# 21.3 Kirchhoff’s rules  (Page 2/10)

 Page 2 / 10

## 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 ${I}_{1}={I}_{2}+{I}_{3}$ (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.

## 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, $V$ , rather than potential energy, but the two are related since ${\text{PE}}_{\text{elec}}=\text{qV}$ . 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.

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.

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