Branch current method
The first and most straightforward network
analysis technique is called the Branch Current Method.
In this method, we assume directions of currents in a
network, then write equations describing their relationships
to each other through Kirchhoff's and Ohm's Laws. Once we
have one equation for every unknown current, we can solve
the simultaneous equations and determine all currents, and
therefore all voltage drops in the network.
Let's use this circuit to illustrate the
method:
The first step is to choose a node (junction
of wires) in the circuit to use as a point of reference for
our unknown currents. I'll choose the node joining the right
of R_{1}, the top of R_{2}, and the left of
R_{3}.
At this node, guess which directions the
three wires' currents take, labeling the three currents as I_{1},
I_{2}, and I_{3}, respectively. Bear in mind
that these directions of current are speculative at this
point. Fortunately, if it turns out that any of our guesses
were wrong, we will know when we mathematically solve for
the currents (any "wrong" current directions will show up as
negative numbers in our solution).
Kirchhoff's Current Law (KCL) tells us that
the algebraic sum of currents entering and exiting a node
must equal zero, so we can relate these three currents (I_{1},
I_{2}, and I_{3}) to each other in a single
equation. For the sake of convention, I'll denote any
current entering the node as positive in sign, and
any current exiting the node as negative in sign:
The next step is to label all voltage drop
polarities across resistors according to the assumed
directions of the 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, remain as
they were according to their symbology (short end negative,
long end positive). It is okay if the polarity of a
resistor's voltage drop doesn't match with the polarity of
the nearest battery, so long as the resistor voltage
polarity is correctly based on the assumed direction of
current through it. In some cases we may discover that
current will be forced backwards through a battery,
causing this very effect. The important thing to remember
here is to base all your resistor polarities and subsequent
calculations on the directions of current(s) initially
assumed. As stated earlier, if your assumption happens to be
incorrect, it will be apparent once the equations have been
solved (by means of a negative solution). The magnitude of
the solution, however, will still be correct.
Kirchhoff's Voltage Law (KVL) tells us that
the algebraic sum of all voltages in a loop must equal zero,
so we can create more equations with current terms (I_{1},
I_{2}, and I_{3}) for our simultaneous
equations. To obtain a KVL equation, we must tally voltage
drops in a loop of the circuit, as though we were measuring
with a real voltmeter. I'll choose to trace the left loop of
this circuit first, starting from the upperleft corner and
moving counterclockwise (the choice of starting points and
directions is arbitrary). The result will look like this:
Having completed our trace of the left loop,
we add these voltage indications together for a sum of zero:
Of course, we don't yet know what the
voltage is across R_{1} or R_{2}, so we
can't insert those values into the equation as numerical
figures at this point. However, we do know that all
three voltages must algebraically add to zero, so the
equation is true. We can go a step further and express the
unknown voltages as the product of the corresponding unknown
currents (I_{1} and I_{2}) and their
respective resistors, following Ohm's Law (E=IR), as well as
eliminate the 0 term:
Since we know what the values of all the
resistors are in ohms, we can just substitute those figures
into the equation to simplify things a bit:
You might be wondering why we went through
all the trouble of manipulating this equation from its
initial form (28 + E_{R2} + E_{R1}). After
all, the last two terms are still unknown, so what advantage
is there to expressing them in terms of unknown voltages or
as unknown currents (multiplied by resistances)? The purpose
in doing this is to get the KVL equation expressed using the
same unknown variables as the KCL equation, for this
is a necessary requirement for any simultaneous equation
solution method. To solve for three unknown currents (I_{1},
I_{2}, and I_{3}), we must have three
equations relating these three currents (not
voltages!) together.
Applying the same steps to the right loop of
the circuit (starting at the chosen node and moving
counterclockwise), we get another KVL equation:
Knowing now that the voltage across each
resistor can be and should be expressed as the
product of the corresponding current and the (known)
resistance of each resistor, we can rewrite the equation as
such:
Now we have a mathematical system of three
equations (one KCL equation and two KVL equations) and three
unknowns:
For some methods of solution (especially any
method involving a calculator), it is helpful to express
each unknown term in each equation, with any constant value
to the right of the equal sign, and with any "unity" terms
expressed with an explicit coefficient of 1. Rewriting the
equations again, we have:
Using whatever solution techniques are
available to us, we should arrive at a solution for the
three unknown current values:
So, I_{1} is 5 amps, I_{2}
is 4 amps, and I_{3} is a negative 1 amp. But what
does "negative" current mean? In this case, it means that
our assumed direction for I_{3} was opposite
of its real direction. Going back to our original
circuit, we can redraw the current arrow for I_{3}
(and redraw the polarity of R_{3}'s voltage drop to
match):
Notice how current is being pushed backwards
through battery 2 (electrons flowing "up") due to the higher
voltage of battery 1 (whose current is pointed "down" as it
normally would)! Despite the fact that battery B_{2}'s
polarity is trying to push electrons down in that branch of
the circuit, electrons are being forced backwards through it
due to the superior voltage of battery B_{1}. Does
this mean that the stronger battery will always "win" and
the weaker battery always get current forced through it
backwards? No! It actually depends on both the batteries'
relative voltages and the resistor values in the
circuit. The only sure way to determine what's going on is
to take the time to mathematically analyze the network.
Now that we know the magnitude of all
currents in this circuit, we can calculate voltage drops
across all resistors with Ohm's Law (E=IR):
Let us now analyze this network using SPICE
to verify our voltage figures. We could analyze current as
well with SPICE, but since that requires the insertion of
extra components into the circuit, and because we know that
if the voltages are all the same and all the resistances are
the same, the currents must all be the same, I'll opt
for the less complex analysis. Here's a redrawing of our
circuit, complete with node numbers for SPICE to reference:
network analysis example
v1 1 0
v2 3 0 dc 7
r1 1 2 4
r2 2 0 2
r3 2 3 1
.dc v1 28 28 1
.print dc v(1,2) v(2,0) v(2,3)
.end
v1 v(1,2) v(2) v(2,3)
2.800E+01 2.000E+01 8.000E+00 1.000E+00
Sure enough, the voltage figures all turn
out to be the same: 20 volts across R_{1} (nodes 1
and 2), 8 volts across R_{2} (nodes 2 and 0), and 1
volt across R_{3} (nodes 2 and 3). Take note of the
signs of all these voltage figures: they're all positive
values! SPICE bases its polarities on the order in which
nodes are listed, the first node being positive and the
second node negative. For example, a figure of positive (+)
20 volts between nodes 1 and 2 means that node 1 is positive
with respect to node 2. If the figure had come out negative
in the SPICE analysis, we would have known that our actual
polarity was "backwards" (node 1 negative with respect to
node 2). Checking the node orders in the SPICE listing, we
can see that the polarities all match what we determined
through the Branch Current method of analysis.

REVIEW:

Steps to follow for the "Branch Current"
method of analysis:

(1) Choose a node and assume directions of
currents.

(2) Write a KCL equation relating currents
at the node.

(3) Label resistor voltage drop polarities
based on assumed currents.

(4) Write KVL equations for each loop of
the circuit, substituting the product IR for E in each
resistor term of the equations.

(5) Solve for unknown branch currents
(simultaneous equations).

(6) If any solution is negative, then the
assumed direction of current for that solution is wrong!

(7) Solve for voltage drops across all
resistors (E=IR).
