Kirchhoffs Current Law (KCL) is Kirchhoff’s first law that deals with the conservation of charge entering and leaving a junction.

## Kirchhoffs Current Law

To ascertain the magnitude of electrical current within an electrical or electronic circuit, it is essential to apply specific laws or principles that enable the representation of these currents in the form of equations. The governing equations employed are based on Kirchhoff’s laws, particularly focusing on **Kirchhoffs Current Law** (KCL) when analyzing circuit currents.

Gustav Kirchhoff’s Current Law stands as a fundamental principle in circuit analysis. According to this law, the total current entering a junction in a parallel path equals precisely the total current leaving the same junction. This equivalence is attributed to the absence of alternative paths for the current, preventing any loss of charge.

In simpler terms, the algebraic sum of all currents entering and exiting a junction must be zero, expressed as Σ IIN = Σ IOUT. Kirchhoff’s concept, known as the Conservation of Charge, underscores the idea that current is conserved around the junction without any loss. Let’s examine a straightforward example of **Kirchhoffs Current Law** (KCL) when applied to a single junction.

### A Single Junction

In this straightforward example involving a single junction, the total current (I_{T}) leaving the junction is the algebraic sum of the two currents, I_{1} and I_{2}, entering the same point. This relationship can be expressed as I_{T} = I_{1} + I_{2}.

Alternatively, it can be correctly formulated as the algebraic sum: I_{T} – (I_{1} + I_{2}) = 0.

For instance, if I_{1} is 3 amperes and I_{2} is 2 amperes, the total current, I_{T}, leaving the junction will be 3 + 2 = 5 amperes. This basic law can be applied to any number of junctions or nodes, as the sum of currents entering and leaving remains constant.

Moreover, if we reverse the directions of the currents, the equations still hold true for I_{1} or I_{2}. For example, I_{1} = I_{T} – I_{2} = 5 – 2 = 3 amps, and I_{2} = I_{T} – I_{1} = 5 – 3 = 2 amps. In this context, currents entering the junction are considered positive (+), while those leaving are negative (-).

Therefore, the mathematical sum of currents, whether entering or leaving the junction and regardless of direction, always equals zero. This principle forms the foundation of Kirchhoff’s Junction Rule, more commonly known as **Kirchhoffs Current Law** (KCL).

### Resistors in Parallel

Let’s examine how Kirchhoffs current law can be employed in the context of parallel resistors, whether the resistances in these branches are identical or different. Take a look at the presented circuit diagram:

In this straightforward parallel resistor illustration, there are two separate junctions for current flow. The first junction is located at node B, and the second junction is situated at node E. Consequently, we can apply Kirchhoff’s Junction Rule to the electrical currents at both of these distinct junctions, considering currents entering and leaving each junction.

Initially, all the current (I_{T}) departs from the 24-volt supply, reaches point A, and then enters node B. Node B serves as a junction where the current can divide into two distinct paths. Some of the current flows downward, passing through resistor R_{1}, while the remaining portion continues its path through resistor R_{2} via node C. It is worth noting that the currents entering and leaving a node point are commonly referred to as branch currents.

We can use Ohm’s Law to determine the individual branch currents through each resistor as: I = V/R, thus:

**For current branch B to E through resistor R _{1}:**

\mathbf{I_{B-E} = I_1 = \frac{V}{R_1} =\frac{24}{8}=3A}

**For current branch C to D through resistor R _{2}:**

\mathbf{I_{C-D} = I_2 = \frac{V}{R_2} =\frac{24}{12}=2A}

Based on the principles of Kirchhoffs current law, it asserts that the total current entering a junction must be equal to the total current leaving the junction. In the aforementioned example, there is one current, I_{T}, entering the junction at node B, while two currents, I_{1} and I_{2}, are leaving the junction.

Upon calculation, it is determined that the currents leaving the junction at node B are I_{1} (3 amps) and I_{2} (2 amps). Consequently, the sum of the currents entering the junction at node B is required to be 3 + 2 = 5 amps, denoted as Σ_{IN} = I_{T} = 5 amperes.

In our scenario with distinct junctions at node B and node E, the value for I_{T} is reaffirmed as the two currents recombine at node E. To satisfy Kirchhoff’s junction rule, the sum of the currents into point F must equal the sum of the currents flowing out of the junction at node E.

Considering the currents entering junction E as 3 amps and 2 amps respectively, the sum of the currents entering point F is established as 3 + 2 = 5 amperes. Consequently, Σ_{IN} = I_{T} = 5 amperes, reaffirming the validity of Kirchhoffs current law, as this aligns with the current leaving point A.

We could have solved the circuit of example two simply and easily just using Ohm’s Law, but we have used * Kirchhoffs Current Law* here to show how it is possible to solve more complex circuits when we can not just simply apply Ohm’s Law.

Read more Tutorials on DC Circuits | |
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1. | DC Circuit Theory |

2. | Ohms Law and Power |

3. | Electrical Units of Measure |

4. | Kirchhoffs Circuit Law |

5. | Kirchhoffs Current Law |

6. | Kirchhoffs Voltage Law |

7. | Mesh Current Analysis |