According to this theorem, if there are a number of
emfs acting simultaneously in any linear bilateral network, then emf acts
independently of the others i.e. as if the other emf did not exist. The value
of current in any conductors is the algebraic sum of the current due to each
emf. Similarly, voltage across any conductor is the algebraic sum of the
voltages which each emf would have produced while acting singly. In other words,
current in or voltage across any conductor in the network is obtained by
superimposing the currents and voltages due to each emf in the network. It is important
to keep in mind that this theorem is appreciable only to linear network where
current is linearly related to voltage as per Ohm’s Law.
emfs acting simultaneously in any linear bilateral network, then emf acts
independently of the others i.e. as if the other emf did not exist. The value
of current in any conductors is the algebraic sum of the current due to each
emf. Similarly, voltage across any conductor is the algebraic sum of the
voltages which each emf would have produced while acting singly. In other words,
current in or voltage across any conductor in the network is obtained by
superimposing the currents and voltages due to each emf in the network. It is important
to keep in mind that this theorem is appreciable only to linear network where
current is linearly related to voltage as per Ohm’s Law.
Hence, this theorem may be stated as follows:
In a network of linear resistances containing more
than one generator (or source emf), the current which flows at any point is the
sum of all the current which would flow at that point of each generator were
considered separately and all the other generators replaced for the time being
by resistance equal to their internal resistances
than one generator (or source emf), the current which flows at any point is the
sum of all the current which would flow at that point of each generator were
considered separately and all the other generators replaced for the time being
by resistance equal to their internal resistances
Solution
Disconnect the 12V battery leaving
the 1Ω internal resistance as shown in the figure below
the 1Ω internal resistance as shown in the figure below
Example 2
By using superposition theorem, find the current in
resistance R shown in the figure below
resistance R shown in the figure below
From the figure above, the resistances seen by the
2.15V source are 1Ω and 0.05Ω parallel resistance in series with a 0.04Ω
resistance. Therefore resistance seen by the 2.15V is
2.15V source are 1Ω and 0.05Ω parallel resistance in series with a 0.04Ω
resistance. Therefore resistance seen by the 2.15V is
I1 + I2 = 0.896 + 1.16
=
2.056A
2.056A
