# Bridge Networks

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.
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 R2 = R3 and R1 = R4. 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 R5, and the voltage across R4 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 R4 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 V5Ω = 0 V or I5Ω = 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 $$V_1 = V_2$$ 0and $$I_1R_1 = I_2R_2$$ or $$I_1 = {I_2R_2 \over R_1}$$ 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 I1 from above yields $$({I_2R_2 \over R_1}) R_3 = I_2R_4$$ or, rearranging, we have $$\bbox[5px,border:1px solid blue] {\color{blue}{{R_1 \over R_3} = {R_2 \over R_4}}} \tag{1}$$ This conclusion states that if the ratio of R1 to R3 is equal to that of R2 to R4, 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.