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
$$V_1 = V_2$$
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 $I_1$ from above yields
$$({I_2R_2 \over R_1}) R_3 = I_2R_4$$
or, rearranging, we have
$$\bbox[5px,border:1px solid blue] {\color{blue}{{R_1 \over R_3} = {R_2 \over R_4}}} \tag{1}$$
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.
