The series and parallel combination of resisters can be applied to simple circuits but they are not applicable for complex electric circuits. So, we need a new way to solve our electric circuit problems which is exactly done by Kirchhoff's Laws.
For example, in the figure below, there is no way you can solve this by the combination of resistors. Simply examine the circuit shown below which can not be reduced to simple series and parallel combination of resistors. Even if you tried to use the combination, the resister \(R_3\) comes in your way.
Kirchhoff's laws can solve even a very complex looking electric circuit. There are two laws in Kirchhoff's laws. The hard or confusing part of Kirchhoff's laws is considered to be choosing the right sign but it is quiet easy to if you are once familiar and practiced with.
In Kirchhoff's laws you must be familiar with two terms. One is the junction also called node and another is the loop. A junction is a point in an electric circuit where three or more wires are connected.
Junction does not have two wires connected. If only two wires are connected, the connected wires as a whole become a single wire. If you need to connect two wires to create a junction, one wire must be connected to another in such a way that three wires come out of the point of connection, that is they should not be connected end to end. Either way at least three wires are connected at a junction.
A loop is a closed path like a circle in an electric circuit. We either move clockwise or anticlockwise in a loop to solve problems with Kirchhoff's laws. All you need to solve complex looking problems related to electric circuit is two laws by the German physicist Gustav Robert Kirchhoff.
Kirchhoff's Current Law (Junction Rule)
The first law is the current law also often known as junction rule. The statement of this law is given below.
KIRCHHOFF'S CURRENT LAW (JUNCTION RULE): The sum of currents at a junction is zero.
The Figure 3 shows a junction where three wires are connected at a point. The current \(I\) is directed into the junction and split into \(I_1\) and \(I_2\). It's obvious that the total incoming current \(I\) is the sum of the split outgoing currents \(I_1\) and \(I_2\).
It's similar to water flowing into the junction from one pipe and split into two pipes as shown in figure below. Mathematically,
\[\begin{align*} I &= I_1 + I_2 \\ \text{or,}\quad I - I_1 - I_2 &= 0\\ \text{or,}\quad I + (-I_1) + (-I_2) &= 0\\ \text{or,}\quad \sum I &= 0 \tag{1} \label{1} \end{align*}\]
This rule is based on the conservation of charge, that is no charge is lost or destroyed or disappeared anywhere at the junction. One thing to note (sign convention also discussed later) while applying junction rule to solve problems is that the currents directed into the junction are taken as \(+I\) (positive) and the currents directed out of the junction are taken as \(-I\)(negative) when you sum those currents up.
The conservation of electric charge enforces that no charge is accumulated at the junction which verifies the junction rule. The Kirchhoff's current law has other names such as Kirchhoff's first law, Kirchhoff's point rule and Kirchhoff's nodal rule.
Kirchhoff's Voltage Law (Loop Rule)
Now it's the time to know what Kirchhoff's voltage law is. This law is also often known as loop rule. The statement of the loop rule is given below.
KIRCHHOFF'S VOLTAGE LAW (LOOP RULE): The sum of all potential differences or voltages around a loop is zero.
The idea behind this is simple. This law is based on the conservation of energy, that is the sum of energy around a closed loop is zero. If the sum of energy is zero, where is the used energy? Of course the used energy in the loop is converted to other forms of energy such as thermal, mechanical etc. In other words the the sum of potential rises is equal to the sum of potential drops in a loop.
For example in the figure above a simple one loop circuit is shown. There is a potential rise within the source of emf and drop across the internal resistance \(r\) and also there are potential drops across external resistances \(R_1\) and \(R_2\). The potential rise must be equal to the sum of the potential drops across resistances, that is \(\mathcal{E} = Ir + IR_1 + IR_2\) or \(\mathcal{E} + -(Ir) + (-IR_1) + -(IR_2)\) which leads us to conclude that
\[\sum V = 0 \tag{2} \label{2}\]
In electromotive force we learned that the potential decreases as the charge moves across resistance and there must be a device in a circuit which increases the potential again, that is acts like a charge pump (review aticle of emf). What Kirchhoff's loop rule suggests is that the sum of the increases in potential must be equal to the sum of decreases in potential in a loop.
The Kirchhoff's voltage law is also known by other names such as Kirchhoff's first law, Kirchhoff's loop rule, Kirchhoff's mesh rule etc. Basically the word "law" and "rule" are used interchangeably with different variations. You don't need to say all the names just pick one and stick to it!
You've not mastered Kirchhoff's laws yet. You still need to understand the sign conventions in Kirchhoff's laws.
Sign Conventions in Kirchhoff's Laws
Junction rule is straightforward. First thing is choose the current direction. Don't worry about choosing the right current direction. If the direction you choose is not the actual direction of current, the current will get the negative sign after you solve the equation which means the current is directed to the opposite direction you assumed initially.
You can at least try to give the correct current direction examining the polarity of a battery and it does not matter if you are failed to do so. Remember that the currents entering a junction are positive (+I) and currents leaving the junction are negative (-I) when you sum up these currents.
Now it's time to move along a loop. You can move in any direction in a loop, that is either clockwise or anticlockwise. You'll get the same results for both clockwise and anticlockwise directions. When you move from the position of lower potential to the position of higher potential, that is from negative terminal to positive terminal across a source of emf (battery), take the potential difference positive.
And if you move from the position of higher potential to the position of lower potential, that is from positive terminal to negative terminal across a source of emf, take the potential difference negative.
If you are moving in the same direction as the assumed direction of current across a resistance, take the potential difference negative (charge flows towards the direction of decreasing potential), that is if \(I\) is the current and \(R\) is the resistance, the potential difference across the resistance is \(-IR\).
Current does not have a direction like a vector quantity has. We are saying "direction of current" only in the sense of which way the charge flow is directed. Note that the direction of current means the direction of the flow of positive charge.
And if you are moving in opposite direction to the assumed direction of current across a resistance, take the potential difference positive (\(+IR\)). The final task is to add all potential differences in a loop and assign the sum to zero. Now let's go ahead and learn an example.
Kirchhoff's Laws: A Multiloop Circuit
In this example we determine the resistance \(R\) and the internal resistance \(r\) in an electric circuit shown in figure below. You must understand the steps and sign conventions clearly before considering to solve this electric circuit. Once you are familiar with it, solving with Kirchhoff's laws is the most easiest task!
In the circuit above you first consider to apply junction rule to find the unknown current \(I\). There are two junctions at points \(a\) and \(b\). You can choose any point for your convenience and here we choose point \(a\) to apply junction rule. The currents entering the junction is a \(3A\) current and the currents leaving the junction are \(1A\) current and unknown current \(I\). After applying junction rule we get,
\[3A - 1A - I = 0\]
Note that the currents entering a junction are taken as positive and those leaving the junction as negative, so the negative sign is used for known \(1A\) and unknown \(I\) currents. The unknown current \(I\) from above equation is \(I = 2A\).
Next we apply loop rule to find the external resistance \(R\) and internal resistance \(r\). You can construct three loops in the circuit. One loop is the outer loop (loop 1) and two loops are inner loops (loop 2 and loop 3). For now we move anticlockwise along loop 2 to find \(R\).
\[8V - (2\Omega)(3A) - R = 0\]
The value for \(R\) from above equation is \(R = 2\Omega\). Next we move anticlockwise along loop 3 to find the internal resistance \(r\), and you get
\[(2\Omega)(1A) - (2A)r + 4V = 0\]
So the value of internal resistance is \(r = 3\Omega\). Note that you could move anticlockwise along the bigger loop, that is loop 1 to directly get the value of internal resistance, for example
\[8V - (2\Omega)(3A) - (2A)r + 4V = 0\]
and the value for the internal resistance from the above equation is \(r = 3\Omega\). Instead of moving anticlockwise you could also move in clockwise direction around the loops and the result would be the same. You can practice more until you are completely familiar with it.
