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Problem-solving strategies for kirchhoff’s rules

  1. Make certain there is a clear circuit diagram on which you can label all known and unknown resistances, emfs, and currents. If a current is unknown, you must assign it a direction. This is necessary for determining the signs of potential changes. If you assign the direction incorrectly, the current will be found to have a negative value—no harm done.
  2. Apply the junction rule to any junction in the circuit. Each time the junction rule is applied, you should get an equation with a current that does not appear in a previous application—if not, then the equation is redundant.
  3. Apply the loop rule to as many loops as needed to solve for the unknowns in the problem. (There must be as many independent equations as unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then carefully and consistently determine the signs of the potential changes for each element using the four bulleted points discussed above in conjunction with [link] .
  4. Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking.
  5. Check to see whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither exceedingly large nor vanishingly small. The signs should be reasonable—for example, no resistance should be negative. Check to see that the values obtained satisfy the various equations obtained from applying the rules. The currents should satisfy the junction rule, for example.

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.

Can 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?

Kirchhoff'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.

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Section summary

  • Kirchhoff’s rules can be used to analyze any circuit, simple or complex.
  • Kirchhoff’s first rule—the junction rule: The sum of all currents entering a junction must equal the sum of all currents leaving the junction.
  • Kirchhoff’s second rule—the loop rule: The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.
  • The two rules are based, respectively, on the laws of conservation of charge and energy.
  • When calculating potential and current using Kirchhoff’s rules, a set of conventions must be followed for determining the correct signs of various terms.
  • The simpler series and parallel rules are special cases of Kirchhoff’s rules.

Conceptual questions

Can all of the currents going into the junction in [link] be positive? Explain.

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The diagram shows a T junction with currents I sub one, I sub two, and I sub three entering the T junction.

Apply the junction rule to junction b in [link] . Is any new information gained by applying the junction rule at e? (In the figure, each emf is represented by script E.)

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The diagram shows a complex circuit with four voltage sources: E sub one, E sub two, E sub three, E sub four and several resistive loads, wired in two loops and two junctions. Several points on the diagram are marked with letters a through g. The current in each branch is labeled separately.

(a) What is the potential difference going from point a to point b in [link] ? (b) What is the potential difference going from c to b? (c) From e to g? (d) From e to d?

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Apply the loop rule to loop afedcba in [link] .

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Apply the loop rule to loops abgefa and cbgedc in [link] .

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Problem exercises

Apply the loop rule to loop abcdefgha in [link] .

I 2 R 2 + emf 1 I 2 r 1 + I 3 R 3 + I 3 r 2 - emf 2 = 0 size 12{ {underline {-I rSub { size 8{2} } R rSub { size 8{3} } + "emf" rSub { size 8{1} } - ital " I" rSub { size 8{2} } r rSub { size 8{1} } + ital " I" rSub { size 8{3} } r rSub { size 8{3} } + ital " I" rSub { size 8{3} } r rSub { size 8{2} } +- "emf" rSub { size 8{2} } =" 0"}} } {}
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Apply the loop rule to loop aedcba in [link] .

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Verify the second equation in [link] by substituting the values found for the currents I 1 size 12{I rSub { size 8{1} } } {} and I 2 size 12{I rSub { size 8{2} } } {} .

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Verify the third equation in [link] by substituting the values found for the currents I 1 size 12{I rSub { size 8{1} } } {} and I 3 size 12{I rSub { size 8{3} } } {} .

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Apply the junction rule at point a in [link] .

The diagram shows a complex circuit with four voltage sources E sub one, E sub two, E sub three, E sub four and several resistive loads, wired in two loops and many junctions. Several points on the diagram are marked with letters a through j. The current in each branch is labeled separately.
I 3 = I 1 + I 2 size 12{I rSub { size 8{3} } = ital " I" rSub { size 8{1} } + ital " I" rSub { size 8{2} } } {}
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Apply the loop rule to loop abcdefghija in [link] .

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Apply the loop rule to loop akledcba in [link] .

emf 2 - I 2 r 2 - I 2 R 2 + I 1 R 5 + I 1 r 1 - emf 1 + I 1 R 1 = 0 size 12{ {underline { "emf" rSub { size 8{2} } +- ital " I" rSub { size 8{2} } r rSub { size 8{2} } +- ital " I" rSub { size 8{2} } R rSub { size 8{2} } + ital " I" rSub { size 8{1} } R rSub { size 8{5} } +I rSub { size 8{1} } r rSub { size 8{1} } +- "emf" rSub { size 8{1} } + ital " I" rSub { size 8{1} } R rSub { size 8{1} } = 0}} } {}
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Find the currents flowing in the circuit in [link] . Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors .

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Solve [link] , but use loop abcdefgha instead of loop akledcba. Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors .

(a) I 1 = 4.75 A size 12{I rSub { size 8{1} } =4 cdot "75 A"} {}

(b) I 2 = - 3 . 5 A size 12{I rSub { size 8{"2 "} } = +- 3 "." "5 A"} {} {}

(c) I 3 = 8 . 25 A size 12{I rSub { size 8{3} } =8 "." "25"" A"} {}

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Find the currents flowing in the circuit in [link] .

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Unreasonable Results

Consider the circuit in [link] , and suppose that the emfs are unknown and the currents are given to be I 1 = 5 . 00 A , I 2 = 3 .0 A size 12{I rSub { size 8{2} } =3 "." 0" A"} {} , and I 3 = –2 . 00 A size 12{I rSub { size 8{3} } "=-"2 "." "00"" A"} {} . (a) Could you find the emfs? (b) What is wrong with the assumptions?

The diagram shows a complex circuit with two voltage sources E sub one and E sub two, and three resistive loads, wired in two loops and two junctions. Several points on the diagram are marked with letters a through h. The current in each branch is labeled separately.

(a) No, you would get inconsistent equations to solve.

(b) I 1 I 2 + I 3 size 12{I rSub { size 8{1} }<>I rSub { size 8{2} } +I rSub { size 8{3} } } {} . The assumed currents violate the junction rule.

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Questions & Answers

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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