{split} (Z_1 + Z_3) I_1 - Z_1 I_2 - Z_3 I_3 &= E\\ (Z_1 + Z_2 + Z_3) I_2 - Z_1 I_1 - Z_3 I_3 &= 0\\ (Z_3 + Z_4+ Z_5) I_3 - Z_3 I_1 - Z_5 I_2 &= 0\\ \end{split}$$
\begin{align*} +(Z_1 + Z_3) I_1 & - Z_1 I_2 & - Z_3 I_3 &= E\\ - Z_1 I_1 & +(Z_1 + Z_2 + Z_3) I_2 & - Z_3 I_3 &= 0\\ - Z_3 I_1 & - Z_5 I_2 & (Z_3 + Z_4+ Z_5) I_3 &= 0\\ \end{align*}
For balance, $I_{Z_{5}} = 0 A$, And
$$I_{Z_{5}} = I_2 - I_3 = 0 A$$
From the above equations,
$$I_2 = {E(Z_1Z_3 + Z_1Z_4 + Z_1 Z_5 + Z_3Z_5) \over \Delta}$$
and
$$I_3 = {E(Z_1Z_3 + Z_3Z_2 + Z_1 Z_5 + Z_3Z_5) \over \Delta}$$
Where
$$I_{Z_{5}} = I_2 - I_3 = { E(Z_1Z_4 - Z_3Z_2) \over \Delta} = 0$$
$$Z_1Z_4 = Z_3Z_2$$
This condition will be analyzed in greater depth later in this section.
Applying nodal analysis to the network of [Fig. 3] will result in the configuration of [Fig. 4], where
Fig. 3: Hay bridge.
Fig. 4: Assigning the nodal voltages and subscripted impedances for the network of Fig. 3.
$$Y_1 = {1 \over Z_1} \, \, Y_2 = { 1 \over Z_2}$$
$$Y_3 = {1 \over Z_3} \, \, Y_4 = { 1 \over Z_4}\, \, Y_5 = { 1 \over Z_5}$$
and
$$\begin{split} (Y_1 + Y_2)V_1 - (Y_1)V_2 - (Y_2)V_3 &= I \\ (Y_1 + Y_3 + Y_5)V_2 - (Y_1)V_1 - (Y_5)V_3 &= 0 \\ (Y_2 + Y_4 + Y_5)V_3 - (Y_2)V_1 - (Y_5)V_2 &= 0 \\ \end{split}$$
For balance, $V_{Z_{5}} = 0 V$, and
$$V_{Z_{5}} = V_2 - V_3 = 0$$
From the above equations,
$$V_2 = { I(Y_1Y_3 + Y_1Y_4 + Y_1Y_5 + Y_3Y_5) \over \Delta}$$
Similarly,
$$V_3 = { I(Y_1Y_3 + Y_3Y_2 + Y_1Y_5 + Y_3Y_5) \over \Delta}$$
For $V_{Z_{5}} = 0 V$, the following must be satisfied for a finite $\Delta$ not equal to zero:
$$\bbox[10px,border:1px solid grey]{Y_1Y_4 = Y_3Y_2}$$
However, substituting values of Y, we have
$${1 \over Z_1Z_4} = {1 \over Z_2Z_3}$$
or
$$\bbox[10px,border:1px solid grey]{Z_1Z_4 = Z_2Z_3} \tag{1}$$
Let us now investigate the balance criteria in more detail by considering the network of [Fig. 5], where it is specified that I and V = 0.
Fig. 5: Investigating the balance criteria for an ac bridge configuration.
Since I = 0,
$$\bbox[10px,border:1px solid grey]{ I_1 = I_3 }$$
$$\bbox[10px,border:1px solid grey]{ I_2 = I_4 }$$
In addition, for V = 0,
$$\bbox[10px,border:1px solid grey]{I_1Z_1 = I_2Z_2} \tag{2}$$
$$\bbox[10px,border:1px solid grey]{I_3Z_3 = I_4Z_4} \tag{3}$$
Substituting the preceding current relations into Eq. 3, we have
$$I_1Z_3 = I_2Z_4$$
and
$$I_2 = { Z_3 \over Z_4 } I_1$$
Substituting this relationship for $I_2$ into Eq. (2) yields
$$I_1Z_1 = ({Z_3 \over Z_4}I_1)Z_2$$
and
$$Z_1Z_4 = Z_2Z_3$$
as obtained earlier. Rearranging, we have
$${Z_1 \over Z_3} = {Z_2 \over Z_4}$$
For the network of [Fig. 3], which is referred to as a Hay bridge when $Z_5$ is replaced by a sensitive galvanometer,
$$\begin{split} Z_1 &= R_1 - jX_C\\ Z_2 &= R_2\\ Z_3 &= R_3\\ Z_4 &= R_4 + jX_L\\ \end{split}$$
This particular network is used for measuring the resistance and inductance of coils in which the resistance is a small fraction of the reactance $X_L$. Substitute into Eq. (1) in the following form:
$$\begin{split} Z_2Z_3 &= Z_1Z_4 \\ R_2R_3 &= (R_4 + j X_L)(R1 - j X_C) \\ R_2R_3 &= R_1R_4 + j (R_1X_L - R_4X_C) + X_C X_L\\ \text{ so that}\\ R_2R_3 + j 0 &= (R_1R_4 + X_C X_L) + j (R_1X_L - R_4X_C)\\ \end{split}$$
For the equations to be equal, the real and imaginary parts must be equal. Therefore, for a balanced Hay bridge,
$$\bbox[10px,border:1px solid grey]{R_2R_3 = R_1R_4 + X_C X_L}$$
and
$$\bbox[10px,border:1px solid grey]{0 = R_1X_L - R_4X_C}$$
or substituting
$$X_L = w L \,\,\text{and} \,\, X_C = { 1 \over w C}$$
we have
$$X_C X_L = ({ 1 \over w C})(wL) = { L \over C}$$
and
$$R_2R_3 = R_1R_4 +{ L \over C}$$
with
$$R_1wL = {R_4 \over wC}$$
Solving for $R_4$ in the last equation yields
$$R_4= w^2LCR_1$$
and substituting into the previous equation, we have
$$R_2R_3 = R_1(w^2LCR_1) +{ L \over C}$$
Multiply through by C and factor:
$$CR_2R_3 = L(w^2C^2R_1^2 +1)$$
and
$$\bbox[10px,border:1px solid grey]{L = { C R_2 R3 \over 1+ w^2C^2R_1^2}} \tag{4}$$
$$\bbox[10px,border:1px solid grey]{R_4 = { w^2C^2R_1R_2R_3 \over 1+ w^2C^2R_1^2}} \tag{5}$$