**Hey, you're in the
**__
power s____upply
zone__*--*YOUR HANDYMAN ZONE!

**
Power Supply 1-2-3: Direct Current (DC)**

# Chapter 10

# DC
NETWORK ANALYSIS

Let's see how this method works on the
same example problem:

The first step in the Mesh Current
method is to identify “loops” within the circuit encompassing all
components. In our example circuit, the loop formed by B_{1},
R_{1}, and R_{2} will be the first while the loop
formed by B_{2}, R_{2}, and R_{3} will be
the second. The strangest part of the Mesh Current method is
envisioning circulating currents in each of the loops. In fact, this
method gets its name from the idea of these currents meshing
together between loops like sets of spinning gears:

The choice of each current's direction
is entirely arbitrary, just as in the Branch Current method, but the
resulting equations are easier to solve if the currents are going
the same direction through intersecting components (note how
currents I_{1} and I_{2} are both going “up” through
resistor R_{2}, where they “mesh,” or intersect). If the
assumed direction of a mesh current is wrong, the answer for that
current will have a negative value.

The next step is to label all voltage
drop polarities across resistors according to the assumed directions
of the mesh currents. Remember that the “upstream” end of a resistor
will always be negative, and the “downstream” end of a resistor
positive with respect to each other, since electrons are negatively
charged. The battery polarities, of course, are dictated by their
symbol orientations in the diagram, and may or may not “agree” with
the resistor polarities (assumed current directions):

Using Kirchhoff's Voltage Law, we can
now step around each of these loops, generating equations
representative of the component voltage drops and polarities. As
with the Branch Current method, we will denote a resistor's voltage
drop as the product of the resistance (in ohms) and its respective
mesh current (that quantity being unknown at this point). Where two
currents mesh together, we will write that term in the equation with
resistor current being the *sum* of the two meshing currents.

Tracing the left loop of the circuit,
starting from the upper-left corner and moving counter-clockwise
(the choice of starting points and directions is ultimately
irrelevant), counting polarity as if we had a voltmeter in hand, red
lead on the point ahead and black lead on the point behind, we get
this equation:

Notice that the middle term of the
equation uses the sum of mesh currents I_{1} and I_{2}
as the current through resistor R_{2}. This is because mesh
currents I_{1} and I_{2} are going the same
direction through R_{2}, and thus complement each other.
Distributing the coefficient of 2 to the I_{1} and I_{2}
terms, and then combining I_{1} terms in the equation, we
can simplify as such:

At this time we have one equation with
two unknowns. To be able to solve for two unknown mesh currents, we
must have two equations. If we trace the other loop of the circuit,
we can obtain another KVL equation and have enough data to solve for
the two currents. Creature of habit that I am, I'll start at the
upper-left hand corner of the right loop and trace
counter-clockwise:

Simplifying the equation as before, we
end up with:

Now, with two equations, we can use
one of several methods to mathematically solve for the unknown
currents I_{1} and I_{2}:

Knowing that these solutions are
values for *mesh* currents, not *branch* currents, we must
go back to our diagram to see how they fit together to give currents
through all components:

The solution of -1 amp for I_{2}
means that our initially assumed direction of current was incorrect.
In actuality, I_{2} is flowing in a counter-clockwise
direction at a value of (positive) 1 amp:

This change of current direction from
what was first assumed will alter the polarity of the voltage drops
across R_{2} and R_{3} due to current I_{2}.
From here, we can say that the current through R_{1} is 5
amps, with the voltage drop across R_{1} being the product
of current and resistance (E=IR), 20 volts (positive on the left and
negative on the right). Also, we can safely say that the current
through R_{3} is 1 amp, with a voltage drop of 1 volt (E=IR),
positive on the left and negative on the right. But what is
happening at R_{2}?

Mesh current I_{1} is going
“up” through R_{2}, while mesh current I_{2} is
going “down” through R_{2}. To determine the actual current
through R_{2}, we must see how mesh currents I_{1}
and I_{2} interact (in this case they're in opposition), and
algebraically add them to arrive at a final value. Since I_{1}
is going “up” at 5 amps, and I_{2} is going “down” at 1 amp,
the *real* current through R_{2} must be a value of 4
amps, going “up:”

A current of 4 amps through R_{2}'s
resistance of 2 Ω gives us a voltage drop of 8 volts (E=IR),
positive on the top and negative on the bottom.

The primary advantage of Mesh Current
analysis is that it generally allows for the solution of a large
network with fewer unknown values and fewer simultaneous equations.
Our example problem took three equations to solve the Branch Current
method and only two equations using the Mesh Current method. This
advantage is much greater as networks increase in complexity:

To solve this network using Branch
Currents, we'd have to establish five variables to account for each
and every unique current in the circuit (I_{1} through I_{5}).
This would require five equations for solution, in the form of two
KCL equations and three KVL equations (two equations for KCL at the
nodes, and three equations for KVL in each loop):

I suppose if you have nothing better
to do with your time than to solve for five unknown variables with
five equations, you might not mind using the Branch Current method
of analysis for this circuit. For those of us who *have* better
things to do with our time, the Mesh Current method is a whole lot
easier, requiring only three unknowns and three equations to solve:

Less equations to work with is a
decided advantage, especially when performing simultaneous equation
solution by hand (without a calculator).

Another type of circuit that lends
itself well to Mesh Current is the unbalanced Wheatstone Bridge.
Take this circuit, for example:

Since the ratios of R_{1}/R_{4}
and R_{2}/R_{5} are unequal, we know that there will
be voltage across resistor R_{3}, and some amount of current
through it. As discussed at the beginning of this chapter, this type
of circuit is irreducible by normal series-parallel analysis, and
may only be analyzed by some other method.

We could apply the Branch Current
method to this circuit, but it would require *six* currents (I_{1}
through I_{6}), leading to a very large set of simultaneous
equations to solve. Using the Mesh Current method, though, we may
solve for all currents and voltages with much fewer variables.

The first step in the Mesh Current
method is to draw just enough mesh currents to account for all
components in the circuit. Looking at our bridge circuit, it should
be obvious where to place two of these currents:

The directions of these mesh currents,
of course, is arbitrary. However, two mesh currents is not enough in
this circuit, because neither I_{1} nor I_{2} goes
through the battery. So, we must add a third mesh current, I_{3}:

Here, I have chosen I_{3} to
loop from the bottom side of the battery, through R_{4},
through R_{1}, and back to the top side of the battery. This
is not the only path I could have chosen for I_{3}, but it
seems the simplest.

Now, we must label the resistor
voltage drop polarities, following each of the assumed currents'
directions:

Notice something very important here:
at resistor R_{4}, the polarities for the respective mesh
currents do not agree. This is because those mesh currents (I_{2}
and I_{3}) are going through R_{4} in different
directions. This does not preclude the use of the Mesh Current
method of analysis, but it does complicate it a bit. Though later,
we will show how to avoid the R_{4} current clash. (See
Example below)

Generating a KVL equation for the top
loop of the bridge, starting from the top node and tracing in a
clockwise direction:

In this equation, we represent the
common directions of currents by their *sums* through common
resistors. For example, resistor R_{3}, with a value of 100
Ω, has its voltage drop represented in the above KVL equation by the
expression 100(I_{1} + I_{2}), since both currents I_{1}
and I_{2} go through R_{3} from right to left. The
same may be said for resistor R_{1}, with its voltage drop
expression shown as 150(I_{1} + I_{3}), since both I_{1}
and I_{3} go from bottom to top through that resistor, and
thus work *together* to generate its voltage drop.

Generating a KVL equation for the
bottom loop of the bridge will not be so easy, since we have two
currents going against each other through resistor R_{4}.
Here is how I do it (starting at the right-hand node, and tracing
counter-clockwise):

Note how the second term in the
equation's original form has resistor R_{4}'s value of 300 Ω
multiplied by the *difference* between I_{2} and I_{3}
(I_{2} - I_{3}). This is how we represent the
combined effect of two mesh currents going in opposite directions
through the same component. Choosing the appropriate mathematical
signs is very important here: 300(I_{2} - I_{3})
does not mean the same thing as 300(I_{3} - I_{2}).
I chose to write 300(I_{2} - I_{3}) because I was
thinking first of I_{2}'s effect (creating a positive
voltage drop, measuring with an imaginary voltmeter across R_{4},
red lead on the bottom and black lead on the top), and secondarily
of I_{3}'s effect (creating a negative voltage drop, red
lead on the bottom and black lead on the top). If I had thought in
terms of I_{3}'s effect first and I_{2}'s effect
secondarily, holding my imaginary voltmeter leads in the same
positions (red on bottom and black on top), the expression would
have been -300(I_{3} - I_{2}). Note that this
expression *is* mathematically equivalent to the first one:
+300(I_{2} - I_{3}).

Well, that takes care of two
equations, but I still need a third equation to complete my
simultaneous equation set of three variables, three equations. This
third equation must also include the battery's voltage, which up to
this point does not appear in either two of the previous KVL
equations. To generate this equation, I will trace a loop again with
my imaginary voltmeter starting from the battery's bottom (negative)
terminal, stepping clockwise (again, the direction in which I step
is arbitrary, and does not need to be the same as the direction of
the mesh current in that loop):

Solving for I_{1}, I_{2},
and I_{3} using whatever simultaneous equation method we
prefer:

**Example:**

Use Octave to find the solution for I_{1},
I_{2}, and I_{3} from the above simplified form of
equations. [octav]

**Solution:**

In Octave, an open source Matlab®
clone, enter the coefficients into the A matrix between square
brackets with column elements comma separated, and rows semicolon
separated.[octav] Enter the voltages into the column vector: b. The
unknown currents: I_{1}, I_{2}, and I_{3}
are calculated by the command: x=A\b. These are contained within the
x column vector.

octave:1>A = [300,100,150;100,650,-300;-150,300,-450]

A =

300 100 150

100 650 -300

-150 300 -450

octave:2> b = [0;0;-24]

b =

0

0

-24

octave:3> x = A\b

x =

-0.093793

0.077241

0.136092

The negative value arrived at for I_{1}
tells us that the assumed direction for that mesh current was
incorrect. Thus, the actual current values through each resistor is
as such:

Calculating voltage drops across each
resistor:

A SPICE simulation confirms the
accuracy of our voltage calculations:[spi]

unbalanced wheatstone bridge

v1 1 0

r1 1 2 150

r2 1 3 50

r3 2 3 100

r4 2 0 300

r5 3 0 250

.dc v1 24 24 1

.print dc v(1,2) v(1,3) v(3,2) v(2,0) v(3,0)

.end

v1 v(1,2) v(1,3) v(3,2) v(2) v(3)

2.400E+01 6.345E+00 4.690E+00 1.655E+00 1.766E+01 1.931E+01

**Example:**

(a) Find a new path for current I_{3}
that does not produce a conflicting polarity on any resistor
compared to I_{1} or I_{2}. R_{4} was the
offending component. (b) Find values for I_{1}, I_{2},
and I_{3}. (c) Find the five resistor currents and compare
to the previous values.

**Solution: **[dvn]

(a) Route I_{3} through R_{5},
R_{3} and R_{1} as shown:

Note that the conflicting polarity on
R_{4} has been removed. Moreover, none of the other
resistors have conflicting polarities.

(b) Octave, an open source (free)
matlab clone, yields a mesh current vector at “x”:[octav]

octave:1> A = [300,100,250;100,650,350;-250,-350,-500]

A =

300 100 250

100 650 350

-250 -350 -500

octave:2> b = [0;0;-24]

b =

0

0

-24

octave:3> x = A\b

x =

-0.093793

-0.058851

0.136092

Not all currents I_{1}, I_{2},
and I_{3} are the same (I_{2}) as the previous
bridge because of different loop paths However, the resistor
currents compare to the previous values:

I_{R1} = I_{1} + I_{3} = -93.793 ma + 136.092 ma = 42.299 ma

I_{R2} = I_{1} = -93.793 ma

I_{R3} = I_{1} + I_{2} + I_{3} = -93.793 ma -58.851 ma + 136.092 ma = -16.552 ma

I_{R4} = I_{2} = -58.851 ma

I_{R5} = I_{2} + I_{3} = -58.851 ma + 136.092 ma = 77.241 ma

Since the resistor currents are the
same as the previous values, the resistor voltages will be identical
and need not be calculated again.

**REVIEW:**
- Steps to follow for the “Mesh Current”
method of analysis:
- (1) Draw mesh currents in loops of
circuit, enough to account for all components.
- (2) Label resistor voltage drop polarities
based on assumed directions of mesh currents.
- (3) Write KVL equations for each loop of
the circuit, substituting the product IR for E in each resistor
term of the equation. Where two mesh currents intersect through
a component, express the current as the algebraic sum of those
two mesh currents (i.e. I
_{1} + I_{2}) if the
currents go in the same direction through that component. If
not, express the current as the difference (i.e. I_{1} -
I_{2}).
- (4) Solve for unknown mesh currents
(simultaneous equations).
- (5) If any solution is negative, then the
assumed current direction is wrong!
- (6) Algebraically add mesh currents to
find current in components sharing multiple mesh currents.
- (7) Solve for voltage drops across all
resistors (E=IR).