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Introduction

Kirchhoff’s first law deals with the conservation of charge entering and leaving a junction. We call this law Kirchhoff’s Current Law, or KCL, for short.

What’s a Junction?

A junction is where conductors in a circuit meet, then branch off in different directions within the circuit — kind of like an intersection in a road and how other roads branch off from it. Current follows the path of the conductor until it comes to a junction where current diverges or converges at the junction.

When current comes to a junction in a circuit it initially has some charge. Once that current splits into separate directions at the junction (or at a node) some of the current (or charge) goes one way, and some of the current goes another way in the circuit.

Let’s Make a Rule!

If we were to set up a rule for ourselves and say that any current entering a node we’ll add those values and any current exiting that same node we’ll subtract those values, then set them equal to zero.

Said another way, any current exiting a node we subtract from any current entering the node, and we set this value to zero. Therefore, the sum of the currents entering a node should equal to the sum of the currents exiting the node.

\begin{equation}
\Sigma \; Current \; Entering \; Node \; – \; \Sigma \; Current \; Exiting \; Node \; = \; 0
\end{equation}

\begin{equation}
\Sigma \; Current \; Entering \; Node \; = \; \Sigma \; Current \; Exiting \; Node
\end{equation}

Where Σ is the capital Greek letter “Sigma“, which in mathematics is used to express the summation of values – literally meaning “the sum of”. You can think of the letter Σ telling you, “Just add this stuff together, okay?!”

The conservation of charge (charge is represented by the lowercase letter q) is written as:

\begin{equation}
q_{initial} \; – \; q_{final} \; = \; 0
\end{equation}

\begin{equation}
q_{initial} \; = \; q_{final}
\end{equation}

But, this is not a physics class — we’re talking about current here. Just know that if you have a certain amount of current in a circuit that comes to a node in the circuit where it branches off, that the sum of those currents that branch off or exit the node should equal the current value that entered that node. Don’t worry, next we’ll go into detail with pictures to explain this concept better.

A Single Junction — Current Entering and Exiting a Node

Let’s say we have current I₁ going through a conductor that comes upon a node in a circuit – we’ll call it node A. Current I₁ has a value of 20 milliamps. When this current reaches node A it can split into two separate directions – some current, I₂ for example, can go “down” as seen in the figure above, or some current, I_T for example, can go “straight” as seen in the image below.

If the incoming current I₁ had an initial value of 20mA, then split off some of this value in the form of I₂ and I_T, then, according to the conservation of charge, the sum of the values exiting node A should equal the incoming current I₁ into node A.

We can see from the figure above that the value of I₂ exiting node A has a value of 7mA and the value of I_T, also exiting node A, has a value of 13mA. If we were to add these two values, we should get the value of I₁, the incoming current into node A with a value of 20mA.

\begin{equation}
\Sigma \; Current \; Entering \; Node \; – \; \Sigma \; Current \; Exiting \; Node \; = \; 0
\end{equation}

\begin{equation}
I_1 \; – \; I_2 \; – \; I_T \; = \; 0
\end{equation}

\begin{equation}
I_1 \; = \; I_2 \; + \; I_T
\end{equation}

\begin{equation}
I_1 \; = \; 7mA \; + \; 13mA
\end{equation}

\begin{equation}
\boxed{I_1 \; = \; 20mA}
\end{equation}

Now, let’s say we have current come upon a node in a circuit — we’ll call it node F — where two current values I₁ and I₂ enter the node, as seen in the figure below. The value of I₂ is 9 milliamps and the value of I₁ is 11 milliamps. We can see in the figure below that there is a current I_T that is exiting node F.

According to the conservation of charge, the sum of the currents entering node F — that would be currents I₁ and I₂ — should equal to the value of currents exiting node F — that would be I_T.

\begin{equation}
\Sigma \; Current \; Entering \; Node \; – \; \Sigma \; Current \; Exiting \; Node \; = \; 0
\end{equation}

\begin{equation}
\Sigma \; Current \; Entering \; Node \; = \; \Sigma \; Current \; Exiting \; Node
\end{equation}

\begin{equation}
I_1 \; + \; I_2 \; = \; I_T
\end{equation}

\begin{equation}
11mA \; + \; 9mA \; = \; 20mA
\end{equation}

\begin{equation}
\boxed{20mA \; = \; 20mA} \; \checkmark
\end{equation}

---

Using KCL on a Circuit

If we take a look at the figure of the breadboard circuit above, we can see there are three resistors in parallel powered by a 9 volt battery. The nodes of the circuit are labeled A through F. Also added are 4 ammeters measuring different sections of the circuit, as follows:

• Ammeter 1: 65.3mA – Between the positive terminal of the 9V battery source and the positive side of the circuit.
• Ammeter 2: 8.90mA – Between the positive side of the circuit and the 1kΩ resistor (R1).
• Ammeter 3: 15.9mA – Between the positive side of the circuit and the 560Ω resistor (R2).
• Ammeter 4: 40.5mA – Between the positive side of the circuit and the 220Ω resistor (R3).

Below the image of the breadboard circuit is its schematic showing the 9V battery, the values of each resistor in parallel, and the labeled nodes A through F.

The image below is a closeup of the schematic of the breadboard circuit above, and the currents have been labeled through each section and branch of the circuit. If you’re unfamiliar with circuit branches, I encourage you to check out our post on circuit branches here.

KCL at Node A

As we can see from the schematic image above, if we start from the positive terminal of the 9V battery and go towards node A, in between these two points the battery provides 65.3 milliamps of current – this is the initial circuit current.

This initial current of 65.3mA enters node A and goes into two separate directions from this point. The current either splits from node A and goes toward node F or goes toward node B.

We can see that from node A, current exits the node going towards node F and has a value of 8.9mA – this is the value measured by ammeter 2 in the breadboard circuit shown above – and current I_AB also exits node A going towards node B. If we know that current entering node A is 65.3mA, and current exiting node A is 8.9mA and I_AB, then we can use KCL at node A to calculate the value of I_AB.

\begin{equation}
\Sigma \; Current \; Entering \; Node \; – \; \Sigma \; Current \; Exiting \; Node \; = \; 0
\end{equation}

\begin{equation}
\Sigma \; Current \; Entering \; Node \; = \; \Sigma \; Current \; Exiting \; Node
\end{equation}

\begin{equation}
65.3mA \; = \; 8.9mA \; + \; I_{AB}
\end{equation}

\begin{equation}
65.3mA \; – \; 8.9mA \; = \; I_{AB}
\end{equation}

\begin{equation}
I_{AB} \; = \; 65.3mA \; – \; 8.9mA
\end{equation}

\begin{equation}
\boxed{I_{AB} \; = \; 56.4mA}
\end{equation}

KCL at Node B

Looking at the image above, we can see that current enters node B with the value of 56.4mA – the value we calculated above. We can also see that from node B, current exits the node going toward node E and has a value of 15.9mA – this is the value measured by ammeter 3 in the breadboard circuit shown earlier, above.

We can see above that current I_BC also exits node B going towards node C. If we know that current entering node B is 56.4mA, and current exiting node B is 15.9mA and I_BC, then we can use KCL at node B to calculate the value of I_BC.

\begin{equation}
\Sigma \; Current \; Entering \; Node \; – \; \Sigma \; Current \; Exiting \; Node \; = \; 0
\end{equation}

\begin{equation}
\Sigma \; Current \; Entering \; Node \; = \; \Sigma \; Current \; Exiting \; Node
\end{equation}

\begin{equation}
56.4mA \; = \; 15.9mA \; + \; I_{BC}
\end{equation}

\begin{equation}
56.4mA \; – \; 15.9mA \; = \; I_{BC}
\end{equation}

\begin{equation}
I_{BC} \; = \; 56.4mA \; – \; 15.9mA
\end{equation}

\begin{equation}
\boxed{I_{BC} \; = \; 40.5mA}
\end{equation}

KCL at Node C

Looking at the image above, we can see that current enters node C with the value of 40.5mA – the value we calculated above. We can also see that from node C, current exits the node going toward node D.

Notice that node C only has one exit branch to go through, and that’s through resistor R3 towards node D. We can think of this section of the circuit as being in series – current coming from node B, going toward node C, through node C to node D. Current through a series circuit is the same value throughout – meaning that if current is 40.5mA between nodes B and C, then the current through resistor R3 has to be 40.5mA as well – which is what was measured by ammeter 4 shown in the breadboard circuit below.

KCL at Node D

Looking at the image above, we can see that current enters node D with the value of 40.5mA. We can also see that from node D, current exits the node going toward node E.

Again, like node C, node D only has one exit path for current to go through, and that’s through the conductor (wire) of the circuit towards node E. We can think of this section of the circuit as being in series – current coming from node C, going through R3 toward node D, out of node D through the wire going towards node E. Current through a series circuit is the same value throughout – meaning that if current is 40.5mA between nodes C and D, then the current I_DE through the conductor (wire) between nodes D and E has to be 40.5mA as well – there’s no other way for the 40.5mA of current through this part of the circuit to go, except through node D, toward node E.

\begin{equation}
\boxed{I_{DE} \; = \; 40.5mA}
\end{equation}

KCL at Node E

Looking at the previous image above, we can see that current enters node E from node D with the value of 40.5mA. We can also see that from node B to node E, current enters node E with a value of 15.9mA. Current I_EF exits node E, going towards node F.

If we know that current entering node E is 40.5mA and 15.9mA, and current exiting node E is I_EF, then we can use KCL at node E to calculate the value of I_EF.

\begin{equation}
\Sigma \; Current \; Entering \; Node \; – \; \Sigma \; Current \; Exiting \; Node \; = \; 0
\end{equation}

\begin{equation}
\Sigma \; Current \; Entering \; Node \; = \; \Sigma \; Current \; Exiting \; Node
\end{equation}

\begin{equation}
40.5mA \; + \; 15.9mA \; = \; I_{EF}
\end{equation}

\begin{equation}
I_{EF} \; = \; 40.5mA \; + \; 15.9mA
\end{equation}

\begin{equation}
\boxed{I_{EF} \; = \; 56.4mA}
\end{equation}

KCL at Node F

Looking at the image above, we can see that current enters node F from node E with the value of 56.4mA – the value we calculated above. We can also see that from node A to node F, current enters node F with a value of 8.9mA. Current I_F exits node F, going towards the negative terminal of the battery – completing the circuit.

If we know that current entering node F is 56.4mA and 8.9mA, and current exiting node F is I_F, then we can use KCL at node F to calculate the value of I_F.

\begin{equation}
\Sigma \; Current \; Entering \; Node \; – \; \Sigma \; Current \; Exiting \; Node \; = \; 0
\end{equation}

\begin{equation}
\Sigma \; Current \; Entering \; Node \; = \; \Sigma \; Current \; Exiting \; Node
\end{equation}

\begin{equation}
56.4mA \; + \; 8.9mA \; = \; I_{F}
\end{equation}

\begin{equation}
I_{F} \; = \; 56.4mA \; + \; 8.9mA
\end{equation}

\begin{equation}
\boxed{I_{F} \; = \; 65.3mA} \; \checkmark
\end{equation}

The current I_F is 65.3mA, which is the current we started with – the initial circuit current leaving the positive terminal of the 9V battery, as measured by ammeter 1 shown below.

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Kirchhoff’s Voltage Law (KVL) here.

We hope that you’ve enjoyed this lesson on KCL and that it was beneficial to you. Let us know in the comments if our explanations are helpful. Remember, stay motivated and keep at it!