To work out the two
equal value resistor in parallel R_{total} =
(R1 or R2)/2

So
R_{total}= (1000)/2 R_{total}=500
ohms

Now to work out
the total resistance RT in the circuit when
resistor are in series add together

RT = R_{total}
+ R3

So RT= 500 +
12000 = 12500.

To find the
total current flow from the battery and at points
C and D using Ohms law.

I_{total}
= V1/ RT so I_{total} = 12/12500 = 960uA

To workout the
voltage drop crossed R3 calling it V3 using Ohms
law

V3 = R3*I_{total}

V3 = 12000*960uA =
11.52V

so V3 = 11.52 Volts

To workout the
voltage drop across R1 and R2 which is V2 there
are a number of ways.

First way
because we have worked out the voltage drop across
R3

V2= V1-V3

so V2= 12 - 11.52 =
0.48V

V2 = 0.48V.

The second way
because we have worked out the two resistor in
parallel R1 and R2 which was called R_{total}
and also the total Current flowing in R_{total}
is I_{total} using Ohms law V2 = R_{total}
x I_{total}

So V2 = 500 x
960uA = 0.48V

So to sum up the
voltage drop across R1 and R2 is V2 = 0.48v, the
voltage drop across V3 is 11.52v

The current
through R3 is 960uA

But what about
the current flowing through R1 to work this out
using Ohms law because the voltage drop across R1
has be worked out V2 the current flow through R1
is V2/R1

So 0.48/1000 =
480uA

To work out
current flowing through R2 is the same way has
working the current flowing in R1 so V2/R2 =
0.48/1000 = 480uA

Calculate
the combined resistance of two or three
resistors in series. Calculate the combined
resistance of two or three **equal**
resistors in parallel.

####
SERIES RESISTORS

Resistance
is the opposition to current flow. The bigger the
resistance the smaller the current that can flow.

Before we look
at the topic in electronic terms let's consider
a large number of people all lined up in one
long straight line.

The
effort that you would need, as an electron!!,
to push through would be quite a bit but not
impossible.

Now what
would happen if the crowd broke into two
lines you would now have to push your way
through the first and then the second line
and you would find it more difficult and you
would have used up more energy.

So if the
crowd broke into three lines you would now
have to push your way through all three and
that would be more difficult than the other
two previous examples and even more energy
would have been used up.

The same
is the case with resistors.

When there
is just one resistor in a circuit as shown
below then the amount of current flowing for
a know voltage through a known resistor can
be calculated based on the formula V = I x R
and would be shown on the **amp meter**
or **ammeter** (a ammeter in series with
the circuit to measure current).

which
re-arranged give you the formula shown
below.

####
I = V / R

When
we have two resistors in series, as shown
below,

the
current has to flow through both and finds
it more difficult.

In fact the value
of the total resistance is simply to add
up the values of the individual
resistors.

####
"SERIES Resistor TOTAL" = R1 + R2

Even
though the two resistors are separated
by the meter they are still connected
in series.

So to
find the **current flowing in a
circuit, **through the meter,**
**with two resistors in series you
first work out the value of the
total resistance and from that
calculate the current from

#### I =
V / R where R = R1 + R2

So when we
have three resistors all lined
up one behind the other the task
of the current passing is still
greater and less flows.

As
with the two resistors
mentioned above the total
resistance is simply to add up
the values of the individual
resistors.

####
"SERIES Resistor TOTAL" = R1
+ R2 + R3

####

Current flow

Above
the three resistors are
"in SERIES" as the
current has to flow
though each as there is
no other way. also it
does not matter where in
the circuit the meter is
to measure the current.

So
to find the current
flowing in a circuit
with three resistors
in series you first
work out the value of
the total resistance
and from that
calculate the current
from

####
I = V / R where R =
R1 + R2 + R3

####
Voltage across
each resistor

If
the voltage
is measured
across each
resistor then
the sum of
these
individual values
will be the same
as that across
the battery as
the set of
resistors could
be replaced by a
single resistor
equal to the
total value of
the individual
resistors (as
they are in
series)

**PARALLEL
RESISTORS**

The
other way that
resistors can
be coupled
together is
for each end
of one
resistor to be
linked to the
same end of
the next
resistor and
the same with
the other end.

We
now have a
different
situation to
the "SERIES"
as the
electrical
path, as we
call it, has
two routes to
chose from. If
each of the
resistors is
of equal value
then it is as
difficult for
the current to
pass through
each resistor
and therefore
the current
divides
equally
between the
resistors.

If
they are not
of similar
value .......
well that is
for a later
training
course so
let's just
think in terms
of similar
values

####
R1 = R2

Value
of Two EQUAL
resistors in
parallel =
Value of one
resistor
divided by 2

####
"**Parallel
Resistance
Total" = R1 /
2**

####

R1 = R2 = R3

Value
of Three EQUAL
resistors in
parallel =
Value of one
resistor
divided by 3

####
"**Parallel
Resistance
Total R" = R1
/ 3**

This
may look
strange but if
you think of
it this way
.... there are
in the first
example two
paths that the
current can
take therefore
with equal
values of
resistor the
resistance is
halved as
there are two
paths

and
with three
resistors of
equal value
the resistance
is a third as
there are
three paths!!

Students
are not
understanding
that when
faced with a
question of a
current in a
simple circuit
that you must
first
determine the
value of the
total
resistance
whether that
is resistors
in series (as
outlined
above) or
resistors in
parallel as
outlined
below. It is
only then that
you can go
forward to
calculate the
current
flowing for a
particular
applied
voltage but
the use of
Ohm's law. so
let's look in
a bit more
details about
this type of
question.

**Current
in the circuit**

Find
the current in
a circuit with
resistors in
parallel

You
must first
find the
actual total
amount of
resistance and
from that
calculate the
current
flowing in the
circuit. The
meter measure
the total
amount in the
circuit and
MUST be placed
to make this
reading
outside the
resistor
network else
it would only
read the
current
passing
through the
single
resistor with
which it is in
series.

####
Practical
example - Make
a DUMMY LOAD.

Let's
take a
practical
example. If
you want to
make a dummy
load then how
many carbon
resistors will
you need and
of what value
to make a 40W
dummy load ?

There
would be
several ways
of going about
this :-

50
0.8W 1 ohms
resistors in
series. That
would make a
fairly bulky
dummy load and
probably only
1W resistors
available so
the unit would
be over rated.

So
look at it
another way.
2W resistor of
the carbon
type are
available and
thus 20
resistors
would give us
the power
capability we
require and
assuming we
are using them
in PARALLEL,
what value of
resistor is
needed ?

From
the equation
above :-

**Resulting
value = Value
of one
resistor /
Number of
resistors**

rearranging
the equation
we get :-

**Value
of one
resistor =
Resulting
value x Number
of resistors**

Value
of one
resistor = 50
x 20 = 1000
ohms or 1k ohm

So
to make a
simple dummy
load we need
20 1k 2w
resistors
linked in
parallel.