5.8 Linear algebra: circuit analysis  (Page 2/2)

In an electrical circuit, one node is usually chosen as a reference node and is considered to have a voltage of zero. Then the voltage at every other node is measured with respect to the reference node. This saves us the trouble of always specifying pairs of nodes for voltage measurements and marking + and – signs for each voltage. Other names for the reference node are common and ground .

A constant voltage source is a device that always forces the voltage between its two terminals to be a constant value. In Figure 1 the circle at the left represents a constant voltage source of 5 volts, so that the voltage at the upper $(+)$ end is always exactly 5 volts higher than the voltage at the lower (-) end. A voltage source is something like a battery, but idealized. Real batteries do not maintain a constant output voltage under all conditions.

Resistance is measured in ohms and is denoted by $R$ . A resistor is shown as a zig-zag line in circuit diagrams and labeled with the value of its resistance in ohms. In this chapter we will consider only devices whose resistance is positive and the same in both directions. Ohm's law , also called the resistor law , relates the voltage and current in a resistor. For the resistor shown in Figure 2 , with reference directions assigned to $v$ and $i$ as shown, Ohm's law is

Note that current flows from + to - through the resistor.

Ohm's law and Kirchhoff's current law are the only principles we need to write equations that will allow us to find the voltages and currents in the resistive circuit of Figure 1 . We begin by choosing a reference node and assigning variables to the voltages at every other node (with respect to the reference node). These choices are shown in Figure 3 .

The constant voltage source forces $v_1$ to be exactly 5 volts higher than the reference node. Thus

Next we write equations by applying Kirchhoff's current law to each node in the circuit (except the reference node and $v_1$ , whose voltages we already know). At the node labeled $v_2$ are three paths for leaving current. The current leaving through the 50 ohm resistor can be found by Ohm's law,where the voltage across that resistor is $v_2-v_1$ :

For current leaving through the 300 ohm resistor, the voltage is $v_2$ . Pay careful attention to the sign; since we are interested in the current leaving the node labeled $v_2$ , Figure 4.14 indicates that to apply Ohm's law we should take the voltage as $+v_2-$ reference $=v_2-0=v_2$ . So

For the 100 ohm resistor, we can write

According to Kirchhoff's current law, the sum of these three leaving currents is zero:

Notice that when we wrote the equation for the node labeled $v_2$ , the variable $v_2$ had $a+sign$ each time it occurred in the equation, while the others had a -sign. This is always the case, and watching for it can help you avoid signerrors. Now we apply Kirchhoff's current law at the node labeled $v_3$ to get the equation

Note that this time it is $v_3$ that always shows up with $a+\text{sign}$ .

Equations 2 , 6 , and 7 give us a system of three equations in the three unknown variables $v_1,v_2$ , and $v_3$ . We now write them in matrix form as

We can determine the current flowing through the lamp from $v_3$ to ground in Example 1 by Ohm's law:

The visible effect will, of course, depend on the lamp. Let us assume that the specifications for our lamp indicate that 0.05 ampere or more is requiredbefore it will glow, and more than 0.075 ampere will cause it to burn out. In this case, our circuit would not make the lamp glow.