L3 4 1Superposition

Demonstrate Superpostion method of analyzing electrical circuits

Hi.
My name is Lee Brinton.

I'm an electrical engineering instructor
at Salt Lake Community College.

We're going to continue our
introduction to electrical and

computer engineering by considering
now the principle of superposition.

Depending on where you are in your
math sequence, you may have already

seen the principle of superposition as it
applies to linear mathematical operators.

For example,
is L is a linear operator, and by that

we mean it's something that performs
operations that are linear in nature.

I know that sounds a little bit circular,
but

let's just follow me through on this and
you'll see what I mean.

If it's a linear operator,
that means that if you operate on

two Different quantities, the sum of two
different quantities you'll get the same

result as you would if you operated
on each of them individually

and then added the individual
results together.

For example,
Differentiation is a linear operator.

We know that if you take
the derivative with respect to time

of two different functions
u(t)+v(t) that's equal

to the derivative with respect
to time of the first one.

Plus the derivative with respect
to time of the second one.

A little more concrete example.

Let's say that u(t) equals, say,

e to the a t, and
v(t) equals the sine of omega t.

And we want to take the derivative
of the sum of those two things.

In other words, the derivative of respect
to t of e to the at plus sine omega t.

Because the derivative is
a linear operator, we know that.

That is simply equal to the derivative
with respect to t of e to the at.

plus the derivative with respect
to t of the sin of omega t.

Which is equal to ae to the at
plus omega Cosine of omega t.

Thus the principle of linearity or
the principle of superposition, rather,

says that if we want to take
the derivative of two things,

the sum of two things, we can take
the derivative of one thing And

add to it the derivative
of the other thing.

When it comes to circuits and
circuit analysis in linear circuits,

there is a similar
principle of superposition.

In a circuit containing more
than one independent source,

More than one independent
input to the circuit.

The total response to the circuit can
be found by combining the responses

to the circuit to each
independent source separately.

What does that mean?

Here we have a linear circuit.

Consisting of two sources,

and it's important to understand
it's independent sources.

Two independent sources, here we
have a current source of 6 amps and

over here we have a voltage
source of 45 volts.

Superposition says that we can determine
some quantity within the circuit,

say for example, the current flowing
through this ten ohm resister.

We can determine the total
current resulting from

both of these sources combined,

by determining the current through this
resistor due to the current source alone.

And the current through this due
to the voltage source alone.

How do we then separate, or how we do
turn off one source and then the other?

Well, sometimes this turning off is
referred to as deactivating the source,

or turning the source to zero.

In this first circuit here,
we have just the current source.

The voltage source has been turned to 0.

Which means that we've got
a voltage source here of 0 volts.

So from this side of the circuit,
this side of the source

to this side of the source
We increase zero volts.

In other words, the voltage here
is the same as the voltage there.

Well that is as though we had
taken that voltage source and

replaced it with a wire or
a short circuit.

To deactivate a voltage source
we replace it simply with.

Of straight wire or short circuit.

On the other hand,

when it comes to analyzing the circuit
with just the voltage source in play,

we need to deactivate the current
source and return it to zero.

If we turn the current source to zero,

that's saying that there are zero
amps flowing through this branch.

It's as though this branch were
unplugged or open circuited.

And we replace that current
source then with an open circuit.

LEt's go ahead now and calculate i the
total i due to both of these two sources

by calculating the current i
one due to the current source.

And the current I2 did the voltage
source and then combine those two.

All right, first of all, I1 can be
calculated using a current divider,

these two resistances are in series and
we have a total current I coming in.

I1 then is going to equal 6 amps times
The five volts divided by five plus ten.

Well, five over 15 is two thirds
therefore two thirds of six,

I'm sorry five over 15 is one third.

One third of six is two amps.

So the component of I due to
independent current source Is two amps.

Now lets determine the current over here,
due to the voltage source.

Well with the current source open
circuited, their is no current coming this

way and so any current that is flowing
through the ten ohm resistor will also be

flowing through the five ohm resistor,
and those two resistors are in series.

Thus we can say then that I2 is equal to,

now you'll notice that
I2 is referenced left or

right but this voltage source we'll be
putting the current going right to left,

so I2, there's gonna be a minus sign here,
I2 is equal

to negative 45 volts divided by the series
combination of those two, five plus Ten

which gives us 45 divided by ten is three
amps with a minus sign in front of it.

So I2 then is equal to
negative three amps.

Now we can determine the total I,
the total I is just equal to I1,

the current due to the current source.

Plus I2 the current
digital voltage source,

and that is equal to 2 amps minus 3 amps

equals -1 amp.