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Water manifold - water out = water in: current node.

Not to disrespect Gustav Kirchhoff's accomplishments, but his two eponymous laws seem so obvious that you might wonder why they even need to be stated, much less be named after someone. The equivalent for the oft-used water in a hose analogy would be as follows: (#1) The volume of water entering the common port of a manifold is equal to the sum of the water exiting all the other ports. (#2) In a closed system, the sum of the the pressure drops (e.g., constricted pipe) and pressure rises (e.g., water pump) is zero. In fairness, when Mr. Kirchhoff formulated his laws in the middle 19th century not many people were familiar with electric circuits, and instruments for making certified measurements with enough precision to validate the claims were not readily available. This 1962 Radio-Electronics magazine article introduces Kirchhoff's current law and voltage law. Remembering and applying them does make writing equations for multi-loop circuits much simpler.

## Solve Your Problems with Kirchhoff's Laws

Fig. 1 - This typical bridge circuit looks really hard to calculate when unbalanced.

It's simple to figure out networks that look hopelessly complicated

By John Collins

Can you solve the bridge problem in Fig. 1? It will probably give you trouble. Why. Because most radio men are thoroughly familiar with Ohm's law but have little more than a nodding acquaintance with Kirchhoff's laws. This is unfortunate. Kirchhoff extended Ohm's theory; and his method makes it easier to analyze complicated circuits like the bridges, filters and impedance-matching networks so common in radio and electronics. So the technician who can apply Kirchhoff's method has a leg up over one whose knowledge stops with Ohm.

Kirchhoff's laws themselves (there are two) are pretty self-evident:

1. The sum of the currents flowing into a junction is equal to the sum of the currents flowing out of it.

2. In any closed circuit, the sum of the voltage drops is equal to the applied voltage.

Fig. 2 illustrates these concepts.

Fig. 2 - Kirchhoff's two laws illustrated.

Fig. 3 - How Kirchhoff's laws are applied to a simple series-parallel circuit.

Fig. 4 - Solution of the Fig. 3 problem.

Fig. 5 - Kirchhoff's laws work both ways.

Fig. 6 - In fact, they work three ways.

Fig. 7 - Fig. 6's way of getting the answer.

Fig. 8 - Back to the bridge circuit.

Fig. 9 - Kirchhoff's laws make this as easy to solve as the simpler circuits.

In applying Kirchhoff's laws to network problems, follow these rules:

1. A separate current is assumed for each closed circuit, or "loop," and is called its "loop" current. (Since the bridge circuit in Fig. 1 has three closed circuits, three loop currents are needed for the solution.)

2. In a given loop, the IR drops caused by its loop current are always positive.

3. In a given loop, the IR drops caused by the current from an adjacent loop may be either positive or negative, depending on whether the direction of flow is the same or opposite to the direction of the loop current. (Don't worry if you don't understand this; it will become clear from the examples that follow.)

4. A -voltage source is given a positive sign if its polarity is such that it aids the loop current - that is, if the loop current flows from negative to positive in the external circuit. It is negative if it opposes the loop current.

A Practical Example

The above rules are logical consequences of the two laws of Kirchhoff. To demonstrate how to apply them to practical circuits, we will start with an analysis of the series-parallel circuit shown in Fig. 3-a. Since there are two loops, we begin by assuming two currents, In flowing through loop A and Ib through loop B. There is nothing in the rules about the direction of the assumed currents. We have shown both flowing in a clockwise direction simply as a matter of choice.

Loop A is redrawn in detail in Fig. 3-b, so that we can concentrate on the factors in that loop without being distracted by the rest of the circuit. Ia flows through both the 2-ohm and 6-ohm resistors, producing IR drops of 2Ia and 6Ia. Since Ia is the "loop" current, both these IR drops are positive, according to Rule 2 above. However, Ib, from the adjacent loop, also flows through the 6-ohm resistor. Since its direction is opposite to L, it causes a negative IR drop, -6Ia. in accordance with Rule 3. The polarity of the battery is such as to aid in and hence, by Rule 4, it is given a positive sign.

Putting all these facts together we can now write the equation describing loop A:

2Ia + 6Ia - 6Ib = 12     (A)

When the two terms involving Ia are added, the equation becomes:

8Ia - 6Ia = 12              (A)

We next examine loop B, as redrawn in Fig. 3-c. You will observe that Ib, the so-called "loop" current for loop B, flows through both the 6-ohm and the 3-ohm resistors, creating two positive IR drops, 6Ib and 3Ib, (Rule 2). Since Ia flows in the opposite direction through the 6-ohm resistor, it produces a negative IR drop, -6Ia, (Rule 3). Finally, since there is no battery or generator in the circuit, the IR drops are equal to zero.

Expressing these observations as an equation, we write:

-6Ia + 6Ib + 3Ib = 0      (B)

We add the two terms involving Ib, and the equation becomes:

-6Ia + 9Ib = 0               (B)

We now have two equations with two unknown quantities, Ia and Ib. Any number of values can be found for Ia and Ib that will satisfy either equation taken by itself. Equation B, for example, can be solved with any of the following sets of values: Ia = 3, Ib = 2; Ia = 6, Ib = 4; Ia = 9, Ib = 6; Ia = 12, Ib = 8; etc. However, only one set of values will satisfy both Equation A and Equation B. Since both must be considered at the same time to find this single solution, they are called "simultaneous" equations.

Simultaneous equations are solved by adding or subtracting them in such a way as to obtain one equation with one unknown. The first step is to modify the equations by multiplying or dividing them so that a term in one becomes identical with a term in the other. As long as each of its terms is multiplied or divided by the same number the equality of the equation is not changed.

To illustrate, we first divide Equation A by 2, and obtain:

4Ia - 3Ib = 6                  (A)

Next, we divide Equation B by 3, obtaining:

-2Ia + 3Ib = 0                (B)

Then we add the two equations:

Ib is obtained by substituting the value found for Ia into either of the original equations. If we substitute 3 for Ia in Equation B, we obtain:

(-2 x 3) + 3Ib = 0

-6 + 3Ib = 0

3Ib = 6

Ib = 2

To prove the answer, you can substitute 3 for Ia and 2 for Ib in Equation A. Since they are simultaneous both equations are solved by the same set of values.

The final solution is obtained by going back to the original problem and filling in the values we have found for the currents, adding or subtracting them as necessary where more than one current flows through a branch of the circuit. The result is shown in Fig. 4. Ia, a current of 3 amperes, flows through the 2-ohm resistor; Ib, a current of 2 amperes, flows through the 3-ohm resistor, and 1 ampere, the difference between Ia and Ib, flows through the 6-ohm resistor.

It's Hard to Go Wrong

In the foregoing example, current flow was in the direction we originally assumed. In more complicated circuits it is not always easy to tell beforehand which way the current will flow through a particular branch. Before leaving this example, therefore, let's see what would happen if we had guessed wrong on the direction of current flow and had assumed Ib to flow counterclockwise as shown in Fig. 5.

With the detailed diagram in Fig. 5-b before us, we write the equation for Loop A:

2Ia + 6Ia + 6Ib = 12            (A)

You will notice that this is identical with the equation we previously found for loop A, except that the voltage drop caused by Ib now has a positive instead of a negative sign. This is because Ib

now is assumed to flow in the same direction as Ia through the 6-ohm resistor, and Rule 3 applies. Adding the two terms involving Ia and dividing the entire equation by 2, we obtain:

4Ia + 3Ib = 6                       (A)

Referring to Fig. 5-c, we write the equation for Loop B:

6Ia + 6Ib + 3Ib = 0              (B)

This equation is the same as in the original example except that the IR drop caused by Ia in the 6-ohm resistor is now positive, since Ia is now assumed to flow in the same direction as Ib.

When the terms involving Ib are added and the entire equation is divided by 3, we obtain:

2Ia + 3Ib = 0                        (B)

We then eliminate Ib by subtracting the two equations:

Ib is found next by substituting 3 for Ia in either of the equations. If we substitute in Equation B, we obtain:

(2 x 3) + 3Ib = 0

6 + 3Ib = 0

3Ib = -6

Ib = -2

A negative sign in the solution means that we have assumed current flow in the wrong direction, and that it actually flows in the opposite direction. In other words, our assumption that Ib flows counterclockwise is false, and it actually flows in a clockwise direction. If we make this correction, we wind up with the same result as in the original solution, even though we started with a false assumption.

The fact that answers are always correct, no matter what preliminary assumptions are made, is one of the more attractive features of the Kirchhoff technique. At the risk of working this example to death, let's try one more set of assumptions, as shown in Fig. 6. We again assume that both Ia and Ib flow in a clockwise direction, but this time we assume that Ia takes the long route, though the 3-ohm resistor.

The details of loop A are shown in Fig. 6-b. Observing the rules previously stated, we write the equation for that loop:

2Ia + 3Ia + 3Ib = 12                (A)

5Ia + 3Ib = 12                         (A)

Referring to Fig. 6-c, we next write the equation for loop B:

3Ia + 6Ib + 3Ib = 0                  (B)

3Ia + 9Ib = 0                           (B)

Equation B is then reduced by dividing it by 3, and we obtain:

Ia + 3Ib = 0                             (B)

The solution is completed as in the previous examples, by subtracting the two equations to eliminate Ib and thus find the value of Ia. That value is then substituted into either of the first equations to obtain the value of Ib. This procedure yields Ia = 3, Ib = -1.

The negative value means that Ib actually flows counterclockwise. The result is illustrated in Fig. 7. Ia flows downward through the 2-ohm resistor, and Ib flows downward through the 6-ohm resistor. In the 3-ohm resistor, the current is the difference between Ia and Ib, or a current of 2 amperes in a downward direction. So the final solution is the same as in Fig. 4, even though we started with a different set of assumptions.

The several other possible combinations will all give the same answer - you can't go wrong!

Solving a Real Problem

The real benefits of the Kirchhoff method show up in problems that are hard to handle with Ohm's law. Now that we are familiar with the ground rules, we can tackle the bridge circuit of Fig. 1. It is redrawn in Fig. 8 to show the three loop currents needed for the solution and to provide detailed drawings of the individual loops. We can write the equations easily:

5Ia - 2Ib - 3Ic = 10                (A)

-2Ia + 8Ib - 2Ic = 0                (B)

-3Ia - 2Ib + 6Ic = 0                (C)

No matter how complicated a circuit may be, it is easy to write the equations describing it. Solving the equations can be harder. When there are only three equations, it is not hard to combine, say, the first and third to eliminate one unknown, and then the second and third to eliminate the same unknown. This will leave two equations with two unknowns, which can then be solved as in the previous examples.

Multiplying the first equation by 2 and adding it to the third equation:

The second equation is then multiplied by 3 and added to the third:

We now multiply Equation AC by 11 and Equation BC by 3 and add them:

By substituting 4.4 for Ia in either equation AC or BC, we find that Ib = 1.8. Similarly, by substituting the values found for Ia and Ib in any of the original equations, we find that Ic = 2.8.

We get the final solution by going back to the circuit diagram and filling in the values for the currents, adding or subtracting them as necessary. The result is shown in Fig. 9.

If you want to experiment, you can assume different directions or paths for the currents. In any case, you should get the same result.

The above examples involve only resistance. Kirchhoff's method works just as well with impedances in ac circuits. In such problems, the instantaneous current at some part of the cycle can be used to establish direction. Once you have mastered the mechanics, you will find Kirchhoff's laws useful in analyzing a variety of circuits that don't respond to other methods.

Posted June 7, 2024