Series and Parallel Circuits
Series
Circuits
In
a series circuit, the electrons in the current have to pass through
all the components, which are arranged in a line.
There are two key points about a series circuit:
The current throughout the circuit is the same
The voltages add up to the battery voltage.
Therefore:
From
Ohm’s Law we know:
V_{T} = IR_{T};
V_{1} = IR_{1};
V_{2} = IR_{2};
V_{3} = IR_{3 }
So we can write:
IR_{T} = IR_{1} + IR_{2} + IR_{3}
Therefore, since the current, I, is the same throughout the circuit, we can write:
R_{T} = R_{1 } + R_{2} + R_{3}
This is true for any number of resistors in series.
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
Resistors
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.
The voltage across each branch
is the same
The currents in each branch add
up to the total current.
From this we can write:
I_{T} = I_{1} + I_{2} + I_{3}
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. 
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.
Work
out the total resistance of the parallel combination.
Work
out the total resistance of the circuit by adding your answer in the
previous step to the values of the series resistors.
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 R_{t} R_{1} R_{2} 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:
Make
sure that the voltages across each part of the circuit add up to the battery
voltage.
Make
sure the currents in the parallel part of the circuit add up to the battery
current.
If they don’t, go back and check what you’ve done wrong!
We
often refer to the total resistance of a circuit as a single resistor
equivalent
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
I_{3} = I_{1}
+ I_{2}.
Mathematically we can write this as:
Notice that I_{3} 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.
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
ℇ.
The curly E is the battery voltage.
We will look at emf later.
From A to B the p.d change is
IR_{1} volts
From B to C the p.d. change is
IR_{2} volts
From C to A the pd. change is
ℇ
volts.
IR_{1} + IR_{2} = ℇ
or
IR_{1} + IR_{2} + ℇ = 0
or
S(p.d.) = 0
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 rollercoaster 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 rollercoasters 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.
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 R_{1} = 220 W and R_{2} = 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 Alevel students)
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.
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 = (I_{1} + I_{2}) × 4 + I_{1} × 2 = 4
Sum the IR for loop 2:
Sum = (I_{1} + I_{2}) × 4 + I_{2} = 2
Therefore: 4I_{1} + 2I_{1} + 4I_{2} = 4 Þ 6I_{1} + 4I_{2} = 4
Also: 4I_{1} + 4I_{2} + I_{2} = 2 Þ 4I_{1} + 5I_{2} = 2
So we have two simultaneous equations: 6I_{1} + 4I_{2} = 4 4I_{1} + 5I_{2} = 2
Multiply the first by 2 and the second by 3
12I_{1} + 8I_{2} = 8 12I_{1} + 15I_{2} = 6
Subtract the two equations from each other to get rid of 12I_{1}
Now substitute into the second equation to get I_{1}:
Total Current = 0.858 – 0.286 = 0.572 A = 0.57 A (2 s.f.)
