Tutorial 7 A - Kirchhoff's Laws  and Circuit Analysis

Learning Objective

To understand and use Kirchhoff’s Laws;

Key Question

What are Kirchhoff’s Laws?

How do we use Kirchhoff's Laws?

 

Kirchhoff's Laws

When we started this course, we looked at Ohm’s Law and simple series and parallel circuits.  In this topic, we are going to see how more complex circuit can be analysed using two simple laws that were formulated by the German physicist Gustav Robert Kirchhoff (1824 – 1887).  He did a lot of work on electrical engineering, as well as spectroscopy work with Robert Bunsen (1811 – 1899) who invented our old friend, the Bunsen burner.

 

 

Kirchhoff I is to do with the current and it states:

At any junction in an electric circuit the total current flowing towards that junction is equal to the total current flowing away from that junction.

 

Kirchhoff II is to do with voltages.  It states:

In any closed loop in a network, the algebraic sum of the potential differences (i.e. the products of current and resistance) is equal to the resultant EMF acting in that loop.

 

Kirchhoff came up with these laws as a twenty year old student.  He doesn't look it in the picture.

 

Kirchhoff I

Currents going into a junction are the same as currents coming out.  By convention the currents leaving a junction are negative.

The law says

S I = 0

 

The strange looking symbol, S,  is “Sigma”, a Greek Capital letter ‘S’.  It means “sum of”.  Therefore:

 

 

So we can write:

 

Worked example

Find the unknown values in this circuit:

 

Answer

Junction B: I1 = 50 – 20 = 30 A

Junction C: I2 = 20 + 15 = 35 A

Junction D: I3 = 30 + -120 = -90 A (the 120 A is flowing away from the junction) 

 

 

Question 1

Now work out I4 and I5.

Answer

 

Kirchhoff II

Voltages add up to the sum of the EMFs.  EMF stands for electromotive force.  It’s not a force at all, but a voltage (energy per unit charge). 

Conventional currents go from positive to negative.

Follow the current round anticlockwise, as current from E1 goes from positive to negative.  As we are going the wrong way through E2, E2 is negative.  As we go through the resistors, the voltage drop is negative, as we are going down the potential hill.  So let’s sum the voltages:

We can write this more meaningfully as:

 

 

Using Kirchhoff's Laws

Kirchhoff laws allow us to analyse some scary looking circuits.  Adopt this problem solving strategy:

  • Break the circuit into loops.

  • Work out the EMF for each loop.

  • Work out the current in each loop.

  • Sum the currents using Kirchhoff I;

  • Sum the voltages using Kirchhoff II.

Example

Find the EMF in this circuit:

Answer

 

Start from A and do a clockwise loop summing the EMFs. 

 

Sum of EMF = 3 + 6 + E – 4 = 5 + E

 

Sum of IR = (2 × 2) + (2 × 2.5) + (2 × 1.5) + (2 × 1) = 4 + 5 + 3 + 2 = 14 V

 

Equate the two:

5 + E = 14

 

Ž E = 9 V

 

 

In this example we only used Kirchhoff II.  Let’s do another example that requires is to use both laws.

 

Worked example

Use Kirchhoff’s Laws to work out the currents flowing in each branch of this network.

 

 


 

Answer

Identify loops 1 and 2.  Go with the conventional current, so that Loop 1 is clockwise and loop 2 is anticlockwise.

 

EMF for Loop1 is 4 V, EMF for Loop 2 is 2 V (Kirchhoff II)

 

Sum the IR for loop 1:

 

Sum = (I1 + I2) × 4 + I1 × 2 = 4

 

Sum the IR for loop 2:

 

Sum = (I1 + I2) × 4 + I2 = 2

 

Therefore:

4I1 + 2I1 + 4I2 = 4

Ž 6I1 + 4I2 = 4

 

Also:

4I1 + 4I2 + I2 = 2

Ž 4I1 + 5I2 = 2

 

So we have two simultaneous equations:

6I1 + 4I2 = 4

4I1 + 5I2 = 2

 

Multiply the first by 2 and the second by 3

 

12I1 + 8I2 = 8

12I1 + 15I2 = 6

 

Subtract the two equations from each other to get rid of 12I1

 

 

Now substitute into the second equation to get I1:

 

Total Current = 0.858 – 0.286 = 0.572 A

 

 

 

 

 

Do not be tempted simply to split the two loops and do separate calculations on each loop, then adding the currents.  This will give you different voltages across the 4 ohm resistor and that will make the calculation go wrong.  The voltage across the 4 ohm resistor is the same for both the loops.

 

 

So what is the point of all of this, other than to make a complicated problem to annoy students?

 

Some complex electronic circuits may have multiple power supplies feeding in to different points in the circuit, especially in computers. 

 

Another use of network theory is the national grid that consists on many loads (transformers that give power supplies to industry and homes), and many sources (power stations).

 

The picture shows the distribution of French power stations.  There is a similar network in the UK.

 

Now you can have a go…

 

Question 2

1.      This is a circuit diagram of a potential divider:

The potential divider is a useful circuit.  It is often called the voltage divider and is useful to electronic engineers.  The voltage divider has a formula for the output voltage:

a.       Show that this formula is consistent with Kirchhoff II.

b.      Calculate the output voltage if R1 = 220 W and R2 = 470 W

c.       A third resistor of 150 W is added in the position shown with the dashed lines.  What is the output voltage now?  Refer to Kirchhoff I in your answer.

 

Answer

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