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.