Tutorial 3  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:
Therefore:
V_{T} = V_{1 }+ V_{2} + V_{3}
From
Ohm’s Law we know:
V_{T} = IR_{T};
V_{1} = IR_{1};
V_{2} = IR_{2};
V_{3} = IR_{3 }
Þ IR_{T} = IR_{1} + IR_{2} + IR_{3}
Therefore:
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.
From this we can write:
I_{tot} = I_{1} + I_{2} + I_{3}
From
Ohm’s Law,
I = V/R,
we can write:
I
_{T} = V ; I_{1}
= V; I_{2}
= V; I_{3} = V
R_{T } R_{1} R_{2} R_{3}
Þ
V =
V + V +
V
R_{T } R_{1}
R_{2} R_{3}
Þ
This is true for any number of parallel resistors.
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:
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
E.
The curly E (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
E
volts.