Kirchhoff's Junction Rule

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.

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.

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:\[Q=1.5\,\mathrm{V}\cdot 5\times10^{-9}\,\mathrm{F}=7.6\,\mathrm{nC}.\]