Series and Parallel Circuits


Series circuits

Parallel Circuits

Kirchhoff's Laws

Using Kirchhoff's Laws


Series Circuits

In a series circuit, the electrons in the current have to pass through all the components, which are arranged in a line.  Consider a typical series circuit in which there are three resistors of value R1, R2, and R3.  The values may be the same, or different.


There are two key points about a series circuit:



From Ohm’s Law we know:

So we can write:


IRT = IR1 + IR2 + IR3


Therefore, since the current, I, is the same throughout the circuit, we can write:


RT = R1 + R2 + R3


This is true for any number of resistors in series.


Question 1

This question refers to the circuit below in which the current is 100 mA:


(a) What is 100 mA in amps?

(b) What is the current in each resistor?

(c) What is the voltage across each resistor?

(d) What is the total resistance?

(e) What is the battery voltage?



Parallel Circuits

Parallel circuits have their components in parallel branches so that an individual electron can go through one of the branches, but not the others.  The current splits into the number of branches there are.  Look at this circuit:


In this case, the current will split into three.   For a parallel circuit we know two things:

From this we can write:

IT = I1 + I2 + I3


From Ohm’s Law, I = V/R, we can write:


We can write:

Since the voltage across each resistor is the same, we can write:

This is true for any number of parallel resistors.

We can also write this as:

We are adding up the reciprocals of the resistance.


Once you have done the addition of the reciprocals, remember to take the reciprocal of the answer.  This will give you the resistance.


Question 2

This question refers to the circuit below.

(a) What is the total resistance of the circuit?  (Watch out for the bear trap.)

(b) What is the current through each resistor?

(c) What is the total current? 



We can combine resistors in both series and parallel.  Tackle the problem step by step.

Here is a worked example:

Worked Example

Look at this circuit:

What is the single resistor equivalent?

What is the total current?

What is the voltage across  the 6 ohm resistor?

What is the current in each resistor?


What is the single resistor equivalent?

We will do the parallel combination first:

1 = 1   + 1   = 1   + 1   = 3

Rt  R1    R2    4      8       8


Rt = 8/3 = 2.67 ohms


Now we can work out the overall resistance:


The overall resistance = 6 ohms + 2.67 ohms = 8.67 ohms

What is the total current?

I = V/R = 12 volts ÷ 8.67 ohms = 1.38 amps

What is the voltage across  the 6 ohm resistor?

V= IR = 1.38 amps × 6 ohms = 8.30 volts

What is the current in each resistor?

We need to know the voltage across the parallel resistors:

Voltage = 12 volts - 8.30 volts = 3.70 volts


Now we can work out the current in each branch because the voltage across each resistor is the same.

For the 4 ohm resistor:

I = V/R = 3.70 volts ÷ 4 ohms = 0.93 A

For the 8 ohm resistor:

I = V/R = 3.70 volts ÷ 8 ohms = 0.46 A


These two currents add up to 1.39 amps, but there are rounding errors.  Watch out for this, but don't worry too much about them.


Take care with such problems:


We often refer to the total resistance of a circuit as a single resistor equivalent.  Resistors are available in certain values.  One example is the E24 series, in which there are 24 values available in each decade (1, 10, 100, 1000, 10 000, etc.). 


Question 3

What is the single resistor equivalent of this circuit below?



Kirchhoff's Laws

These two simple laws were drawn up in the Nineteenth Century by Gustav Robert Kirchhoff.  They explain all observations we see in electric circuits.  We can explain everything we have looked at in series and parallel circuits in terms of the two laws.  They can also be used to explain more difficult circuits which cannot be explained in terms of simple series and parallel circuits.


Kirchhoff I

This states that:

 the total current flowing into a point is equal to the current flowing out of that point


In other words, the charge does not leak out or accumulate at that point.  Charge that flows away must be replaced.  It is conserved.

From this diagram we can easily see that I3 = I1 + I2.


Mathematically we can write this as:

  I1 + I2 + -I3 = 0


Notice that I3 has a minus sign. This means that the current going out is regarded as negative while current coming in is positive.  At no point is there any reference to charge pooling at the junction, for the simple reason that it does not.


In some text books you will see written SI = 0.  The strange looking symbol S is Sigma, a Greek capital letter S, which means "sum of".  So the sum of currents is zero, as we have seen above.


Question 4

Suppose we had a high voltage junction where was a fault.  A spark was jumping as well as current flowing away (i.e. not all the current was in the spark.)  How is that consistent with Kirchhoff I?  The fault is shown in the diagram:




Kirchhoff II

Kirchhoff’s Second Law is not quite so easy to grasp.  It states:


The potential differences around a circuit add up to zero.


Provided the charge returns to the same place as it started, the gains and losses are equal, no matter what route is taken by the charge. The battery in this circuit has an emf (electromotive force or open terminal voltage) of .  The curly E is the battery voltage.  We will look at emf later.


  Let us do a journey around the circuit from A to B to C, and back to A.

  If we add up all the voltages, we can write:

 IR1 + IR2 =


IR1 + IR2 + - = 0


S(p.d.) = 0


 This is another way of saying that the voltages add up to the battery voltage.  


Although the statement above may seem obvious enough to some, Kirchhoff II does throw a lot of students, so let's have a further look.


The picture shows a simple roller-coaster railway.  The cars leave the station.  They can go down hill the straight way or the wiggly way.  It doesn't matter.  They then have to go back up the potential hill to reach the station they started from.


Some roller-coasters have several hills the cars go up.  As they go up their potential increases.  It's like a circuit with several batteries.  You go up the potential hill as you go pass each battery.  When you go up the potential hill, the sign is negative.


Question 5

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? 

d.       Work out the currents and show that they are consistent with Kirchhoff I.



Using Kirchhoff's Laws (Extension for A-level and IB students)

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

Worked example

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





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



4I1 + 2I1 + 4I2 = 4

Ž 6I1 + 4I2 = 4



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 = 0.57 A (2 s.f.)