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Transcript
```Superposition Theorem
Superposition theorem is one of the important
method that takes a complex circuit and
simplifies it in a way that makes a perfect
sense and make that circuit a simple one and
easy to understand.
Superposition theorem helps to analyze a
complex circuit with multiple sources to
determine the net current/voltage in a desire
component when all sources are connected
Superposition Theorem………
• The strategy used in the Superposition
Theorem is to eliminate all but one
source of power within a network at a
time, using series/parallel analysis to
determine voltage drops and/or currents
within the modified network for each
power source separately.
Superposition Theorem………
• Then, once voltage drops and/or currents have
been determined for each power source
working separately, the values are all
“superimposed” on top of each other (added
/subtracted algebraically) to find the actual
voltage drops/currents with all sources active
at a time.
• Let's look at our example circuit again and
apply Superposition Theorem to it:
Superposition Theorem………
Since we have two sources of power (28V and 7V) in
this circuit, we will have to calculate two sets of
values for voltage drops and/or currents flowing
through each component, one for the circuit with
only the 28 volt battery in effect and other with the 7
volt battery in effect . . .
Superposition Theorem………
Analysis when B1 is connected and B2 is replaced by its equivalent.
Analyzing the circuit with only the 28 volt battery, we obtain the
following values for voltage and current:
Complete mathematical analysis for 28 Volt battery when
7volts battery is short.
R2 || R3 = 2 || 1 = 0.667
RT = R1 + (R2 || R3) = 4 + 0.667
= 4.667
IT = VT / RT = 28V / 4.667 = 5.99A  6A
V1 = I1 * R1 = 4 * 6A
= 24V
Using Current Divider Rule we can calculate the branch
currents I2 and I3 as follows:
I2 = (R3 / (R2 + R3)) * IT = (1 /3) * 6A
I3 = (R2 / (R2 + R3)) * IT = (2 /3) * 6A
= 2A
= 4A
Superposition Theorem………
Now V2 and V3 can be calculated by using Ohm’s Law as:
V2 = I2 * R2 = 2A * 2 = 4V and
V3 = I3 * R3 = 4A * 1 = 4V
V2 = V3 (because R2 || R3) and
IT = I1 = I2 + I3 = 2A + 4A
= 6A
and finally VT = V1 + (V2 = V3) = 24V + 4V = 28V
Superposition Theorem………
Analysis when B2 is connected and B1 is replaced by its equivalent.
When re-drawing the circuit for series/parallel analysis with one
source, all other voltage sources are replaced by wires (shorts),
and all current sources with open circuits (breaks). Since we only
have voltage sources (batteries) in our example circuit, we will
replace every inactive source during analysis with a wire.
Complete mathematical analysis for 7 Volt battery when
28volts battery is short.
R2 || R1 = 2 || 4 = 1.333
RT = R3 + (R2 || R1) = 1 + 1.333 = 2.333
IT = VT / RT
= 7V / 2.333 = 3.00A
V3 = IT * R3
= 3A * 1 = 3V
Using Current Divider Rule we can calculate the branch currents I1
and I2 as follows:
I1 = (R2 / (R1 + R2)) * IT = (2 /6) * 3A
I2 = (R1 / (R1 + R2)) * IT = (4 /6) * 3A
= 1A
= 2A
Now V1 and V2 can be calculated by using Ohm’s Law as:
V1 = I1 * R1 = 1A * 4 = 4V and
V2 = I2 * R2 = 2A * 2 = 4V
V1 = V2 (because R1 || R2) and
IT = I3 = I1 + I2 = 1A + 2A = 3A
and finally VT = V3 + (V1 = V2)
= 3V + 4V
= 7V
Important point:
• When superimposing these values of
voltage and current, we have to be very
careful to consider polarity of each
(voltage drop) and direction of current
(electron/hole flow), as the values have to
be algebraically added or subtracted
depending upon the direction of current.
Applying these superimposed voltage figures to the
circuit, the end result looks something like this:
• Currents add up algebraically as well, and can either
be superimposed as done with the resistor voltage
drops, or simply calculated from the final voltage
drops and respective resistances (I =E/R).
• Either way, the answers will be the same. Here I will
show the superposition method applied to current:
• Once again applying these superimposed figures to our circuit:
Superposition Theorem………
• Quite simple and elegant, don't you think?
• It must be noted, though, that the Superposition Theorem
works only for circuits that are reducible to series/parallel
combinations for each of the power sources at a time.
• Thus, this theorem is useless for analyzing an unbalanced
bridge circuit), and it only works where the underlying
equations are linear (no mathematical powers or roots).
• The requisite of linearity means that Superposition
Theorem is only applicable for determining voltage and
current, not power!!!
Superposition Theorem………
• Power dissipations, being nonlinear functions, do
not algebraically add to an accurate total when
only one source is considered at a time.
• The need for linearity also means this Theorem
cannot be applied in circuits where the resistance
of a component changes with voltage or current.
• Hence, networks containing components like
lamps (incandescent or gas-discharge) or varistors
could not be analyzed.
Superposition Theorem………
• Another prerequisite for Superposition
Theorem is that all components must be
“bilateral”, meaning that they behave the same
with electrons flowing either direction through
them.
• Resistors have no polarity-specific behavior,
and so the circuits we've been studying so far
all meet this criterion.
Superposition Theorem………
• The Superposition Theorem finds use in the study of
alternating current (AC) circuits, and semiconductor
(amplifier) circuits, where sometimes AC is often mixed
(superimposed) with DC.
• Because AC voltage and current equations (Ohm's Law)
are linear just like DC, we can use Superposition to
analyze the circuit with just the DC power source, then
just the AC power source, combining the results to tell
what will happen with both AC and DC sources in effect.
• For now, though, Superposition will suffice as a break
from having to do simultaneous equations to
Review…….
• The Superposition Theorem states that a circuit
can be analyzed with only one source of power at
a time, the corresponding component voltages and
currents algebraically added to find out what
they'll do with all power sources in effect.
• To negate all but one power source for analysis,
replace any source of voltage (batteries) with a
wire; replace any current source with an open
(break).
Question
Answer
```