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6.3: Kirchhoff's Rules

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    100362
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    Learning Objectives

    By the end of the section, you will be able to:

    • State Kirchhoff’s junction rule.
    • State Kirchhoff’s loop rule.
    • Use Kirchhoff's rules and Ohm's law to calculate the potential changes and current in basic resistor circuit.

    As we start to develop our strategy for analyzing DC resistor circuits, we will need some rules or "laws" that will help us to calculate potential changes and currents in the circuits.

    Even though this circuit cannot be analyzed using the methods already learned, two circuit analysis rules can be used to analyze any circuit, simple or complex. The rules are known as Kirchhoff’s rules, after their inventor Gustav Kirchhoff (1824–1887).

    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: \[\sum I_{in} = \sum I_{out}.\]
    • Kirchhoff’s second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero:
      \[\Delta V_{\mathrm{loop}} = \sum_i (\Delta V)_i = 0.\]

    When combined with Ohm's Law \(\Delta V = IR\), which relates potential changes to current, we will have a sufficient set of tools in our circuit analysis "toolbox" to be able to start doing analysis of some basic DC resistor circuits. 

    Kirchhoff’s Junction Rule

    Kirchhoff’s junction rule applies to the charge entering and leaving a junction (Figure \(\PageIndex{1}\)). As stated earlier, a junction, or node, is a connection of three or more wires. Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out.

    The figure shows a junction with six current branches, four with incoming currents and two with outgoing currents. The sum of incoming currents is equal to sum of outgoing currents.
    Figure \(\PageIndex{1}\): Charge must be conserved, so the sum of currents into a junction must be equal to the sum of currents out of the junction.

    Although it is an over-simplification, an analogy can be made with water pipes connected in a plumbing junction. If the wires in Figure \(\PageIndex{1}\) were replaced by water pipes, and the water was assumed to be incompressible, the volume of water flowing into the junction must equal the volume of water flowing out of the junction.

    Kirchhoff’s Loop Rule

    Kirchhoff’s loop rule applies to potential differences. The loop rule is stated in terms of potential \(V\) rather than potential energy, but the two are related since \(U = qV\). In a closed loop, whatever energy is supplied by a voltage source, the energy must be transferred into other forms by the devices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. Kirchhoff’s loop rule states that the algebraic sum of potential differences, including voltage supplied by the voltage sources and resistive elements, in any loop must be equal to zero.  

    To be able to use Kirchhoff's Loop Rule, we need to have a method to determine which potential changes are positive and which are negative so that the sum around the loop can be zero.

    Tactics: Using Kirchhoff’s Loop Rule with Ohm's Law
    1. Create a circuit diagram. Label points in the circuit diagram using lowercase letters a, b, c, …. These labels simply help with orientation.
    2. Assign a direction to the current.  
      • If the direction of the current is known, then you can choose that direction.
      • If the direction of the current is unknown, then pick a direction.  If the actual direction of the current is opposite to the chosen direction, then the current will simply turn out to be negative. 
    3. Move around the loop in the assigned direction of travel.  Choose a starting point and step through the various components of the circuit in the direction of travel using the rules shown in Fig. \(\PageIndex{2}\) to determine the change in potential across each component. In particular, Ohm's Law is used to determine the potential change across the resistors.
       
      Part a shows voltage difference across a resistor when direction of travel is same as current direction. Part b shows voltage difference across a resistor when direction of travel is opposite to current direction. Part c shows voltage difference across a voltage source when direction of travel is same as current direction. Part d shows voltage difference across a voltage source when direction of travel is opposite to current direction.
      Figure \(\PageIndex{2}\): Each of these resistors and voltage sources is traversed from a to b. (a) When moving across a resistor in the same direction as the current flow, subtract the potential drop. (b) When moving across a resistor in the opposite direction as the current flow, add the potential drop. (c) When moving across a voltage source from the negative terminal to the positive terminal, add the potential drop. (d) When moving across a voltage source from the positive terminal to the negative terminal, subtract the potential drop.
    4. Apply the loop rule.  Add up all the potential changes according to \( \sum_i (\Delta V)_i = 0 \).

    Basic Circuit Analysis

    Example \(\PageIndex{1}\) shows an example of a basic circuit that illustrates how Kirchhoff's loop rule can be used along with the tactics described above.

    Example \(\PageIndex{1}\): Analyzing a Basic Resistor Circuit

    A simple loop with no junctions includes a 12.00 V battery, a 1.00 ohm resistor, a 2.00 ohm resistor, and a 3.00 ohm resistor connected in series. What is the current flowing in the circuit?

    Solution
    PLAN

    We will model the battery and connecting wires between the battery and resistors as ideal (negligible resistance).  We will then use Kirchhoff's rules and Ohm's Law to calculate the current.

    SKETCH

    1. Create a circuit diagram. 

    Figure \(\PageIndex{2}\) shows an example of a schematic diagram of the circuit.

    alt
    Figure \(\PageIndex{2}\): A simple loop with no junctions. Kirchhoff’s loop rule states that the algebraic sum of the voltage differences is equal to zero.

    The circuit consists of a voltage source and three external load resistors. The labels a, b, c, and d serve as references, and have no other significance. The usefulness of these labels will become apparent soon. The loop is designated as Loop abcda, and the labels help keep track of the voltage differences as we travel around the circuit.

    2. Assign a direction to the current.  

    In this circuit, we can see that the current (conventional current or flow of positive charge) should flow from the positive side of the battery to the negative side, resulting in a clockwise current flow as seen from the front side of the circuit.  We therefore choose the direction of the current to be clockwise.

    3. Move around the loop in the assigned direction of travel. 

    We can start anywhere in the circuit.  We choose to start at Point a and travel clockwise in the loop:

    • Traveling from Point a to Point b: According to Fig. \(\PageIndex{2}\mathrm{(c)}\), the potential change of the battery is positive (\(\Delta V_{\mathrm{battery}} = +V\)) as we go from the negative side of the battery to the positive side of the battery in the direction of travel.   According to Fig. \(\PageIndex{2}\mathrm{(a)}\), the potential change across the resistor \(R_1\) is a negative (\(\Delta V_1 = -IR_1\)).
    • Traveling from Point b to Point c: According to Fig. \(\PageIndex{2}\mathrm{(a)}\), the potential change across the resistor \(R_2\) is a negative (\(\Delta V_2 =-IR_2\)). 
    • Traveling from Point c to Point d: According to Fig. \(\PageIndex{2}\mathrm{(a)}\), the potential change across the resistor \(R_3\) is a negative (\(\Delta V_3 =-IR_3\)). 
    • Traveling from Point d to Point a: The connecting wire is modeled as ideal so there is no change in potential.

    Figure \(\PageIndex{3}\) shows a graph of the voltage as we travel around the loop. Voltage increases as we cross the battery, whereas voltage decreases as we travel across each resistor. The potential drop, or change in the electric potential, is equal to the current through the resistor times the resistance of the resistor following Ohm's Law. In this example, we are modeling both the battery and connecting wires as ideal, taking them to have negligible (zero) resistance. Under these assumptions, the change in voltage across the battery's terminal is the battery's listed source voltage (\(V = +12\,\mathrm{V}\)), and the voltage remains constant as we cross the wires connecting the components \(\Delta V_{\mathrm{wire}} = 0\).

    The graph shows voltage at different points of a closed loop circuit with a voltage source and three resistances. The points are shown on x-axis and voltages on y-axis
    Figure \(\PageIndex{4}\): A voltage graph as we travel around the circuit. The voltage increases as we cross the battery and decreases as we cross each resistor. Since the resistance of the wire is quite small, we assume that the voltage remains constant as we cross the wires connecting the components (ideal wire model).

    4. Apply the loop rule.  We will do this in the following calculation.

    CALCULATE

    A. Kirchhoff's Junction Rule

    We first observe that the circuit does not have any junctions, so that means that the current must be a constant value throughout the circuit.  In other words, \(I_1 = I_2 = I_3 =I\).

    B. Kirchhoff's Loop Rule

    The circuit has one loop. Kirchhoff’s loop rule then states that 

    \[\Delta V_{\mathrm{battery}} + \Delta V_1 + \Delta V_2 + \Delta V_3 = 0.\]

    C. Ohm's Law

    Using the tactics box and Ohm's Law, it follows from Kirchhoff's Loop Rule that 

    \[V - IR_1 - IR_2 - IR_3 = 0.\]

    The loop equation can be used to find the current through the loop:

    \[I = \frac{V}{R_1 +R_2 +R_3} = \frac{12.00 \, \mathrm{V}}{1.00 \, \Omega + 2.00 \, \Omega + 3.00 \, \Omega} = 2.00 \, \mathrm{A}.\]

    CHECK

    We can check to see that the Loop Rule is satisfied by calculating the potential drop across each resistor using Ohm's Law:

    \[ \Delta V_1 = -(2.00 \, \mathrm{A})(1.00\,\Omega) = -2.00\,\mathrm{V} \]

    \[ \Delta V_2 = -(2.00 \, \mathrm{A})(2.00\,\Omega) = -4.00\,\mathrm{V} \]

    \[ \Delta V_3 = -(2.00 \, \mathrm{A})(3.00\,\Omega) = -6.00\,\mathrm{V} \]

    We can then see that \(\Delta V_{\mathrm{battery}} + \Delta V_1 + \Delta V_2 + \Delta V_3 = +12.00\,\mathrm{V} - 2.00\,\mathrm{V} - 4.00\,\mathrm{V} - 6.00\,\mathrm{V} = 0 \), and Kirchhoff's Loop Rule is satisfied.

    We can see from this example that by using each of the tools in our toolbox, we can analyze the circuit and thereby calculate the current through and potential drop across each component in the circuit.

    Exercise \(\PageIndex{1}\)

    A simple resistor circuit includes a \(9.00\, \mathrm{V}\) battery, a \(150\text{-}\Omega\) resistor and a \(250\text{-}\Omega\) ohm resistor connected in series. 
    (a) What is the current flowing in the circuit? 
    (b) What are the potential drops across each resistor?

    Answer

    PLAN 

    Model the battery and connecting wires as ideal.  Use Kirchhoff's rules and Ohm's law to analyze the circuit.

    SKETCH

    Create a circuit diagram like Fig. \(\PageIndex{2}\) but with only the battery and two resistors in series.  The current should flow in the direction from the positive side of the battery to the negative side.  Use the tactics box to determine the potential changes across each circuit element.

    CALCULATE   

    A. Kirchhoff's Junction Rule

    We first observe that the circuit does not have any junctions, so that means that the current must be a constant value throughout the circuit.  In other words, \(I_1 = I_2  =I \).

    B. Kirchhoff's Loop Rule

    The circuit has one loop. Kirchhoff’s loop rule then states that 

    \[\Delta V_{\mathrm{battery}} + \Delta V_1 + \Delta V_2  = 0.\]

    C. Ohm's Law

    Using the tactics box and Ohm's Law, it follows from Kirchhoff's Loop Rule that 

    \[V - IR_1 - IR_2 = 0.\]

    (a) The loop equation can be used to find the current through the loop:

    \[I = \frac{V}{R_1 +R_2} = \frac{9.00 \, \mathrm{V}}{150 \, \Omega + 250 \, \Omega } = 0.0225 \, \mathrm{A}.\]

    (b) The potential drops across each resistor can be calculated by Ohm's Law:

    \[ \Delta V_1 = -(0.0225 \, \mathrm{A})(150\,\Omega) = -3.375\,\mathrm{V}, \]

    \[ \Delta V_2 = -(0.0225 \, \mathrm{A})(250\,\Omega) = -5.625\,\mathrm{V}. \]

    CHECK

    The Loop Rule is satisfied because \(\Delta V_{\mathrm{battery}} + \Delta V_1 + \Delta V_2 = +9.00\,\mathrm{V} - 3.375\,\mathrm{V} - 5.625\,\mathrm{V} = 0 \).

    We will see an example of a basic circuit with junctions in the next section.

    Contributors and Attributions

    Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).


    6.3: Kirchhoff's Rules is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ronald Kumon.