Current due to this additional source of 1.11 V in the 3 Ω branch i
a
is,
i
a
=
1.11
(1+3+(6∥4))
i
a
=
1.11
(1+3+2.4)
i
a
=0.17A
This current flows in the opposite direction to that of the original current i through the 3 Ω
branch(i.e. i
a
is opposite to i)
Hence Ammeter reading = i
a
-i=(1.11 – 0.17) = 0.94 A
Millman’s Theorem:
Millman’s Theorem is a theorem which helps in simplifying electrical
networks
with a bunch of parallel branches. The utility of this theorem that, any number of parallel voltage
sources can be reduced to one equivalent one.
Millman’s Theorem Statement:
The Millman’s Theorem states that – when a number of voltage sources (V
1
, V
2
,
V
3
……… V
n
) are in parallel having internal resistance (R
1
, R
2
, R
3
………….R
n
) respectively, the
arrangement can replace by a single equivalent voltage source V in
series with an equivalent
series resistance R.
As per Millman’s Theorem,
V =
±V
1
G
1
± V
2
G
2
± ⋯ … … … ± V
n
G
n
G
1
+ G
2
+ ⋯ … … … + G
n
;
𝑅 =
1
𝐺
=
1
G
1
+ G
2
+ ⋯ … … … + G
n
A DC network of numerous parallel voltage sources with internal resistances supplying power to
a load resistance RL as shown in the figure below.
Let I represent the resultant current of the parallel current sources
while G the equivalent
conductance as shown in the figure below.
I = I
1
+ I
2
+ I
3
… … … … … ;
G = G
1
+ G
2
+ G
3
… … … … …
(or)
I =
±I
1
R
1
±I
2
R
2
±⋯………±I
n
R
n
R
1
+R
2
+⋯………+R
n
;
𝐺 =
1
𝑅
=
1
R
1
+ R
2
+ ⋯ … … … + R
n
The resulting current source is converted to an equivalent voltage source as shown in the fig.
𝑉 =
𝐼
𝐺
=
±I
1
± I
2
± ⋯ … … … ± I
n
G
1
+ G
2
+ ⋯ … … … + G
n
𝑅 =
1
𝐺
=
1
G
1
+ G
2
+ ⋯ … … … + G
n
And as we know,I = V/R, and we can also write R = 1/G as G = 1/R
So the equation can be
written as,
𝑉 =
±
V
1
R
1
±
V
2
R
2
± ⋯ … … … ±
V
n
R
n
1
R
1
+
1
R
2
+ ⋯ … … … +
1
R
n
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