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  • Analyze a complex circuit using Kirchhoff’s rules, using the conventions for determining the correct signs of various terms.

Many complex circuits, such as the one in [link] , cannot be analyzed with the series-parallel 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    , after their inventor Gustav Kirchhoff (1824–1887).

A complicated circuit diagram shows multiple resistances and voltage sources wired in series and in parallel. The circuit has three arms. The first has a cell of e m f script E sub one and internal resistance r sub one in series with a resistor R sub two. The second has a cell of e m f script E sub two and internal resistance r sub two in series with resistor R sub three. The third arm has a resistor R sub one. The three arms are connected in parallel.
This circuit cannot be reduced to a combination of series and parallel connections. Kirchhoff’s rules, special applications of the laws of conservation of charge and energy, can be used to analyze it. (Note: The script E in the figure represents electromotive force, emf.)

Kirchhoff’s rules

  • 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.

Explanations of the two rules will now be given, followed by problem-solving hints for applying Kirchhoff’s rules, and a worked example that uses them.

Kirchhoff’s first rule

Kirchhoff’s first rule (the junction rule    ) is an application of the conservation of charge to a junction; it is illustrated in [link] . Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out. Kirchhoff’s first rule requires that I 1 = I 2 + I 3 size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {} (see figure). Equations like this can and will be used to analyze circuits and to solve circuit problems.

Making connections: conservation laws

Kirchhoff’s rules for circuit analysis are applications of conservation laws    to circuits. The first rule is the application of conservation of charge, while the second rule is the application of conservation of energy. Conservation laws, even used in a specific application, such as circuit analysis, are so basic as to form the foundation of that application.

This schematic drawing shows a T-junction, with one current I sub one flowing into the T and two currents I sub two and I sub three flowing out of the T junction.
The junction rule. The diagram shows an example of Kirchhoff’s first rule where the sum of the currents into a junction equals the sum of the currents out of a junction. In this case, the current going into the junction splits and comes out as two currents, so that I 1 = I 2 + I 3 size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {} . Here I 1 size 12{I rSub { size 8{1} } } {} must be 11 A, since I 2 size 12{I rSub { size 8{2} } } {} is 7 A and I 3 size 12{I rSub { size 8{3} } } {} is 4 A.

Kirchhoff’s second rule

Kirchhoff’s second rule (the loop rule    ) is an application of conservation of energy. The loop rule is stated in terms of potential, V size 12{V} {} , rather than potential energy, but the two are related since PE elec = qV size 12{ ital "PE" rSub { size 8{"elec"} } = ital "qV"} {} . Recall that emf is the potential difference of a source when no current is flowing. In a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. [link] illustrates the changes in potential in a simple series circuit loop.

Questions & Answers

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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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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