MrAl
Flashlight Enthusiast
Re: CPF University: EE Course (20)
Hello again!
Today we are going to talk about a shortcut method of analyzing
a circuit with more then one source in it. This method is so
handy we cant go without it!
Serial: EE-Ch0002-p106-{A939C0B7-558E-4687-9271-AB92E4935FD0}
In fig 12a we again have a circuit in which there are no series or
parallel combinations of resistors that will simplify the circuit.
This is the same circuit we used last time, except the node at the
top is labeled "V" and the node on the bottom is zero.
We can easily analyze this circuit using nodal analysis or mesh
analysis like we did last time, but this time we will try a very
different method. This new method might be called:
"Kill all sources" but officially it's called the
"Superposition Theorem".
The idea is to 'kill' all sources in the circuit and then one at
a time remove the 'kills' and analyze each node as we would any
other circuit. The nice thing about this method is that for
any given circuit the resulting set of circuits that remain to
be analyzed are often GREATLY simplified. This means we can usually
get fast answers even with circuits with many sources in them
because as each source is killed the circuit gets more and more
simple.
What it means to 'kill' a source is very simple:
For a voltage source it gets short circuited, and for
a current source (which we havent covered yet anyway) it
gets open circuited. To 'unkill' the source you simply
put it back the way it was originally.
Now lets look at an actual example...
If you did the mesh analysis on this circuit last time the first
thing you will note when you follow this analysis is how much
simpler it is and how much faster we get the answer, except
maybe for the drawing of the pictures
In fig 12a we'll set the values as follows:
V1=3v, V2=11v, R1=10 ohms, R2=5 ohms, R3=5 ohms
Now to start the analysis first we would 'kill' both sources,
V1 and V2 by short circuiting both of them. Then, we would
'unkill' say V1 first and start analyzing. The shorted V2
circuit is shown in the top of fig 12b and the resulting circuit
is shown right below it in 12b also. The thing most notable
now is that two of the resistors R2 and R3 now appear in
parallel, whereas before we shorted V2 this wasnt true.
Also, i've called the top node Va (the bottom is still zero).
The reason for the new name is because Va is a new node, more
or less, and will not be the same voltage as V in fig 12a because
we've changed the circuit by shorting V2.
Ok, now we must find out the voltage of Va in the bottom of fig 12b
so we start by simply finding the value of R2 and R3 in parallel:
Rp=1/(1/R2+1/R3)=1/(1/5+1/5)=1/0.4)=2.5
so we replace R2 and R3 with a 2.5 ohm resistor. This leads
to a circuit with R1 in series with a 2.5 ohm resistor driven
by voltage source V1 so we use voltage division:
Va=V1*2.5/(10+2.5), note R1=10
Since V1 is 3 volts this means
Va=3*2.5/12.5 which equals 0.6 volts, so
Va=0.6 volts and we have the answer to the value of the
node that comes from the contribution from source V1.
Now we have to do the same thing with the other source as shown
in fig 12c.
R1 in parallel with R3 is:
Rp=1/(1/R1+1/r3)=1/(1/10+1/5)=1/(0.1+0.2)=10/3
Next we can again use voltage division with values 10/3 and R2
so we get:
Vb=V2*(10/3)/(5+10/3)
so
Vb=11*10/(15+10)=11*10/25
so
Vb=4.4 volts.
Now we have the two answers, Va and Vb.
To get the final answer of what V is, we simply add Va and Vb:
V=Va+Vb
and we get
V=5 volts
Since we know R3=5 ohms we can get the circuit using Ohm's Law:
iR3=5/5=1 amp
Which agrees with the analysis from last time.
It's very interesting that by changing the circuit a little we
could reduce the circuit itself into two smaller circuits
(fig 12d) that could both be analyzed easier then the original
circuit and so provided us with another means of finding the answer.
Pretty cool if you ask me!!
With all good things there are some drawbacks, and this method
is no exception. In order to use this method on a given circuit
he first requirement is that ALL elements of the circuit are
what's known as "linear" elements. A linear element is defined
pretty intensely, but let's try to attach some form of definition
to what a linear element is.
A linear element has a linear voltage current relationship, so
that a multiplication by it's current of a factor K results
in a multiplication of it's voltage by the same factor K.
Now i dont know how much good that will do you at this point, so
let's look at a simple example in another field...
You drive two strong nails into the wall approximately on the same
level line, spacing them about 12 inches apart. You stretch a
rubber band across both nails so it stretches across horizontally.
Now you begin hanging weights in the middle of the band, so that
for each weight you measure a deflection of the horizontal
band in the center which now forms a vee shape.
Let's say the first weight of 1 ounce causes a deflection of
1 inch. Now you hang a second weight of 1 ounce and it too
causes another defection of 1 inch so the total now is 2 inches.
You hang a third weight of 1 ounce and another 1 inch results so
now it deflects by 3 inches.
In each case we hung the same weight on the band the same result
occurred: one additional inch of deflection.
The rubber band has thus responded linearly, in that a change
of input of a certain degree led to a change in output of
the same amount as with the original input amount.
On the other hand, if we keep hanging weights on it eventually
it breaks and so it 'stretches' far more for that last one ounce of weight.
This second kind of response would make it 'non linear', which is
simply put, "not linear".
Soon we'll encounter the LED again to see just why this is called
a nonlinear device. Things will make much more sense then too,
and why it's important to know a nonlinear device from a linear one.
For now, for the purposes of what we are doing at this point it will
probably be good enough to know that at least a resistor is a
linear element. When we come to other elements, we'll look a little
more at this too. It's interesting however, to note that in most
analysis the capacitor and inductor are considered linear too, so
we'll find this method is of use later as well.
Thanks for joining in today!
Take care,
MrAl
Last minute note:
Please dont try to short out your voltage sources
in circuits that you or someone else has built up.
This is a theoretical method, and if you want to
try it in practice you need to first open circuit
each voltage source (like a battery) before you
short the circuit where the battery originally
connected. Please remember this!
Today's Self Test Question
Take the circuit of fig 12a and make the following changes:
V1=11v, V2=3v, R1=5 ohms, R2=10 ohms, R3=5 ohms
Find the voltage at the node labeled "V".
If you see a shortcut, that's very good, but you should
try this method at least once anyway.
Scroll down for the answer.
V=5 volts.
All we did was swap V1 and V2, and R1 and R2, from the previous circuit
so the analysis comes out the same! If you didnt notice this it's ok
and you should try this method anyway.
Hello again!
Today we are going to talk about a shortcut method of analyzing
a circuit with more then one source in it. This method is so
handy we cant go without it!
Serial: EE-Ch0002-p106-{A939C0B7-558E-4687-9271-AB92E4935FD0}
In fig 12a we again have a circuit in which there are no series or
parallel combinations of resistors that will simplify the circuit.
This is the same circuit we used last time, except the node at the
top is labeled "V" and the node on the bottom is zero.
We can easily analyze this circuit using nodal analysis or mesh
analysis like we did last time, but this time we will try a very
different method. This new method might be called:
"Kill all sources" but officially it's called the
"Superposition Theorem".
The idea is to 'kill' all sources in the circuit and then one at
a time remove the 'kills' and analyze each node as we would any
other circuit. The nice thing about this method is that for
any given circuit the resulting set of circuits that remain to
be analyzed are often GREATLY simplified. This means we can usually
get fast answers even with circuits with many sources in them
because as each source is killed the circuit gets more and more
simple.
What it means to 'kill' a source is very simple:
For a voltage source it gets short circuited, and for
a current source (which we havent covered yet anyway) it
gets open circuited. To 'unkill' the source you simply
put it back the way it was originally.
Now lets look at an actual example...
If you did the mesh analysis on this circuit last time the first
thing you will note when you follow this analysis is how much
simpler it is and how much faster we get the answer, except
maybe for the drawing of the pictures
In fig 12a we'll set the values as follows:
V1=3v, V2=11v, R1=10 ohms, R2=5 ohms, R3=5 ohms
Now to start the analysis first we would 'kill' both sources,
V1 and V2 by short circuiting both of them. Then, we would
'unkill' say V1 first and start analyzing. The shorted V2
circuit is shown in the top of fig 12b and the resulting circuit
is shown right below it in 12b also. The thing most notable
now is that two of the resistors R2 and R3 now appear in
parallel, whereas before we shorted V2 this wasnt true.
Also, i've called the top node Va (the bottom is still zero).
The reason for the new name is because Va is a new node, more
or less, and will not be the same voltage as V in fig 12a because
we've changed the circuit by shorting V2.
Ok, now we must find out the voltage of Va in the bottom of fig 12b
so we start by simply finding the value of R2 and R3 in parallel:
Rp=1/(1/R2+1/R3)=1/(1/5+1/5)=1/0.4)=2.5
so we replace R2 and R3 with a 2.5 ohm resistor. This leads
to a circuit with R1 in series with a 2.5 ohm resistor driven
by voltage source V1 so we use voltage division:
Va=V1*2.5/(10+2.5), note R1=10
Since V1 is 3 volts this means
Va=3*2.5/12.5 which equals 0.6 volts, so
Va=0.6 volts and we have the answer to the value of the
node that comes from the contribution from source V1.
Now we have to do the same thing with the other source as shown
in fig 12c.
R1 in parallel with R3 is:
Rp=1/(1/R1+1/r3)=1/(1/10+1/5)=1/(0.1+0.2)=10/3
Next we can again use voltage division with values 10/3 and R2
so we get:
Vb=V2*(10/3)/(5+10/3)
so
Vb=11*10/(15+10)=11*10/25
so
Vb=4.4 volts.
Now we have the two answers, Va and Vb.
To get the final answer of what V is, we simply add Va and Vb:
V=Va+Vb
and we get
V=5 volts
Since we know R3=5 ohms we can get the circuit using Ohm's Law:
iR3=5/5=1 amp
Which agrees with the analysis from last time.
It's very interesting that by changing the circuit a little we
could reduce the circuit itself into two smaller circuits
(fig 12d) that could both be analyzed easier then the original
circuit and so provided us with another means of finding the answer.
Pretty cool if you ask me!!
With all good things there are some drawbacks, and this method
is no exception. In order to use this method on a given circuit
he first requirement is that ALL elements of the circuit are
what's known as "linear" elements. A linear element is defined
pretty intensely, but let's try to attach some form of definition
to what a linear element is.
A linear element has a linear voltage current relationship, so
that a multiplication by it's current of a factor K results
in a multiplication of it's voltage by the same factor K.
Now i dont know how much good that will do you at this point, so
let's look at a simple example in another field...
You drive two strong nails into the wall approximately on the same
level line, spacing them about 12 inches apart. You stretch a
rubber band across both nails so it stretches across horizontally.
Now you begin hanging weights in the middle of the band, so that
for each weight you measure a deflection of the horizontal
band in the center which now forms a vee shape.
Let's say the first weight of 1 ounce causes a deflection of
1 inch. Now you hang a second weight of 1 ounce and it too
causes another defection of 1 inch so the total now is 2 inches.
You hang a third weight of 1 ounce and another 1 inch results so
now it deflects by 3 inches.
In each case we hung the same weight on the band the same result
occurred: one additional inch of deflection.
The rubber band has thus responded linearly, in that a change
of input of a certain degree led to a change in output of
the same amount as with the original input amount.
On the other hand, if we keep hanging weights on it eventually
it breaks and so it 'stretches' far more for that last one ounce of weight.
This second kind of response would make it 'non linear', which is
simply put, "not linear".
Soon we'll encounter the LED again to see just why this is called
a nonlinear device. Things will make much more sense then too,
and why it's important to know a nonlinear device from a linear one.
For now, for the purposes of what we are doing at this point it will
probably be good enough to know that at least a resistor is a
linear element. When we come to other elements, we'll look a little
more at this too. It's interesting however, to note that in most
analysis the capacitor and inductor are considered linear too, so
we'll find this method is of use later as well.
Thanks for joining in today!
Take care,
MrAl
Last minute note:
Please dont try to short out your voltage sources
in circuits that you or someone else has built up.
This is a theoretical method, and if you want to
try it in practice you need to first open circuit
each voltage source (like a battery) before you
short the circuit where the battery originally
connected. Please remember this!
Today's Self Test Question
Take the circuit of fig 12a and make the following changes:
V1=11v, V2=3v, R1=5 ohms, R2=10 ohms, R3=5 ohms
Find the voltage at the node labeled "V".
If you see a shortcut, that's very good, but you should
try this method at least once anyway.
Scroll down for the answer.
V=5 volts.
All we did was swap V1 and V2, and R1 and R2, from the previous circuit
so the analysis comes out the same! If you didnt notice this it's ok
and you should try this method anyway.
