Bridge Networks

Electrical Circuit Analysis > Methods of Analysis

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What is a bridge network?

The bridge network is a configuration that has a multitude of applications. This type of network is used in both dc and ac meters. Electronics courses introduce these in the discussion of rectifying circuits used in converting a varying signal to one of a steady nature (such as dc).

Fig. 1: Various formats for a bridge network.

What is an example of bridge network?

The bridge network may appear in one of the three forms as indicated in [Fig. 1]. The network in [Fig. 1(c)] is also called a symmetrical lattice network if $R_2 = R_3$ and $R_1 = R_4$. [Fig. 1(c)] is an excellent example of how a planar network can be made to appear nonplanar. For the purposes of investigation, let us examine the network in [Fig. 2] using mesh and nodal analysis.

Fig. 2: Assigning the mesh currents to the network.

Mesh analysis ([Fig. 2]) yields

$$(3 Ω + 4 Ω + 2 Ω)I_1 - (4 Ω)I_2 - (2 Ω)I_3 = 20 V$$

$$(4 Ω + 5 Ω + 2 Ω)I_2 - (4 Ω)I_1 - (5 Ω)I_3 = 0$$

$$(2 Ω + 5 Ω + 1 Ω)I_3 - (2 Ω)I_1 - (5 Ω)I_2 = 0$$

and

$$ \begin{split} 9I_1 - 4I_2 - 2I_3 &= 20 \\ -4I_1 + 11I_2 - 5I_3 &= 0 \\ -2I_1 - 5I_2 + 8I_3 &= 0 \end{split}$$

with the result that

$$ \begin{split} I_1 &= 4 A\\ I_2 &= 2.67 A\\ I_3 &= 2.67 A \end{split}$$

The net current through the 5 Ω resistor is

$$I_{5Ω} = I_2 - I_3 = 2.67 A - 2.67 A = 0 A$$

Fig. 3: Defining the nodal voltages for the network.

Nodal analysis ([Fig.3]) yields

$$ \begin{split} ({1 \over 3 Ω} + {1 \over 4 Ω} +{1 \over 2 Ω})V_1 -({1 \over 4 Ω})V_2 - ({1 \over 2 Ω})V_3 &={20 \over 3} A \\ ({1 \over 4 Ω} + {1 \over 2 Ω} + {1 \over 5 Ω})V_2 - ({ 1 \over 4 Ω})V_1 - ({ 1 \over 5 Ω})V_3 &= 0 \\ ({ 1 \over 5 Ω} + {1 \over 2 Ω} + {1 \over 1 Ω})V_3 - ({ 1 \over 2 Ω})V_1 - ({1 \over 5 Ω})V_2 &= 0 \end{split}$$

and

$$ \begin{split} ({1 \over 3 Ω} + {1 \over 4 Ω}+{1 \over 2 Ω})V_1 - ({1 \over 4 Ω})V_2 - ({1 \over 2 Ω})V_3 &= 6.67 A \\ - ({ 1 \over 4 Ω})V_1 + ({1 \over 4 Ω} + {1 \over 2 Ω} +{1 \over 5 Ω})V_2 - ({1 \over5 Ω})V_3 &= 0\\ - ({1 \over2 Ω})V_1 - ({1 \over 5 Ω})V_2 + ({1 \over 5 Ω} + {1 \over 2 Ω} +{1 \over 1 Ω})V_3 &= 0 \end{split}$$

Solving for voltages,

$$\begin{split} V_1 = 8.02 V \\ V_2 = 2.67 V \\ V_3 = 2.67 V \end{split}$$

and the voltage across the $5 Ω$ resistor is

$$V_{5Ω} = V_2 - V_3 = 2.67 A - 2.67 A = 0 V$$

Since $V5Ω = 0 V$, we can insert a short in place of the bridge arm without affecting the network behavior. (Certainly $V = IR = I (0) = 0 V$.)

(a)

(b)

Fig.4: (a)Substituting the short-circuit equivalent for the balance arm of a balanced bridge. (b) Redrawing the network

In [Fig. 4(a)], a short circuit has replaced the resistor $R_5$, and the voltage across $R_4$ is to be determined. The network is redrawn in [Fig. 4(b)], and

$$ \begin{split} V_{1Ω} &= {(2 Ω || 1 Ω)20 V \over (2 Ω || 1 Ω) + (4 Ω || 2 Ω) + 3 Ω} \\ &={{ 2 \over 3}(20 V) \over {2 \over 3} + {8 \over 6} + 3} \\ &={40 V \over 15} \\ &= 2.67 V \end{split}$$

as obtained earlier.

Fig. 5: Substituting the open-circuit equivalent for the balance arm of a balanced bridge.

We found through mesh analysis that $I_{5Ω} = 0 A$, which has as its equivalent an open circuit as shown in [Fig. 5(a)]. (Certainly $I = V>R = 0>(0/\infty) = 0 A.$) The voltage across the resistor $R_4$ is again determined and compared with the result above. The network is redrawn after combining series elements as shown in [Fig. 5(b)], and

$$ \begin{split} V_{3Ω} = {(6 Ω || 3 Ω)(20 V) \over 6 Ω || 3 Ω + 3 Ω} \\ = {2 Ω(20 V) \over 2 Ω + 3 Ω} = 8 V \end{split}$$

and

$$ \begin{split} V_{1Ω} &= {1 Ω(8 V) \over 1 Ω + 2 Ω} \\ &= {8 V \over 3} = 2.67 V \end{split} $$

as above.

Fig. 6: Establishing the balance criteria for a bridge network.

The condition $V_{5Ω} = 0 V$ or $I_{5Ω} = 0$ A exists only for a particular relationship between the resistors of the network. Let us now derive this relationship using the network in [Fig. 6], in which it is indicated that $I = 0 A$ and $V = 0 V$.
The bridge network is said to be balanced when the condition of $I = 0 A$ or $V = 0 V$ exists.
If $V = 0 V$ (short circuit between a and b), then

$$V_1 = V_2$$

and

$$ I_1R_1 = I_2R_2 $$

$$I_1 = {I_2R_2 \over R_1}$$

In addition, when $V = 0 V$,

$$V_3 = V_4$$

and

$$I_3R_3 = I_4R_4$$

If we set I = 0 A, then $I_3 = I_1$ and $I_4 = I_2$, with the result that the above equation becomes

$$I_1R_3 = I_2R_4$$

Substituting for $I_1$ from above yields

$$({I_2R_2 \over R_1}) R_3 = I_2R_4$$

or, rearranging, we have

$$\bbox[5px,border:1px solid grey] {{R_1 \over R_3} = {R_2 \over R_4}} \tag{1}$$

This conclusion states that if the ratio of $R_1$ to $R_3$ is equal to that of $R_2$ to $R_4$, the bridge is balanced, and $I = 0 A$ or $V = 0 V$. A method of memorizing this form is indicated in [Fig. 7].

Fig. 7: A visual approach to remembering the balance condition.

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