Ok, continuing on with looking at a different circuit calculation in ohm's law. We are going
to look at the parallel circuit. So here we have a 13.2 volt battery supplied voltage
and then the circuit breaks actually into two sections now. So like the last diagram
that we took a look at which was a series circuit this one has a different characteristic
and a parallel circuit is defined as having more than one path and more than one load.
So any time we see a split in the circuit it shows that we have more than one path and
this particular diagram shows that we have more than load and again it returns back to
ground. So I am going to show you a couple of different ways to calculate this circuit.
There is two ways that it can be done. One is using the ONE OVER method and I will explain
that. And then the other one that I will show you is called PRODUCT OVER SUM. And the product
over sum only works when we have two resistors in parallel in a circuit. It can't be applied
to a series circuit and it can't be applied to any more than two resistors at a time in
a parallel branch. So how we are going to calculate this out first of all is that we
need to take a look at RT and how we go about determining RT and I will just start right
down here. RT equals 1 over and this is the formula that I was talking about the over
formula. So what we are going to do is 1 over 1 over R1 plus 1 over 1 over R2 and we will
just move down and start plugging in numbers. So 1 over R1 is 1 over 6 plus 1 over 5 and
we'll just move over here to equal again and if we take the 1 over 6 we need to convert
this into a decimal to add them up or common denominators. So 1 over 6 actually calculates
out to so we will install the 1 and divided by 6 equals .166 plus 1 divided by 5 equals .2
So now if we do the 1 over and we add that value up we end up with 1 over .366.
And we end up dividing the top into the bottom it gives us 2.73 ohms. Now one of the other
characteristics of a parallel circuit is that the resistance total in a parallel circuit
will equal less than the lowest resistance in the circuit. In this particular case we
have 5 ohms and if you were to ask me what it would be I know that it will be roughly
about half of that value which gets us close to 2.73 ohms. So we can get this a different
way also by using the product over sum. Now understanding, product over sum or resistance
total equals product over sum. And if we define product its the multiplication of numbers.
So if we take a look at the product it would be 6 times 5 and then the sum is the addition
of the numbers. So then we have 6 plus 5 giving us 11 on the bottom and 30 on the top. And
if we divide 30 by 11 we end up with 2.72 ohms. And if we take a look at these two right
here and I said to you already, it depends on rounding decimal places we are gonna end
up pretty close or within a decimal point. So if I was to cross this one off and I was
to cross this one off we would still end up with 2.7 ohms and 2.7 ohms whether we are
using the one over formula or the product over sum formula. I wanted to show that you
could actually get both of these values within that decimal place so that they both still
are applicable for calculating out parallel circuits. So next thing we have to take a
look at in this particular circuit is finding out how much current is flowing. So we are
going to take a look at It and It equals E over RT and if we plug in the numbers here
13.2 volts divided by the resistance total of 2.7 and we divide the top into the bottom
13.2 divided by 2.7 will equal 4.8 amps. So numbers we need to take a look at here is
our amperage, our resistance based on product over sum, our resistance based on the one
over formula. Now one thing that a lot of technicians get a misunderstanding about is
do we have the same amount of current flowing in each path of a parallel circuit. Well if
we were to put like I had drawn in other diagrams, if I was to put an amp meter right here then
our amp meter would read 4.8 amps . And if I was to put it right here our amp meter would
equal 4.8 amps. But if I put one right here and I put one right here they are gonna be of
different numbers and that is based upon the amount of resistance that is in each path
of the parallel circuit. So that is going to determine how much current will flow based
on the resistance of each path. So we are going to take a look at the amount of current
flowing through R1 and R2. So I am going to call this I1 and I2. So I1 equals E over R1
which in this particular case in a parallel circuit we are going to have 13.2 volts applied
here and 13.2 volts applied there. So we take 13.2 volts divided by the resistance of R1
which is 6 ohms which will give us 2.2 amps. And then if we move down to I2 its E over
R2 which equals 13.2 volts applied with 5 ohms of resistance calculates out to 2.64 amps.
And if we add the two of these together, 2.2 amps and 2.64 amps it should calculate
out to 4.84 amps. And if we again take a look at the numbers that we have here, and we knock
that decimal place off we have the exact same thing right here 4.8 amps. So we know that
we have done the calculation right based on the voltage and the overall resistance then
we can break it down into individual paths of current flow based on the resistance in
each path and then see if we add up to the exact same amount that we started with here.
Ended up with and broke down based on the resistance of 6 ohms in R1, 5 ohms in R2.
So 2.64 amps here and 2.2 amps over here .