214. Kirchhoff’s RulesLearning Objectives
Many complex circuits, such as the one in Figure 1, cannot be analyzed with the seriesparallel techniques developed in Resistors in Series and Parallel and Electromotive Force: Terminal Voltage. There are, however, two circuit analysis rules that can be used to analyze any circuit, simple or complex. These rules are special cases of the laws of conservation of charge and conservation of energy. The rules are known as Kirchhoff’s Rules
Explanations of the two rules will now be given, followed by problemsolving hints for applying Kirchhoff’s rules, and a worked example that uses them. Kirchhoff’s First RuleKirchhoff’s first rule (the Making Connections: Conservation LawsKirchhoff’s rules for circuit analysis are applications of Kirchhoff’s Second RuleKirchhoff’s second rule (the Kirchhoff’s second rule requires $\text{emf}\text{Ir}{\text{IR}}_{1}{\text{IR}}_{2}=0$. Rearranged, this is $\text{emf}=\text{Ir}+{\text{IR}}_{1}+{\text{IR}}_{2}$, which means the emf equals the sum of the $\text{IR}$ (voltage) drops in the loop. Applying Kirchhoff’s RulesBy applying Kirchhoff’s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, or resistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can be solved. There are two decisions you must make when applying Kirchhoff’s rules. These decisions determine the signs of various quantities in the equations you obtain from applying the rules.
Figure 4 and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs are traversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to be consistent for the sign of the change in potential. (See Example 1.)
Example 1: Calculating Current: Using Kirchhoff’s RulesFind the currents flowing in the circuit in Figure 5. Strategy This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the seriesparallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled ${I}_{1}$, ${I}_{2}$, and ${I}_{3}$ in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents. Solution We begin by applying Kirchhoff’s first or junction rule at point a. This gives $${I}_{1}={I}_{2}+{I}_{3},$$since ${I}_{1}$ flows into the junction, while ${I}_{2}$ and ${I}_{3}$ flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must be applied. Now we consider the loop abcdea. Going from a to b, we traverse ${R}_{2}$ in the same (assumed) direction of the current ${I}_{2}$, and so the change in potential is ${I}_{2}{R}_{2}$. Then going from b to c, we go from $\u2013$ to +, so that the change in potential is $+{\text{emf}}_{1}$. Traversing the internal resistance ${r}_{1}$ from c to d gives ${I}_{2}{r}_{1}$. Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of ${I}_{1}{R}_{1}$. The loop rule states that the changes in potential sum to zero. Thus, $${I}_{2}{R}_{2}+{\text{emf}}_{1}{I}_{2}{r}_{1}{I}_{1}{R}_{1}={I}_{2}({R}_{2}+{r}_{1})+{\text{emf}}_{1}{I}_{1}{R}_{1}=0.$$Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives $${3I}_{2}+\text{18}{6I}_{1}=0.$$Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives $$+\phantom{\rule{0.25em}{0ex}}{I}_{1}{R}_{1}+{I}_{3}{R}_{3}+{I}_{3}{r}_{2}{\text{emf}}_{2}\text{= +}{I}_{1}{R}_{1}+{I}_{3}\left({R}_{3}+{r}_{2}\right){\text{emf}}_{2}=0.$$Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes $$+\phantom{\rule{0.25em}{0ex}}{6I}_{1}+{2I}_{3}\text{45}=0.$$These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for ${I}_{2}$: $${I}_{2}=6{2I}_{1}.$$Now solve the third equation for ${I}_{3}$: $${I}_{3}=\text{22}\text{.}5{3I}_{1}.$$Substituting these two new equations into the first one allows us to find a value for ${I}_{1}$: $${I}_{1}={I}_{2}+{I}_{3}=(6{2I}_{1})+(\text{22}\text{.}5{3I}_{1})=\text{28}\text{.}5{5I}_{1}.$$Combining terms gives $${6I}_{1}=\text{28}\text{.}\mathrm{5,\; and}$$$${I}_{1}=4\text{.}\text{75 A}.$$Substituting this value for ${I}_{1}$ back into the fourth equation gives $${I}_{2}=6{2I}_{1}=69.50$$ $${I}_{2}=3\text{.}\text{50 A}.$$The minus sign means ${I}_{2}$ flows in the direction opposite to that assumed in Figure 5. Finally, substituting the value for ${I}_{1}$ into the fifth equation gives $${I}_{3}=\text{22.5}{3I}_{1}=\text{22.5}\text{14}\text{.}\text{25}$$ $${I}_{3}=8\text{.}\text{25 A}.$$Discussion Just as a check, we note that indeed ${I}_{1}={I}_{2}+{I}_{3}$. The results could also have been checked by entering all of the values into the equation for the abcdefgha loop. ProblemSolving Strategies for Kirchhoff’s Rules
The material in this section is correct in theory. We should be able to verify it by making measurements of current and voltage. In fact, some of the devices used to make such measurements are straightforward applications of the principles covered so far and are explored in the next modules. As we shall see, a very basic, even profound, fact results—making a measurement alters the quantity being measured. Check Your UnderstandingCan Kirchhoff’s rules be applied to simple series and parallel circuits or are they restricted for use in more complicated circuits that are not combinations of series and parallel? Show/Hide Solution SolutionKirchhoff's rules can be applied to any circuit since they are applications to circuits of two conservation laws. Conservation laws are the most broadly applicable principles in physics. It is usually mathematically simpler to use the rules for series and parallel in simpler circuits so we emphasize Kirchhoff’s rules for use in more complicated situations. But the rules for series and parallel can be derived from Kirchhoff’s rules. Moreover, Kirchhoff’s rules can be expanded to devices other than resistors and emfs, such as capacitors, and are one of the basic analysis devices in circuit analysis. Section Summary
Conceptual QuestionsExercise 1Can all of the currents going into the junction in Figure 6 be positive? Explain. Figure 6 Exercise 2Apply the junction rule to junction b in Figure 7. Is any new information gained by applying the junction rule at e? (In the figure, each emf is represented by script E.) Figure 7 Exercise 3(a) What is the potential difference going from point a to point b in Figure 7? (b) What is the potential difference going from c to b? (c) From e to g? (d) From e to d? Exercise 4Apply the loop rule to loop afedcba in Figure 7. Exercise 5Apply the loop rule to loops abgefa and cbgedc in Figure 7. Problem ExercisesExercise 1Apply the loop rule to loop abcdefgha in Figure 5. Show/Hide Solution Solution$${I}_{2}{R}_{2}+{\text{emf}}_{1}{\text{I}}_{2}{r}_{1}+{\text{I}}_{3}{R}_{3}+{\text{I}}_{3}{r}_{2}{\text{emf}}_{2}=\text{0}$$Exercise 2Apply the loop rule to loop aedcba in Figure 5. Exercise 3Verify the second equation in Example 1 by substituting the values found for the currents ${I}_{1}$ and ${I}_{2}$. Exercise 4Verify the third equation in Example 1 by substituting the values found for the currents ${I}_{1}$ and ${I}_{3}$. Exercise 5Apply the junction rule at point a in Figure 8. Figure 8 Show/Hide Solution Solution$${I}_{3}={\text{I}}_{1}+{\text{I}}_{2}$$Exercise 6Apply the loop rule to loop abcdefghija in Figure 8. Exercise 7Apply the loop rule to loop akledcba in Figure 8. Show/Hide Solution Solution$${\text{emf}}_{2}{\text{I}}_{2}{r}_{2}{\text{I}}_{2}{R}_{2}+{\text{I}}_{1}{R}_{5}+{I}_{1}{r}_{1}{\text{emf}}_{1}+{\text{I}}_{1}{R}_{1}=0$$Exercise 8Find the currents flowing in the circuit in Figure 8. Explicitly show how you follow the steps in the ProblemSolving Strategies for Series and Parallel Resistors. Exercise 9Solve Example 1, but use loop abcdefgha instead of loop akledcba. Explicitly show how you follow the steps in the ProblemSolving Strategies for Series and Parallel Resistors. Show/Hide Solution Solution(a) ${\text{I}}_{1}=\text{4.75 A}$ (b) ${\text{I}}_{\text{2}}=3\text{.}\text{5 A}$$$ (c) ${\text{I}}_{3}=8\text{.}\text{25 A}$ Exercise 10Find the currents flowing in the circuit in Figure 7. Exercise 11Unreasonable Results Consider the circuit in Figure 9, and suppose that the emfs are unknown and the currents are given to be ${I}_{1}=5\text{.}\text{00 A}$, ${I}_{2}=3\text{.0 A}$, and ${I}_{3}=\mathrm{\u20132}\text{.}\text{00 A}$. (a) Could you find the emfs? (b) What is wrong with the assumptions? Figure 9 Show/Hide Solution Solution(a) No, you would get inconsistent equations to solve. (b) ${I}_{1}\ne {I}_{2}+{I}_{3}$. The assumed currents violate the junction rule.
