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Bridge topologies are common structures in circuits because they usually permit current to flow in either direction across the "bridge". This feature allows a motor (for example) connected across the bridge to be spun either clockwise or counterclockwise.

Figure 1

There are no pair of resistors in series or parallel, so we cannot use any shortcuts to help analyze the circuit. Instead, we have to use KVL, KCL and Ohm's law. On the rest of this page, we describe the node method, a systematic method to apply these fundamental circuit laws.

So far, we have always referred to voltage as a drop across an element. Now, we define a zero voltage level called ground, so that every node in a circuit can have a node voltage relative to that ground.

Figure 2

Once a ground symbol is attached to a circuit schematic, all other node voltages are then labeled relative to that zero level.

Figure 3

Now, the voltage drop across any resistor or the current through it can be expressed in terms of the node voltages.

Figure 4

Therefore, we obtain the voltage drop across $R_5$ and the current through it by KVL and Ohm's law.

\begin{align}

V_5 &= V_a-V_b\\

I_5 &= \frac{V_5}{R_5} = \frac{V_a-V_b}{R_5} \label{NDM-ER5}

\end{align}

Notice that the arrow label associated with $I_5$ points from node voltage $V_a$ to $V_b$. Similarly, we express the other resistor currents in terms of the node voltages in Fig. 3.

\begin{align}

I_1 &= \frac{15-V_a}{R_1}\\

I_2 &= \frac{15-V_b}{R_2}\\

I_3 &= \frac{V_a-0}{R_3}\\

I_4 &= \frac{V_b-0}{R_4} \label{NDM-ER4}

\end{align}

There are two unknown node voltages $V_a$ and $V_b$ in Fig. 4. The KCL equations at those two nodes are obtained by equating total currents flowing in with total currents flowing out.

\begin{align}

I_1 &= I_3+I_5\\

I_2+I_5 &= I_4

\end{align}

Since all the currents flow through resistors, we can rewrite these equations in terms of the node voltages by using equations $\eqref{NDM-ER5}$ to $\eqref{NDM-ER4}$.

\begin{align}

\frac{15-V_a}{R_1} &= \frac{V_a-0}{R_3}+\frac{V_a-V_b}{R_5} \label{NDM-KCA}\\

\frac{15-V_b}{R_2}+\frac{V_a-V_b}{R_5} &= \frac{V_b-0}{R_4} \label{NDM-KCB}

\end{align}

Given resistor values $R_1$ to $R_5$, we can solve these two equations for the two unknown node voltages $V_a$ and $V_b$, from which we can determine any voltage drop or current in the circuit. The following table gives examples of calculating $V_a$ and $V_b$ (from equations $\eqref{NDM-KCA}$ and $\eqref{NDM-KCB}$) as well as the bridge current $I_5$ (from equation $\eqref{NDM-ER5}$) given known values of $R_1$ to $R_5$.

Resistor Values | Node Voltages | Bridge Current |
---|---|---|

$\begin{aligned} R_1&=R_2=R_3=R_5=5\text{ k}\Omega \\ R_4&=1\text{ k}\Omega \end{aligned}$ | $\begin{aligned} V_a&=6\text{ V} \\ V_b&=3\text{ V} \end{aligned}$ | $\begin{aligned} I_5 &=0.6\text{ mA} \end{aligned}$ |

$\begin{aligned} R_1&=R_2=R_3=R_5=5\text{ k}\Omega \\ R_4&=5\text{ k}\Omega \end{aligned}$ | $\begin{aligned} V_a&=7.5\text{ V} \\ V_b&=7.5\text{ V} \end{aligned}$ | $\begin{aligned} I_5 &=0\text{ mA} \end{aligned}$ |

$\begin{aligned} R_1&=R_2=R_3=R_5=5\text{ k}\Omega \\ R_4&=9\text{ k}\Omega \end{aligned}$ | $\begin{aligned} V_a&=8\text{ V} \\ V_b&=9\text{ V} \end{aligned}$ | $\begin{aligned} I_5 &=-0.2\text{ mA} \end{aligned}$ |

Observe that increasing the resistor value $R_4$ in the circuit in Fig. 1 changes the bridge current $I_5$ from positive to zero to negative.

When there is just one voltage source in a circuit, you should attach the ground to one of the nodes connected to that voltage source, just as in Fig. 3. In this way, both nodes connected to the voltage source have node voltages that are known. But if there are two or more voltage sources not all connected to a common node, then at least one of them cannot be connected to the ground. We call these these ones floating voltage sources because their adjacent node voltages are unknown.

Figure 5

The problem with the floating voltage source $V_x$ is that the current $I_x$ cannot be expressed directly in terms of the node voltages. Thus, $I_x$ cannot be used in node method KCL equations to solve for the node voltages. We work around this problem by encapsulating the floating voltage source $V_x$ in a supernode. Then we can apply KCL to the supernode without involving $I_x$

Figure 6

At the supernode, KCL gives $I_1+I_2 = I_3+I_4$, which can be rewritten in terms of node voltages as follows.

\begin{align}

\frac{15-V_a}{R_1} + \frac{15-(V_a+5)}{R_2}&= \frac{V_a-0}{R_3}+\frac{(V_a+5)-0}{R_4} \label{NDM-KCS}

\end{align}

Given resistor values $R_1$ to $R_4$, we can solve this equation for the unknown node voltage $V_a$. Knowing all the node voltages, we can determine any resistor's voltage drop or current. But to find the current $I_x$ through the floating voltage source, we need to apply KCL to one of the adjacent nodes (not the supernode). For example, KCL at the node labeled $V_a$ in Fig. 6 gives $I_1=I_3+I_x$. Therefore,

\begin{align}

I_x &= I_1-I_3\\

&= \frac{15-V_a}{R_1} - \frac{V_a-0}{R_3} \label{NDM-IXE}

\end{align}

The following table gives examples of calculating $V_a$ (from equation $\eqref{NDM-KCS}$) as well as the bridge current $I_x$ (from equation $\eqref{NDM-IXE}$) given known values of $R_1$ to $R_4$.

Resistor Values | Node Voltages | Currents | Bridge Current |
---|---|---|---|

$\begin{aligned} R_1&=R_2=R_3=5\text{ k}\Omega \\ R_4&=1\text{ k}\Omega \end{aligned}$ | $\begin{aligned} V_a&=0\text{ V} \\ V_a+5&=5\text{ V} \end{aligned}$ | $\begin{aligned} I_1 &=3\text{ mA} \\ I_3 &=0\text{ mA} \end{aligned}$ | $\begin{aligned} I_x &=3\text{ mA} \end{aligned}$ |

$\begin{aligned} R_1&=R_2=R_3=5\text{ k}\Omega \\ R_4&=5\text{ k}\Omega \end{aligned}$ | $\begin{aligned} V_a&=5\text{ V} \\ V_a+5&=10\text{ V} \end{aligned}$ | $\begin{aligned} I_1 &=2\text{ mA} \\ I_3 &=1\text{ mA} \end{aligned}$ | $\begin{aligned} I_x &=1\text{ mA} \end{aligned}$ |

$\begin{aligned} R_1&=R_2=R_3=5\text{ k}\Omega \\ R_4&=9\text{ k}\Omega \end{aligned}$ | $\begin{aligned} V_a&=6.25\text{ V} \\ V_a+5&=11.25\text{ V} \end{aligned}$ | $\begin{aligned} I_1 &=1.75\text{ mA} \\ I_3 &=1.25\text{ mA} \end{aligned}$ | $\begin{aligned} I_x &=0.5\text{ mA} \end{aligned}$ |

The flowchart shown below efficiently solves a circuit with sources and resistors using the node method. Use this technique when you cannot simplify the circuit any further using circuit analysis shortcuts.

Figure 7

Attach a ground to the node connected to as many voltage sources as possible. Use these voltage sources to label known node voltages.

If there are any floating voltages (ones connected to unknown node voltages), enclose each of them in a supernode.

Label all unknown node voltages using variables. Across a supernode, your variables should include the value of the enclosed floating voltage source.

Write a KCL equation at every supernode and at every other unknown node in terms of node voltages.

Determine the unknown node voltages by simultaneously solving the KCL equations.

Use the node voltages to obtain desired quantities in the circuit.

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