>> Okay, here's a problem
that's a little bit different
than previous problems, where
there's a voltage source,
a 0.6 volt source, where
neither side is ground.
So we're going to see in this
problem how you handle a case
like this.
We already know, and you can see
in this problem I already have,
our reference node,
which is the ground node,
circled in red here.
And we already know
from previous problems
that if we have a voltage
source where one side is ground,
the other side the node
that the other side
of the source is
connected to is known.
So this node here gas 20
volts with respect to ground,
so that's a known node.
And then also over
here, this node, okay,
that's between this one K olm
resistor and the plus side
of this five-volt source,
well, that's also known, okay,
here's five volts at this
node with respect to ground.
So what we have here
is two unknown nodes.
There's an unknown node here,
that I'm going to label A.
And on the other side of this .6
volt source is also an unknown
node, that I'll label B, okay.
You see when you
have a voltage source
where neither side is ground,
the value of that voltage
source, .6 volts in this case,
tells you the difference in
voltage between the two nodes
that it's connected to.
So we know that VA is
.6 volts higher than VB.
And the reason why I know
it's .6 volts higher instead
of .6 volts lower is
the polarity, okay.
Remember the polarity
tells me the higher end
of the voltage difference.
So since node A is
connected to the plus side
of this .6 volt source, node VA
is .6 volts higher than node VB.
But we don't know what the
voltage at node B with respect
to ground is, or what
the voltage at node A
with respect to ground is yet.
But we will after we write
our node voltage expressions.
So to set up our node
voltage equations,
I'm going to define
a current like this,
that I'm going to call I1.
And I'm going to
call this current I2.
This current through the .6 volt
source, I'm going to label IS.
And then this current
through this 1K here,
I'm going to label I3.
Try that again.
So this here is I3.
All right, now I
do my KCL equation.
So KCL at node A, okay, is
going to be I1 flowing in,
and that's going to
equal I2 plus IS,
which is flowing out, okay.
And then KCL at B is just going
to be IS, which is flowing in.
And then I3.
All right, so first step node
voltage method you write your
KCL equations, now I'm going to
write these currents in terms
of voltage using Ohm's law.
So for I1, it's just going
to be 20 minus VA over 1K,
I2 is just VA minus
zero over 1K.
Now, see this is something
we haven't seen before.
And that's, when you
have a current defined
through a voltage
source like this,
you can't use Ohm's law, okay.
Because if you were to take
the ideal internal resistance
of the voltage source, which
is zero, this term would blow
up because you'd
have .6 volts divided
by that ideal internal
resistance
of zero, which is infinity.
So we're just going to
leave that current as IS.
And then you'll see what's --
or how we're going to take care
of it in the next expression.
Because here, we have IS on
this side of the equal sign
for the KCL at node B, but
we can write I3 in terms
of voltages and Ohm's law.
Because I3, which is
this current right here,
that's a three, is the
current through this 1K.
So we can use Ohm's law there.
So this would just be
VB minus 5 over 1K.
All right, so now I can take IS,
plug it into this
expression here, right.
So now after I plug in my IS
into my equation at node A,
we would get 20 minus
VA over 1K, VA.
See, VA divided by 1K, and
then plug it in here for IS.
This would be plus
VB minus 5 over 1K.
So you can see now that we
have one equation in terms
of voltage, but there's two
unknowns, right, VA and VB.
So we need another
equation in order to solve.
And the way we get that
other equation is we go back
to what I was talking
about earlier,
that this voltage source between
these two nodes tell us the
difference between the
two nodes voltages.
See, I know node A and
B differ by .6 volts.
And the polarity tells
me which one is greater.
So node A is .6 volts greater
than node B. So I write that,
EA minus VB is 0.6 volts.
All right, so there's your two
equations, your two unknowns.
What I did is I solved for VA.
And of you solve correctly for
VA, VA comes out to 8.53 volts.
So that means that VB
has to be .6 volts lower.
So VB is 7.93 volts.
Okay, so the key points in
a problem like this is that,
when you have a voltage source
where neither side is ground,
just label the current through
it as IS or some other variable.
Realize that you can't use
Ohm's law to get the current
through the source
in terms of voltage.
You're going to have to do
that with the KCL equation
at a different node, where,
in this case, it was node B.
And then you're going to end up
with an equation or equations
where you have one extra
unknown compared to the number
of equations you have.
So you're going to have to
write another equation based
on knowing that the value
of the voltage source
between the two nodes
that aren't ground
gives you the difference
between those two nodes.
