Component failure
analysis
"I consider that I understand an
equation when I can predict the properties of its
solutions, without actually solving it."
P.A.M Dirac, physicist
There is a lot of truth to that quote from
Dirac. With a little modification, I can extend his wisdom
to electric circuits by saying, "I consider that I
understand a circuit when I can predict the approximate
effects of various changes made to it without actually
performing any calculations."
At the end of the series and parallel
circuits chapter, we briefly considered how circuits could
be analyzed in a qualitative rather than
quantitative manner. Building this skill is an important
step towards becoming a proficient troubleshooter of
electric circuits. Once you have a thorough understanding of
how any particular failure will affect a circuit (i.e. you
don't have to perform any arithmetic to predict the
results), it will be much easier to work the other way
around: pinpointing the source of trouble by assessing how a
circuit is behaving.
Also shown at the end of the series and
parallel circuits chapter was how the table method works
just as well for aiding failure analysis as it does for the
analysis of healthy circuits. We may take this technique one
step further and adapt it for total qualitative analysis. By
"qualitative" I mean working with symbols
representing "increase," "decrease," and "same" instead of
precise numerical figures. We can still use the principles
of series and parallel circuits, and the concepts of Ohm's
Law, we'll just use symbolic qualities instead of
numerical quantities. By doing this, we can gain more
of an intuitive "feel" for how circuits work rather than
leaning on abstract equations, attaining Dirac's definition
of "understanding."
Enough talk. Let's try this technique on a
real circuit example and see how it works:
This is the first "convoluted" circuit we
straightened out for analysis in the last section. Since you
already know how this particular circuit reduces to series
and parallel sections, I'll skip the process and go straight
to the final form:
R3 and R4 are in
parallel with each other; so are R1 and R2.
The parallel equivalents of R3//R4 and
R1//R2 are in series with each other.
Expressed in symbolic form, the total resistance for this
circuit is as follows:
RTotal = (R1//R2)--(R3//R4)
First, we need to formulate a table with all
the necessary rows and columns for this circuit:
Next, we need a failure scenario. Let's
suppose that resistor R2 were to fail shorted. We
will assume that all other components maintain their
original values. Because we'll be analyzing this circuit
qualitatively rather than quantitatively, we won't be
inserting any real numbers into the table. For any quantity
unchanged after the component failure, we'll use the word
"same" to represent "no change from before." For any
quantity that has changed as a result of the failure, we'll
use a down arrow for "decrease" and an up arrow for
"increase." As usual, we start by filling in the spaces of
the table for individual resistances and total voltage, our
"given" values:
The only "given" value different from the
normal state of the circuit is R2, which we said
was failed shorted (abnormally low resistance). All other
initial values are the same as they were before, as
represented by the "same" entries. All we have to do now is
work through the familiar Ohm's Law and series-parallel
principles to determine what will happen to all the other
circuit values.
First, we need to determine what happens to
the resistances of parallel subsections R1//R2
and R3//R4. If neither R3
nor R4 have changed in resistance value, then
neither will their parallel combination. However, since the
resistance of R2 has decreased while R1
has stayed the same, their parallel combination must
decrease in resistance as well:
Now, we need to figure out what happens to
the total resistance. This part is easy: when we're dealing
with only one component change in the circuit, the change in
total resistance will be in the same direction as the change
of the failed component. This is not to say that the
magnitude of change between individual component and
total circuit will be the same, merely the direction
of change. In other words, if any single resistor decreases
in value, then the total circuit resistance must also
decrease, and visa-versa. In this case, since R2
is the only failed component, and its resistance has
decreased, the total resistance must decrease:
Now we can apply Ohm's Law (qualitatively)
to the Total column in the table. Given the fact that total
voltage has remained the same and total resistance has
decreased, we can conclude that total current must increase
(I=E/R).
In case you're not familiar with the
qualitative assessment of an equation, it works like this.
First, we write the equation as solved for the unknown
quantity. In this case, we're trying to solve for current,
given voltage and resistance:
Now that our equation is in the proper form,
we assess what change (if any) will be experienced by "I,"
given the change(s) to "E" and "R":
If the denominator of a fraction decreases
in value while the numerator stays the same, then the
overall value of the fraction must increase:
Therefore, Ohm's Law (I=E/R) tells us that
the current (I) will increase. We'll mark this conclusion in
our table with an "up" arrow:
With all resistance places filled in the
table and all quantities determined in the Total column, we
can proceed to determine the other voltages and currents.
Knowing that the total resistance in this table was the
result of R1//R2 and R3//R4
in series, we know that the value of total current
will be the same as that in R1//R2 and
R3//R4 (because series components
share the same current). Therefore, if total current
increased, then current through R1//R2
and R3//R4 must also have increased
with the failure of R2:
Fundamentally, what we're doing here with a
qualitative usage of Ohm's Law and the rules of series and
parallel circuits is no different from what we've done
before with numerical figures. In fact, it's a lot easier
because you don't have to worry about making an arithmetic
or calculator keystroke error in a calculation. Instead,
you're just focusing on the principles behind the
equations. From our table above, we can see that Ohm's Law
should be applicable to the R1//R2 and
R3//R4 columns. For R3//R4,
we figure what happens to the voltage, given an increase in
current and no change in resistance. Intuitively, we can see
that this must result in an increase in voltage across the
parallel combination of R3//R4:
But how do we apply the same Ohm's Law
formula (E=IR) to the R1//R2 column,
where we have resistance decreasing and current
increasing? It's easy to determine if only one variable is
changing, as it was with R3//R4, but
with two variables moving around and no definite numbers to
work with, Ohm's Law isn't going to be much help. However,
there is another rule we can apply horizontally to
determine what happens to the voltage across R1//R2:
the rule for voltage in series circuits. If the voltages
across R1//R2 and R3//R4
add up to equal the total (battery) voltage and we know that
the R3//R4 voltage has increased while
total voltage has stayed the same, then the voltage across R1//R2
must have decreased with the change of R2's
resistance value:
Now we're ready to proceed to some new
columns in the table. Knowing that R3 and R4
comprise the parallel subsection R3//R4,
and knowing that voltage is shared equally between parallel
components, the increase in voltage seen across the parallel
combination R3//R4 must also be seen
across R3 and R4 individually:
The same goes for R1 and R2.
The voltage decrease seen across the parallel combination of
R1 and R2 will be seen across R1
and R2 individually:
Applying Ohm's Law vertically to those
columns with unchanged ("same") resistance values, we can
tell what the current will do through those components.
Increased voltage across an unchanged resistance leads to
increased current. Conversely, decreased voltage across an
unchanged resistance leads to decreased current:
Once again we find ourselves in a position
where Ohm's Law can't help us: for R2, both
voltage and resistance have decreased, but without knowing
how much each one has changed, we can't use the I=E/R
formula to qualitatively determine the resulting change in
current. However, we can still apply the rules of series and
parallel circuits horizontally. We know that the
current through the R1//R2 parallel
combination has increased, and we also know that the current
through R1 has decreased. One of the rules of
parallel circuits is that total current is equal to the sum
of the individual branch currents. In this case, the current
through R1//R2 is equal to the current
through R1 added to the current through R2.
If current through R1//R2 has
increased while current through R1 has decreased,
current through R2 must have increased:
And with that, our table of qualitative
values stands completed. This particular exercise may look
laborious due to all the detailed commentary, but the actual
process can be performed very quickly with some practice. An
important thing to realize here is that the general
procedure is little different from quantitative analysis:
start with the known values, then proceed to determining
total resistance, then total current, then transfer figures
of voltage and current as allowed by the rules of series and
parallel circuits to the appropriate columns.
A few general rules can be memorized to
assist and/or to check your progress when proceeding with
such an analysis:
-
For any single component failure
(open or shorted), the total resistance will always change
in the same direction (either increase or decrease) as the
resistance change of the failed component.
-
When a component fails shorted, its
resistance always decreases. Also, the current through it
will increase, and the voltage across it may drop.
I say "may" because in some cases it will remain the same
(case in point: a simple parallel circuit with an ideal
power source).
-
When a component fails open, its
resistance always increases. The current through that
component will decrease to zero, because it is an
incomplete electrical path (no continuity). This may
result in an increase of voltage across it. The same
exception stated above applies here as well: in a simple
parallel circuit with an ideal voltage source, the voltage
across an open-failed component will remain unchanged.
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