When first learning the basics of circuit diagrams, we usually consider simple series and parallel circuits. However, if you were to open up any old household appliance, you'd likely see a seemingly indecipherable mess of wires and electrical components. Most appliances are formed from extremely complex circuits that are neither series nor parallel circuits, where calculating quantities with \(V=IR\) would take far too long. So, how are such circuits designed and analyzed in practice? Thankfully, there are two laws of circuitry, known as Kirchhoff's Laws, which can be applied to any circuit diagram no matter how complex. In this article, we'll take a deep dive into the first of these laws, known as Kirchhoff's Junction Rule.
Millions of flashcards designed to help you ace your studies
Sign up for freeThree currents intersect at a junction. If \(I_1=5\,\mathrm{A}\) and \(I_2=-3\,\mathrm{A}\), what is \(I_3\)?
Which of these statements is true about series circuits?
Consider a junction containing four branches. Which of these is not a possible combination of currents?
Consider a branch of a circuit with a junction at either end. If the sign of the current through the branch is \(+\) in one junction, what is the sign of the current through the other junction?
Consider a junction where 5 currents intersect. Is it possible that every current has the same magnitude?
Of which fundamental conservation law is Kirchhoff's Junction Rule a consequence?
Kirchhoff's Loop Rule is a consequence of which fundamental conservation law?
The number of branches carrying current into a junction must equal the number of branches carrying current away from the junction.
Three currents intersect at a junction. If \(I_1=5\,\mathrm{A}\) and \(I_2=-3\,\mathrm{A}\), what is \(I_3\)?
Which of these statements is true about series circuits?
Consider a junction containing four branches. Which of these is not a possible combination of currents?
Consider a branch of a circuit with a junction at either end. If the sign of the current through the branch is \(+\) in one junction, what is the sign of the current through the other junction?
Consider a junction where 5 currents intersect. Is it possible that every current has the same magnitude?
Of which fundamental conservation law is Kirchhoff's Junction Rule a consequence?
Kirchhoff's Loop Rule is a consequence of which fundamental conservation law?
The number of branches carrying current into a junction must equal the number of branches carrying current away from the junction.
Achieve better grades quicker with Premium
Geld-zurück-Garantie, wenn du durch die Prüfung fällst
Review generated flashcards
Start learning or create your own AI flashcards
Team Kirchhoff's Junction Rule Teachers
In 1845, a Prussian physics student, Gustav Kirchhoff, investigated the behavior of currents and potential differences in circuits containing multiple loops and nodes or junctions. Applying Ohm's Law to these circuits, he managed to generalize the results into two simple laws which are, to this day, incredibly useful in the field of electrical engineering. Their main use is in reducing incredibly complex circuit diagrams into sets of simple algebraic expressions.
Fig. 1 - Gustav Kirchoff (1824-1887) was a Prussian physicist who made important contributions to the understanding of electrical circuits, most notably his two laws.
The first of these laws, known as Kirchhoff's Junction Rule, concerns current flowing in and out of a circuit's junction and will be the focus of this article. Recall that the current within a wire is simply the rate at which charge flows through that wire. It is common for circuits to include junctions, as shown in figure 2, whereby the current in a wire splits off into two or more wires. Kirchhoff's Junction Rule ensures that the current flowing into a junction is always the same as the current leaving a junction.
Kirchhoff's Junction Rule states that the sum of the currents meeting at a junction of a circuit must always be zero.
Fig. 2 - An example of a junction in a circuit. Kirchoff's junction rule states that \(i_2+i_3=i_1+i_4\).
If a set of currents \(I_1,I_2,\dots,I_k\) meet at a junction, then Kirchhoff's Junction Rule can be expressed algebraically by assigning a \(\pm\) sign to each current depending on whether they are entering or leaving the junction. By convention, currents entering a junction get a \(+\) sign, and those leaving get a \(-\) sign. Kirchhoff's Junction Rule can then be expressed as the equation
\[\sum_{k,\text{entering}} I_k-\sum_{l,\text{leaving}}I_l=0\,\mathrm{A}.\]
For example, in figure 2, \(i_2\) and \(i_3\) are entering the junction and \(i_1\) and \(i_4\) are leaving the junction, so
\[\begin{align}&i_2+i_3-i_1-i_4=0\,\mathrm{A}\,\\\iff &i_2+i_3=i_1+i_4,\end{align}\]
demonstrating that the sum of the currents entering the junction must be equal to the sum of the currents leaving the junction. Conversely, if the direction of a current is not known, it can be determined by Kirchhoff's Junction Rule by assuming it is entering or leaving the junction and finding its sign. If its value is negative, the assumption was wrong.
As a simple example, consider the currents in a series circuit as shown in figure 3. We have to be mindful of how we define the sign of the current at each junction as the same current can get a different sign depending on which junction we consider. If the current flows clockwise from the positive to the negative terminal, then we can assign a positive sign to the currents entering the junction and negative signs to those leaving the junction. In figure 3, we can say that each corner is a junction, so each 'junction' only has one current entering and one current leaving, so Kirchhoff's Junction rule tells us the following must hold\[\begin{align} &I_1-I_2=0\,\mathrm{A},\\ &I_2-I_3=0\,\mathrm{A},\\ &I_3-I_4=0\,\mathrm{A},\\&I_4-I_1=0\,\mathrm{A},\\\implies &I_1=I_2=I_3=I_4.\end{align}\]
In other words, the current in a series circuit is the same at all points of the circuit.
Fig. 3 - Kirchoff's Junction rule requires the current in a series circuit to be equal everywhere, i.e \(I_1=I_2=I_3=I_4\) in the circuit above.
Let's take a look at some ways we can use Kirchhoff's Junction rule to analyze circuit diagrams and calculate missing currents.
Fig. 4 - Kirchhoff's Junction Rule allows us to solve circuits like the one above without knowing the voltage of the battery.
Consider the circuit diagram given above, where we want to find the missing branch currents \(I_1,I_2,I_3\). Each branch contains a resistor of different resistance. Immediately, Kirchhoff's Junction rule tells us that \(I_1=5\,\mathrm{A}\) as the current flowing into the battery must be the same as the current flowing out of the battery (we can pick the battery as a junction and see this directly). Next, we need to find the currents \(I_2\) and \(I_3\). Looking at the right junction and assuming that both currents \(I_2\) and \(I_3\) go from right to left, the junction rule states that \[\begin{align}I_1-I_2-I_3&=0\,\mathrm{A},\\\implies I_2+I_3&=5\,\mathrm{A}.\end{align}\]
We can use \(V=IR\), noting that the potential in parallel circuits is the same across branches, to find the proportion of current which flows into each branch. As current is inversely proportional to resistance, twice as much current will flow through the \(2\,\mathrm{\Omega}\) resistor as the \(4\,\mathrm{\Omega}\) resistor. Hence, \(I_3=2I_2\) and so \[\begin{align}3I_2&=5\,\mathrm{A},\\\implies I_2&=\frac{5}{3}\,\mathrm{A},\\\implies I_3&=\frac{10}{3}\,\mathrm{A}.\end{align}\]
As we get positive values for the unknown currents, we know that our assumption of their direction was correct, so we conclude that the currents indeed run from right to left.
Fig. 5 - Kirchoff's Junction rule is particularly useful with more complex circuit diagrams where \(V=IR\) becomes trickier to utilise.
Whilst using \(V=IR\) to find missing quantities is usually the simplest method for solving circuits, if a circuit is particularly complex such as the one given in figure 5 then Kirchhoff's Junction Rule is the most effective method. Let's apply it to find the missing currents \(I_1,I_2,I_3\).First, we can apply the Junction Rule to find \(I_1\). As \(I_1\) exits a junction into which a \(3\,\mathrm{A}\) and \(6\,\mathrm{A}\) current enter, we know that \[\begin{align}&-I_1+3\,\mathrm{A}+6\,\mathrm{A}=0\,\mathrm{A},\\\implies &I_1=9\,\mathrm{A}.\end{align}\]
As expected, the positive value we get for \(I_1\) tells us that our assumption was correct so that the current \(I_1\) must flow out of the junction, from right to left.
At the next junction, applying the correct sign for the current's direction as given in the circuit diagram, we find\[\begin{align}I_1-I_2-2\,\mathrm{A}&=0\,\mathrm{A},\\\implies I_2=I_1-2\,\mathrm{A}&=7\,\mathrm{A}.\end{align}\]
Again, as expected, the positive value confirms that \(I_2\) flows out of the junction, downwards.
There are two junctions we can choose to find \(I_3\), let's choose the lower junction. It may seem that we don't know what the third current at this junction is, along with \(I_2\) and \(I_3\). However, note that there are no further junctions between this one and the \(3\,\mathrm{A}\) current coming from the lower battery and so this is the third current involved. Applying the junction rule assuming \(I_3\) is left to right (as indicated in the image) gives\[\begin{align}I_2+I_3-3\,\mathrm{A}&=0\,\mathrm{A},\\\implies I_3=3\,\mathrm{A}-7\,\mathrm{A}&=-4\,\mathrm{A}.\end{align}\]
Hold on, we got a negative value! That means that our assumption of the direction of \(I_3\) was wrong. We conclude that \(I_3=4\,\mathrm{A}\) but its direction is actually right to left. The image is wrong!
This is one of the most useful properties of Kirchhoff's Laws: they can correct any incorrect initial assumptions about the direction of a current.
Kirchhoff's Junction Rule is simply a statement about the conservation of current within a circuit. Conservation of current is itself a consequence of the fundamental law of charge conservation. As a fundamental conservation law, no system can ever violate charge conservation.
The law of conservation of current states that, at a fixed voltage and resistance, current cannot be created or destroyed within the circuit.
The law of conservation of electric charge states that the total electric charge on an isolated system, i.e., the sum of all the negative and positive charges in a system, must always remain constant. For example, the total charge in the universe has remained constant since the Big Bang.
To illustrate the relationship between the conservation of charge and Kirchhoff's Junction Rule, let's imagine a scenario where the junction rule does not hold.
Consider, three currents \(I_1,I_2,I_3\) meeting at a junction, where \(I_1,I_2\) enter the junction and \(I_3\) leaves it. Let's say
\[I_1+I_2-I_3=x\,\mathrm{A},\text{ where } x\neq 0\,\mathrm{A}.\]
By the definition of current, this means that every second, \(x\,\mathrm{C}\) of charge is spontaneously destroyed at the junction. Clearly, this violates the fundamental law of charge conservation, so Kirchhoff's Junction Rule must always hold.
Kirchhoff's Second Law, known as Kirchhoff's Loop Rule, regards the sum of potential differences around a loop in a closed circuit. When used in tandem with Kirchhoff's Junction Rule, it becomes a powerful tool for analyzing complex circuits and finding unknown quantities such as potential differences, resistances, and currents.
Kirchhoff's Loop Rule states that the sum of potential differences around any loop in a circuit must be zero:\[\sum_kV_k=0\,\mathrm{V}.\]
Kirchhoff's Loop rule can be seen as a consequence of the conservation of energy. As most parallel circuits contain multiple loops, there is a freedom to choose the simplest loop to apply Kirchhoff's Loop Rule, which often drastically simplifies problems. To apply the loop rule, we consider batteries as sources of positive potential difference, whilst components, such as resistors, are sources of negative potential difference. Let's take a look at an example problem where we can apply both rules to find missing quantities.
Fig. 6 - Applying both of Kirchoff's laws is required to solve the above circuit.
In this example, we're going to look at a somewhat complex circuit, containing resistors and a capacitor in parallel. Here, the capacitor is at steady-state, meaning no current is flowing through it. There is however a build up of charge \(Q\) on the capacitor given by \[Q=CV,\]
where \(C=5\times10^{-9}\,\mathrm{F}\) is the capacitance and \(V\) is the voltage which we do not know yet.
We can apply both of Kirchhoff's Laws to find the missing potential differences and currents in the circuit above, also allowing us to find the charge on the capacitor.
Firstly, Kirchhoff's Junction rule tells us that the currents \(I_1\) and \(I_2\) entering and the current \(I_3\) leaving the junction on the right must satisfy\[\begin{align}I_1+I_2-I_3&=0\,\mathrm{A},\\\implies I_1+I_2&=I_3.\end{align}\]
The left junction is simply the same case, with the signs of the currents reversed, leading to an equivalent equation.
Kirchhoff's Loop Rule gives us two further conditions, from which we can solve all the unknown variables. We can choose a few different loops, but the simplest option is to split the circuit into two main loops, top and bottom, both bypassing the capacitor. We know that the sum of the potential differences around each loop must be zero, giving the following equations.\[\begin{align}5\,\mathrm{V}-V_1-V_2&=0\,\mathrm{V},\\3\,\mathrm{V}-V_2-V_3&=0\,\mathrm{V}.\end{align}\]
We can express the unknown potential differences in terms of the currents and resistances of the resistors using \(V=IR\), which, when combined with equations from Kirchhoff's Junction Rule, forms a set of solvable simultaneous equations.
\[\begin{align}&I_1+I_2=I_3\tag{1},\\&5\,\mathrm{V}-(3\,\mathrm{\Omega})I_1-(1\,\mathrm{\Omega})I_3=0\,\mathrm{V}\tag{2},\\&3\,\mathrm{V}-(1\,\mathrm{\Omega})I_3-(4\,\mathrm{\Omega})I_2=0\,\mathrm{V}\tag{3}.\end{align}\]
If we divide the last two equations by the unit \(\Omega\), we get three current equations:
\[\begin{align}&I_1+I_2=I_3\tag{1},\\&5\,\mathrm{A}-3I_1-I_3=0\,\mathrm{A}\tag{2},\\&3\,\mathrm{A}-I_3-4I_2=0\,\mathrm{A}\tag{3}.\end{align}\]
Substituting \(I_3\) as given in the first equation into the other two equations gives\[\begin{align}&5\,\mathrm{A}-4I_1-I_2=0\,\mathrm{A}\tag{4},\\&3\,\mathrm{A}-I_1-5I_2=0\,\mathrm{A}\tag{5}.\end{align}\]
We can isolate \(I_2\) by combining \((4)\) and \((5)\) in the following way\[\begin{align}(4)-4\times (5)&\implies -7\,\mathrm{A}+19I_2=0\,\mathrm{A},\\&\implies I_2=0.4\,\mathrm{A}.\end{align}\]Substituting this into \((4)\) gives\[\begin{align}I_1=1.2\,\mathrm{A}.\end{align}\]
Putting all together into equation \((1)\) gives all three currents to be\[I_1=1.2\,\mathrm{A},\, I_2=0.4\,\mathrm{A},\, I_3=1.5\,\mathrm{A}.\]
Using \(V=IR\), we find the three voltages to be\[V_1=3.5\,\mathrm{V},\,V_2=1.5\,\mathrm{V},\,V_3=1.5\,\mathrm{V}.\]
Lastly, we want to find the charge on the capacitor. To do so, we need to find the potential difference across the capacitor. Again, Kirchhoff's Loop rule can be used. Consider the smallest loop of the circuit, containing both the \(1\,\mathrm{\Omega}\) resistor and the capacitor. There are only two potential differences involved here, the one across the capacitor \(V_C\) and \(V_2\). Kirchhoff's loop rule tells us these must add to zero, and so \[V_C=V_2=1.5\,\mathrm{V}.\]
Multiplying the voltage by the capacitance gives the accumulated charge on the capacitor:\[Q=1.5\,\mathrm{V}\cdot 5\times10^{-9}\,\mathrm{F}=7.6\,\mathrm{nC}.\]
Sign up for free to gain access to all our flashcards.
Where do we use KVL and KCL?
Kirchhoff's Current and Voltage laws are used in circuit analysis to find missing currents and voltages. They are particularly useful for analyzing complex circuits where V=IR is not suitable.
What is Kirchhoff's junction rule?
Kirchhoff's junction rule states that the sum of currents meeting at a circuit junction must always be zero.
How to find current using Kirchhoff's law?
Kirchhoff's junction rule gives an equation for the set of currents {I_k} meeting at a junction.
ΣI_k=0
This can be re-arranged to find missing currents if the others are known.
How many Kirchhoff laws are there?
There are two Kirchhoff's Laws. The Junction Rule and The Loop Rule.
What is Kirchhoff's current law formula?
For a set of currents {I_k} meeting at a junction,
ΣI_k=0.
At StudySmarter, we have created a learning platform that serves millions of students. Meet the people who work hard to deliver fact based content as well as making sure it is verified.
Lily Hulatt is a Digital Content Specialist with over three years of experience in content strategy and curriculum design. She gained her PhD in English Literature from Durham University in 2022, taught in Durham University’s English Studies Department, and has contributed to a number of publications. Lily specialises in English Literature, English Language, History, and Philosophy.
Get to know Lily
Gabriel Freitas is an AI Engineer with a solid experience in software development, machine learning algorithms, and generative AI, including large language models’ (LLMs) applications. Graduated in Electrical Engineering at the University of São Paulo, he is currently pursuing an MSc in Computer Engineering at the University of Campinas, specializing in machine learning topics. Gabriel has a strong background in software engineering and has worked on projects involving computer vision, embedded AI, and LLM applications.
Get to know Gabriel