The 10 Percent Rule: This rule is a standard method for selecting R 1 and R 2 that takes into account
2.17 Kirchhoff’s Laws
Often, you will run across a circuit that cannot be analyzed with simple resistor cir- cuit reduction techniques alone. Even if you could find the equivalent resistance by using circuit reduction, it might not be possible for you to find the individual cur- rents and voltage through and across the components within the network. Likewise, if there are multiple sources, or complex networks of resistors, using Ohm’s law, as well as the current and voltage divider equations, may not cut it. For this reason, we turn to Kirchhoff’s laws.
FIGURE 2.61
FIGURE 2.62
Kirchhoff’s laws provide the most general method for analyzing circuits. These laws work for either linear (resistor, capacitors, and inductors) or nonlinear elements (diodes, transistors, etc.), no matter how complex the circuit gets. Kirchhoff’s two laws are stated as follows.
Kirchhoff’s Voltage Law (or Loop Rule): The algebraic sum of the voltages around any loop of a circuit is zero:
V V V Vn
closed path
1 2 0
∑
∆ = + + + = (2.20)In essence, Kirchhoff’s voltage law is a statement about the conservation of energy. If an electric charge starts anywhere in a circuit and is made to go through any loop in a circuit and back to its starting point, the net change in its potential is zero.
To show how Kirchhoff’s voltage law works, we consider the circuit in Fig. 2.63.
We start anywhere we like along the circuit path—say, at the negative terminal of the 5- V battery. Then we start making a loop trace, which in this case we choose to go clockwise, though it doesn’t really matter which direction you choose. Each time we encounter a circuit element, we add it to what we’ll call our ongoing loop equation. To determine the sign of the voltage difference, we apply the loop trace rules shown in the shaded section of the figure. We continue adding elements until we make it back to the start of the loop, at which point we terminate the loop equation with an “= 0.”
As noted, Kirchhoff’s voltage law applies to any circuit elements, both linear and nonlinear. For example, the rather fictitious circuit shown here illustrates Kirchhoff’s voltage law being applied to a circuit that has other elements besides resistors and dc sources—namely, a capacitor, an inductor, and a nonlinear diode, along with a sinusoi- dal voltage source. We can apply the loop trick, as we did in the preceding example, and come up with an equation. As you can see, the expressions used to describe the voltage changes across the capacitor, inductor, and diode are rather complex, not to mention the solution to the resulting differential equation. You don’t do electronics this way (there are tricks), but it nevertheless demonstrates the universality of Kirchhoff’s law.
FIGURE 2.63
Kirchhoff’s Current Law (or Junction Rule): The sum of the currents that enter a junc- tion equals the sum of the currents that leave the junction:
Iin
∑
Iout∑
= (2.21)Kirchhoff’s current law is a statement about the conservation of charge flow through a circuit: at no time are charges created or destroyed.
The following example shows both Kirchhoff’s current and voltage laws in action.
Example: By applying Kirchhoff’s laws to the following circuit, find all the unknown currents, I1, I2, I3, I4, I5, I6, assuming that R1, R2, R3, R4, R5, R6, and V0 are known. After that, the voltage drops across the resistors V1, V2, V3, V4, V5, and V6 can be found using Ohm’s law: Vn = InRn.
Answer: To solve this problem, you apply Kirchhoff’s voltage law to enough closed loops and apply Kirchhoff’s current law to enough junctions that you end up with enough equations to counterbalance the unknowns. After that, it is simply a matter of doing some algebra. Figure 2.66 illustrates how to apply the laws in order to set up the final equations.
In Fig. 2.66, there are six equations and six unknowns. According to the rules of algebra, as long as you have an equal number of equations and unknowns, you can usually figure out what the unknowns will be. There are three ways we can think of to solve for the unknowns in this case. First, you could apply the old “plug and chug”
FIGURE 2.64
FIGURE 2.65
method, better known as the substitution method, where you combine all the equa- tions together and try to find a single unknown, and then substitute it back into another equation, and so forth. A second method, which is a lot cleaner and perhaps easier, involves using matrices. A book on linear algebra will tell you all you need to know about using matrices to solve for the unknown.
A third method that we think is useful—practically speaking—involves using a trick with determinants and Cramer’s rule. The neat thing about this trick is that you do not have to know any math—that is, if you have a mathematical computer program or calculator that can do determinants. The only requirement is that you be able to plug numbers into a grid (determinant) and press “equals.” We do not want to spend too much time on this technique, so we will simply provide you with the equations and use the equations to find one of the solutions to the resistor circuit problem. See Fig. 2.67a.
FIGURE 2.66
FIGURE 2.67a
For example, you can find ∆ for the system of equations from the resistor problem by plugging all the coefficients into the determinant and pressing the “evaluate”
button on the calculator or computer. See Fig. 2.67b.
Now, to find, say, the current through R5 and the voltage across it, you find ∆I5, then use I5 = ∆I5/∆ to find the current. Then you use Ohm’s law to find the voltage. Figure 2.67c shows how it is done.
To solve for the other currents (and voltages), simply find the other ∆I’s and divide by ∆.
The last approach, as you can see, requires a huge mathematical effort to get a single current value. For simplicity, you can find out everything that’s going on in the circuit by running it through a circuit simulator program. For example, using MultiSim, we get the results in Fig. 2.68:
FIGURE 2.67b
FIGURE 2.67c
FIGURE 2.68
Doing long calculations is good theoretical exercise, but in practice it’s a waste of time.
A problem such as this only takes a few min- utes to solve using a simulator. The results from simulation:
V1 = 2.027 V I1 = 2.027 A V2 = 2.351 V I2 = 1.175 A V3 = 2.555 V I3 = 0.852 A V4 = 0.204 V I4 = 0.051 A V5 = 5.622 V I5 = 1.124 A V6 = 5.417 V I6 = 0.903 A
Plugging the results back into the diagram, you can see Kirchhoff’s voltage and current laws in action, as shown in Fig. 2.69. Take any loop, sum the changes in voltage across components, and you get 0 (note that the voltages indicated in black shadow represent point voltages relative to 0- V reference ground). Also, the current that enters any junction will equal the sum of the currents exiting the junction, and vice versa—Kirchhoff’s current law.
Now, before you get too gung ho about playing with equations or become lazy by grabbing a simulator, you should check out a special theorem known as Thevenin’s theorem. This theorem uses some very interesting tricks to analyze circuits, and it may help you avoid dealing with systems of equations or having to resort to simula- tion. Thevenin’s theorem utilizes something called the superposition theorem, which we must consider first.