20.10

Kirchhoff’s Rules

Electric circuits that contain a number of resistors can often be analyzed by combining individual groups of resistors in series and parallel, as Section 20.8 discusses. However, there are many circuits in which no two resistors are in series or in parallel. To deal with such circuits it is necessary to employ methods other than the series-parallel method. One alternative is to take advantage of Kirchhoff’s rules, named after their developer Gustav Kirchhoff (1824–1887).

CONCEPTS AT A GLANCE There are two rules, the junction rule and the loop rule. The Concepts-at-a-Glance chart in Figure 20-27 emphasizes that these rules arise from principles and ideas that we have encountered earlier. The junction rule is an application of the law of conservation of electric charge (see Section 18.2) to the electric current in a circuit. The loop rule is an application of the principle of conservation of energy (see Section 6.8) to the electric potential (see Section 19.2) that exists at various places in a circuit.

Figure 20-27 CONCEPTS AT A GLANCE Kirchhoff s rules are applications to electric circuits of the law of conservation of electric charge and the principle of conservation of energy. The currents and voltages in complex electrical circuits, such as those in this photograph, can be analyzed using Kirchhoff’s rules. (James Schnepf/The Image Bank/Getty Images, Inc.)

Figure 20-28 illustrates in greater detail the basic idea behind Kirchhoff’s junction rule. The picture shows a junction where several wires are connected together. As Section 18.2 discusses, electric charge is conserved. Therefore, since there is no accumulation of charges at the junction itself, the total charge per second flowing into the junction must equal the total charge per second flowing out of it. In other words, the junction rule states that the total current directed into a junction must equal the total current directed out of the junction, or 7 A = 5 A + 2 A for the specific case shown in the picture.

Figure 20-28 A junction is a point in a circuit where a number of wires are connected together. If 7 A of current is directed into the junction, then a total of 7 A (5 A + 2 A) of current must be directed out of the junction.

To help explain Kirchhoff’s loop rule, Figure 20-29 shows a circuit in which a 12-V battery is connected to a series combination of a and a resistor. The plus and minus signs associated with each resistor remind us that, outside a battery, conventional current is directed from a higher toward a lower potential. From left to right, there is a potential drop of 10 V across the first resistor and another drop of 2 V across the second resistor. Keeping in mind that potential is the electric potential energy per unit charge, let us follow a positive test charge clockwise* around the circuit. Starting at the negative terminal of the battery, we see that the test charge gains potential energy because of the 12-V rise in potential due to the battery. The test charge then loses potential energy because of the 10-V and 2-V drops in potential across the resistors, ultimately arriving back at the negative terminal. In traversing the closed-circuit loop, the test charge is like a skier gaining gravitational potential energy in going up a hill on a chair lift and then losing it to friction in coming down and stopping. When the skier returns to the starting point, the gain equals the loss, so there is no net change in potential energy. Similarly, when the test charge arrives back at its starting point, there is no net change in electric potential energy, the gains matching the losses. The loop rule expresses this example of energy conservation in terms of the electric potential and states that for a closed-circuit loop, the total of all the potential drops is the same as the total of all the potential rises, or 10 V + 2 V = 12 V for the specific case in Figure 20-29.

Figure 20-29 Following a positive test charge clockwise around the circuit, we see that the total voltage drop of 10 V + 2 V across the two resistors equals the voltage rise of 12 V due to the battery. The plus and minus signs on the resistors emphasize that, outside a battery, conventional current is directed from a higher potential ( + ) toward a lower potential ( − ).

Kirchhoff’s rules can be applied to any circuit, even when the resistors are not in series or in parallel. The two rules are summarized below, and Examples 14 and 15 illustrate how to use them.

| Kirchhoff’s Rules
| Junction rule. The sum of the magnitudes of the currents directed into a junction equals the sum of the magnitudes of the currents directed out of the junction. Loop rule. Around any closed-circuit loop, the sum of the potential drops equals the sum of the potential rises.

| Example 14 | | | Using Kirchhoff’s Loop Rule

Figure 20-30 shows a circuit that contains two batteries and two resistors. Determine the current I in the circuit.

Figure 20-30 A single-loop circuit that contains two batteries and two resistors.

Reasoning The first step is to draw the current, which we have chosen to be clockwise around the circuit. The choice of direction is arbitrary, and if it is incorrect, I will turn out to be negative.
The second step is to mark the resistors with plus and minus signs, which serve as an aid in identifying the potential drops and rises for Kirchhoff’s loop rule. Remember that, outside a battery, conventional current is always directed from a higher potential ( + ) toward a lower potential ( − ). Thus, we must mark the resistors as indicated in Figure 20-30, to be consistent with the clockwise direction chosen for the current.

Problem solving insight

We may now apply Kirchhoff’s loop rule to the circuit, starting at corner A, proceeding clockwise around the loop, and identifying the potential drops and rises as we go. The potential across each resistor is given by Ohm’s law as V = IR. The clockwise direction is arbitrary, and the same result is obtained with a counterclockwise path.
Solution Starting at corner A and moving clockwise around the loop, there is

1. | A potential drop (+ to −) of across the resistor
2. | A potential drop (+ to −) of 6.0 V across the 6.0-V battery
3. | A potential drop (+ to −) of across the resistor
4. | A potential rise (− to +) of 24 V across the 24-V battery

Setting the sum of the potential drops equal to the sum of the potential rises, as required by Kirchhoff’s loop rule, gives

Solving this equation for the current yields . The current is a positive number, indicating that our initial choice for the direction (clockwise) of the current was correct.

Need more practice?

Interactive LearningWare20.1
The circuit in the drawing contains three resistors and two batteries. Find the currents in the resistor and the resistor.

Related Homework: Problem 74

| Example 15 | | | The Electrical System of a Car

In a car, the headlights are connected to the battery and would discharge the battery if it were not for the alternator, which is run by the engine. Figure 20-31 indicates how the car battery, headlights, and alternator are connected. The circuit includes an internal resistance of for the car battery and its leads and a resistance of for the headlights. For the sake of simplicity, the alternator is approximated as an additional 14.00-V battery with an internal resistance of . Determine the currents through the car battery (IB), the headlights (IH), and the alternator (IA).

Figure 20-31 A circuit showing the headlight(s), battery, and alternator of a car.

The physics of an automobile electrical system.

Reasoning The drawing shows the directions chosen arbitrarily for the currents IB, IH, and IA. If any direction is incorrect, the analysis will reveal a negative value for the corresponding current.
Next, we mark the resistors with the plus and minus signs that serve as an aid in identifying the potential drops and rises for the loop rule, recalling that, outside a battery, conventional current is always directed from a higher potential ( + ) toward a lower potential ( − ). Thus, given the directions selected for IB, IH, and IA, the plus and minus signs must be those indicated in Figure 20-31.

Problem solving insight In problems involving Kirchhoff’s rules, it is always helpful to mark the resistors with plus and minus signs to keep track of the potential rises and drops in the circuit.

Kirchhoff’s junction and loop rules can now be used.
Solution The junction rule can be applied to junction B or junction E. In either case, the same equation results:

In applying the loop rule to the lower loop BEFA, we start at point B, move clockwise around the loop, and identify the potential drops and rises. There is a potential rise (− to +) of across the resistor and then a drop (+ to −) of 12.00 V due to the car battery. Continuing around the loop, we find a 14.00-V rise (− to +) across the alternator, followed by a potential drop (+ to −) of across the resistor. Setting the sum of the potential drops equal to the sum of the potential rises gives the following result:

Since there are three unknown variables in this problem, IB, IH, and IA, a third equation is needed for a solution. To obtain the third equation, we apply the loop rule to the upper loop CDEB, choosing a clockwise path for convenience. The result is

These three equations can be solved simultaneously to show that

The negative answer for IB indicates that the current through the battery is not directed from right to left, as drawn in Figure 20-31. Instead, the 9.0-A current is directed from left to right, opposite to the way current would be directed if the alternator were not connected. It is the left-to-right current created by the alternator that keeps the battery charged.
Note that we can check our results by applying Kirchhoff’s loop rule to the outer loop ABCDEF. If the results are correct, then the sum of the potential drops around this loop will be equal to the sum of the potential rises.

| Reasoning Strategy
| Applying Kirchhoff’s Rules

1. | Draw the current in each branch of the circuit. Choose any direction. If your choice is incorrect, the value obtained for the current will turn out to be a negative number.
2. | Mark each resistor with a plus sign at one end and a minus sign at the other end, in a way that is consistent with your choice for the current direction in Step 1. Outside a battery, conventional current is always directed from a higher potential (the end marked +) toward a lower potential (the end marked −).
3. | Apply the junction rule and the loop rule to the circuit, obtaining in the process as many independent equations as there are unknown variables.
4. | Solve the equations obtained in Step 3 simultaneously for the unknown variables. (See Appendix C.3.)

Check Your Understanding 4

The drawing shows a circuit containing three resistors and three batteries. In preparation for applying Kirchhoff’s rules, the currents in each resistor have been drawn. For these currents, write down the equations that result from applying Kirchhoff’s junction rule and loop rule. Apply the loop rule to loops ABCD and BEFC. (The answers are given at the end of the book.)

Background: In the circuit in the drawing no two resistors are in series or in parallel. Kirchhoff’s rules do apply, however.
For similar questions (including calculational counterparts), consult Self-Assessment Test 20.2, which is described at the end of Section 20.13.

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