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In certain cases, the topology of a circuit lends itself to easier analysis if
the circuit is analyzed in terms of Conductances instead of Resistances.
By definition conductance G, is the reciprocal of resistance:
G = 1/R
Conductance is usually measured in units called mhos, i.e. the reciprocal of ohms.
Ohm's Law can be restated in terms of conductance as follows:
I = GV eq.1
Conductance is particularly useful when analyzing parallel elements such as
resistors. For example, the total resistance of a string of parallel resistors
is calculated as follows:
1/RT = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
But 1/R by definition is Conductance, so using Conductance, the total Conductance
of same circuit is
CT = C1 + C2 + C3+ ... Cn
A much simpler problem to solve.
If an entire circuit can be redrawn as a network of parallel branches of voltages
and conductances, Millman's theorem can be used to simplify analysis.
Millman's Theorem states: The node voltage of a network of parallel branches,
each branch consisting of a voltage source and a conductance in series (Fig 1) can
be calculated by the formula:
V0 = (G1V1 + G2V2 + ... + GNVN)
------------------------------------------- eq.2
(G1 + G2 + ... + GN)
To illustrate the use of Millman's Theorem, let's take a look at Figure 2, from
the article on Thevenin's Theorem.
In that article, we needed to calculate the
Thevenin voltage of the circuit looking back into the Thevenin equivalent circuit.
Figure 3 shows the Thevenin equivalent circuit re-drawn.
Taking Figure 3 one
step more, we redraw the circuit so each branch looks the same - we show a
single Conductance and a voltage source (even if it is 0) for each branch (Fig. 4).
From Kirchoff's Law, we know that the Algebraic Sum of all the currents entering
into the top node is:
i1 + i2 + i3 + i4 + i5 = 0 eq.3
Looking at branch 1 of fig.4, if the voltage drop across G1 is VG1, then
VTH = V1 - VG1
and
VG1 = V1 - VTH
Multiplying through by the conductance of R1 which is G1, we get
G1VG1 = G1V1 - G1VTH eq.4
From eq.1 we can write
i1 = G1V1 - G1VTH eq.5
Similarly, for all branch circuits, we write:
i2 = G2V2 - G2VTH eq.6
i3 = G3V3 - G3VTH eq.7
i4 = G4V4 - G4VTH eq.8
i5 = G5V5 - G5VTH eq.9
Substituting eqs.4 - 8 into eq.3:
(G1V1 + ... + G5V5) - (G1 + ... + G5)VTH = 0
Solving for VTH we get
VTH = (G1V1 + ... + G5V5)
---------------------------- eq.10
(G1 + ... + G5)
Referring to Figure 3, plugging in the values, we get
VTH = (0.046x15 + 0.125x0 + 0.067x24 + 0.125x0 + 0.111x12)
--------------------------------------------------------------------
(0.046 + 0.125 + 0.067 + 0.125 + 0 .111)
= 7.658V
Which, as you may recall is lot easier than solving for 4 mesh currents as we did before.
This relationship can be generalized for any circuit of N parallel branches as shown in Fig. 1. The node Voltage V0 can be calculated by the equation:
V0 = (G1V1 + G2V2 + ... + GNVN)
-------------------------------------
(G1 + G2 + ... + GN)
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