Kirchhoff’s Law

In this section Kirchhoff’s Law will help us understand how to analyse a DC circuit. DC means Direct Current where the current travels in one direction, from the +ve terminal of the voltage source (battery) to the -ve terminal.

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Why to analyse a DC circuit?

And what does it mean?

Well, it means we take the components that we know the value of in a circuit and we calculate what we don’t know. For example a value of a resistor or the flow of current.

Circuits get biiig. Really big. With DC circuit analysis we could simplify a large circuit, so a few components would represent the large circuit. In theory of course, so example we can substitute 100 resistors with 1: which will be the sum of them all. But we will get to this shortly.

Let’s focus on Mr. Kirchhoff. He has 2 important laws:

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1. Kirchhoff’s Current Law

Let’s take this junction for an example. It is part of a random circuit.

• There are 4 wires joined together at a node

• Each wire has a different current in it

• I1 and I2 are flowing toward the node
  • Since current flows from +ve to -ve terminal, the node has a lower potential than whatever we have on the other side of the I1 and I2 wires.

• I3 and I4 are flowing away from the node
  • for them the node is at hire potential than wherever I3 and I4 currents are flowing to.

We can express the currents the following way:

From the equation, it is straightforward why I3 and I4 are negative. But think of it this way: They are in the opposite direction as I1 and I2, in such way that they don’t go towards the node, they go away from it.

The sum of the total current entering a junction is equal to the sum of total current exiting the junction.

In other words, the sum of all the currents is equal to zero. The sigma sign is a mathematical representation for ‘sum’:

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2. Kirchhoff’s Voltage Law

To understand this, let’s start with a small circuit:

• We have 1 voltage source: E and two resistors R1 and R2.

• The red arrows symbolise the voltage drop
  • The arrow points towards the higher potential.

• There is a current I in the circuit.

• The voltage drop across R1: V(R1) = IR1 and across R2: V(R2) = IR2.

• The current is the same across both resistors.

• The voltage drop across R1 plus the voltage drop across R2 is equal to the voltage source E.

• E is the EMF.

We can express the above as:

Kirchhoff’s 2nd law states that the sum of voltages around a closed circuit is equal to zero.

So, the sum of all EMFs (multiple batteries) minus the voltage drops across all elements (resistors) is equal to 0:

Solved Example:

Find the current flowing across R1.

Solution:

1. Use Kirchhoff’s Current Rule

• We have 3 currents of which we only know the value of I. Node X joins them together.

• I1 and the current source I flow towards node X while I2 flows away.

• from this we can write the following equations. Note, I1 and I2 are negative as they flow towards the node.

2. Substitute equation 2 and 3 to equation 2

• Our aim is to find the voltage at node X.

• By knowing Vx we will know the voltage drop across R1. The value of the resistor R1 is given, thus, by using Ohm’s law we can calculate I1.

3. Substituting to the equation:

4. Use equation 2 to find I1 and equation 3 to find I2

Solved Example 2:

Find the current flowing in the circuit as well as the individual voltage drops across each resistor.

Solution:

1. Use Kirchhoff’s Voltage Rule

• There are 4 resistors in the circuit and 2 voltage sources

• All resistors are in series. thus, the same current flows through them

• By Kirchhoff’s Voltage Law the sum of the voltage drops across each resistor is equal to the sum of all the voltage sources.

• VR1 below simple means the voltage drop across R1. Voltage by Ohm’s Law is the resistance x current: V = RI.

• By substitution we can easily find the value of I.

2. Use Ohm’s Law to find the voltage across each resistor