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Kirchhoff's Current Law

Understanding Nodes in a Circuit

tag Circuits KCL Current

Kirchhoff's Current Law:

Understanding Nodes in a Circuit

Introduction

Related Material:
Lumped Parameters Circuits and Nodes
Current

Kirchhoff's current law and Kirchhoff's voltage law are the basis for analysis of lumped parameter circuits. These laws, together with the voltage-current characteristics of the circuit elements in the system, provide us with the ability to perform a systematic analysis of any electrical network. This section presents Kirchhoff's current law.

  • Kirchhoff's current law (commonly abbreviated in these exercises as KCL) states:
    • The algebraic sum of all currents entering (or leaving) a node is zero.
  • A common alternate statement for KCL is:
    • The sum of the currents entering any node equals the sum of the currents leaving the node.
  • Or even:
    • What goes into a node must come out again.

Kirchhoff's current law depends upon the concept of a node. A node is a point in a circuit where two or more circuit elements are interconnected. Nodes are discussed in the link provided to the right. Let's illustrate application of KCL with a couple of examples.

Note

  • Current directions at the node are based on assumed directions of the currents. As long as the assumed directions of the currents are consistent from node to node, the final result of the analysis will reflect the actual current directions in the circuit.

Example 1:

In the figure below, the assumed directions of i1(t), i2(t) and i3(t) are as shown.

If we (arbitrarily) choose a sign convention such that currents entering the node are positive, then currents leaving the node are negative and KCL applied at this node results in:

  • \({i_1}(t) + {i_2}(t) - {i_3}(t) = {\rm{ }}0\)

If, on the other hand, we choose a sign convention that currents entering the node are negative, then currents leaving the node are positive and KCL applied at this node results in:

  • \( - {i_1}(t) - {i_2}(t) + {i_3}(t) = {\rm{ }}0\)

These two equations are the same; the second is just the negative of the first! It doesn't matter what sign convention we use in KCL, as long as it's consistent. Both of the above equations are equivalent to the statement:

  • \({i_1}(t) + {i_2}(t) = {i_3}(t)\)

This amounts to the statement that the total current entering the node is the same as the total current leaving the node. The terms “entering” and “leaving” are based on the assumed directions; the actual directions of the currents will work themselves out when all is said and done.

Example 2:

Use KCL to determine the value of the current, i , in the figure below.

Approach 1: Sum the currents, assuming positive currents are entering the node:

\[4A - ( - 1A) - 2A - i = 0 \Rightarrow 4A + 1A - 2A = 3A\]
  • i = 3A , leaving the node (since our assumed direction of i on the diagram is that it is leaving the node, and our numerical value for i is positive).

Approach 2: Sum the currents, assuming positive currents are leaving the node:

\[ - 4A + ( - 1A) + 2A + i = 0 \Rightarrow i = 4A + 1A - 2A = 3A\]
  • So i = 3A , just as in approach 1. Thus, changing our assumption about the sign convention has not affected our results.

Example 3:

Let's switch our assumed direction for the current i in example 2, and again use KCL to determine the value of the current. The appropriate diagram is shown below.

Now, applying KCL again, assuming positive currents are entering the node:

\[4A - ( - 1A) - 2A + i = 0 \Rightarrow i = - 4A - 1A + 2A = - 3A\]

The negative sign on i means that the current is in the opposite direction to that shown on the diagram. Thus, the current is 3A leaving the node, and we come to the same result as in Example 2.

We can generalize Kirchhoff's current law to include any enclosed portion of a circuit. To illustrate this concept, consider the portion of a larger circuit enclosed by a surface as shown in Fig. 1 below. Since none of the circuit elements within the surface can accumulate charge, the total charge which can be stored within any enclosed surface is zero. Thus, the net charge entering an enclosed surface must be zero. This leads to a generalization of our previous statement of KCL:

  • The algebraic sum of all currents entering (or leaving) any enclosed surface is zero.

Applying this statement to the circuit of Fig. 1 results in:

  • \({i_1} + {i_2} + {i_3} = {\rm{ }}0\)

Figure 1. KCL applied to a closed circuit.

Important Points:

  • Kirchhoff's current law states that the sum of the current entering (or leaving) a node must be zero. A node in a circuit is any point at which two or more circuit elements are interconnected.

  • Kirchhoff's current law can be generalized to state that the sum of the current entering (or leaving) any enclosed surface is zero.

Test Your Knowledge!

  1. What is the current I in the circuit below?
  2. What is the current I in the circuit below? (Hint: this is a trick question.)
  3. What are the currents I1 and I2 in the circuit below?
  4. What are the currents I1 and I2 in the circuit below?

Selected Answers

  1. We will identify the nodes in the circuit as A and B, as shown below.
    • KCL at A, with currents entering the node assumed to be positive, results in: \(2A{\rm{ }} - {I_1} = {\rm{ }}0\). So \({I_1} = {\rm{ }}2A\). (Since the value is positive, the indicated direction is the actual current direction.)
    • KCL at B, with currents entering the node assumed to be positive, results in \({I_1} + {I_2} = {\rm{ }}0\). Since we previously found that \({I_1} = {\rm{ }}2A\), this means that \({I_1} = - 2A\), and the current I2 is actually in the opposite direction to that shown.
  2. We don't have enough information to determine I, since we don't know anything about the current through an ideal voltage source. To do the problem, we would define a current Is through the source and do KCL at node A, as indicated below.
    • KCL at node A, with positive currents assumed to be leaving the node results in \(I + {I_s} = 0\). Since we don't know anything about Is , or about the circuit element represented by the grey box in the figure, we are unable to figure out what I is.
  3. Let's first identify two nodes in the circuit as A and B, as shown on the figure below.
    • Now, applying KCL at node A, and assuming that positive currents leave the node, we get:
      • \(1A{\rm{ }} + {I_1} = {\rm{ }}0\)   So   \({I_1} = {\rm{ }} - 1A.\)
    • Now, apply KCL at node B, again assuming that currents leaving the node are positive. We get:
      • \( - {I_1} - {I_2} - 2A{\rm{ }} = {\rm{ }}0.\)
    • Since we've previously found that \({I_1} = {\rm{ }} - 1A\), this becomes \( - \left( { - 1A} \right) - {I_2} - 2A = 0\), so that \({I_2} = - 1A.\)
    • The above equations are not the only possible approach to the problem. For example, if we do KCL at the bottommost node in the circuit (assuming now—arbitrarily—that positive currents enter the node) we get:
      • \(1A - {I_2} - 2A{\rm{ }} = {\rm{ }}0.\)
    • Solving for I2 gives \({I_2} = - 1A,\) as before.
  4. First, let's identify the nodes in the circuit. There are four of them, which are labeled on the schematic below. When applying KCL for this problem, we will (arbitrarily) set a sign convention that currents leaving the node are positive1.
    • KCL at node A gives: \( - {I_1} + {\rm{ }}4A = 0\),   so   \({I_1} = {\rm{ }}4A.\)
    • KCL at node B gives: \( - 4A - {I_2} - 6A = 0\),   so   \({I_2} = - 10A.\)

    These solutions are fine, as far as they go, but let's try checking our results by applying KCL at nodes C and D:

    • KCL at node C gives: \({I_1} - 2A + 3A = 0\),   so now   \({I_1} = - 1A\)
    • KCL at node D gives: \(2A + {I_2} + 6A - 3A = 0\),   so now   \({I_2} = - 5A\)!

    Our results aren't consistent for all nodes, so there is something wrong with this problem! To help identify what's going on, let's try defining a couple of enclosed surfaces, as shown below.

    KCL applied to either of these surfaces results in: \(2A - 4A - 3A = 0\). This is not true, so the given currents in the problem are fundamentally inconsistent with Kirchhoff's current law—we can't expect to be able to solve the problem!

  • 1Try re-doing the problem with the opposite convention. You should get the same answers.
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