Mesh current method
The Mesh Current Method is quite
similar to the Branch Current method in that it uses
simultaneous equations, Kirchhoff's Voltage Law, and Ohm's
Law to determine unknown currents in a network. It differs
from the Branch Current method in that it does not
use Kirchhoff's Current Law, and it is usually able to solve
a circuit with less unknown variables and less simultaneous
equations, which is especially nice if you're forced to
solve without a calculator.
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 upperleft corner and moving
counterclockwise (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 upperleft hand corner of the
right loop and trace counterclockwise:
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
counterclockwise 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 seriesparallel 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. Normally, we try to
avoid this when establishing our mesh current directions,
but in a bridge circuit it is unavoidable: two of the mesh
currents will inevitably clash through a component. This
does not preclude the use of the Mesh Current method of
analysis, but it does complicate it a bit.
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 righthand node, and
tracing counterclockwise):
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:
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 will confirm the accuracy
of our voltage calculations:
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

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).
