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Singly Excited Magnetic 
System
Page 2


Singly Excited Magnetic 
System
Singly Excited Magnetic System
In this article we will derive the expression for:-
?
 Electrical energy input.
?
 Magnetic energy stored.
?
 Mechanical work done.
Page 3


Singly Excited Magnetic 
System
Singly Excited Magnetic System
In this article we will derive the expression for:-
?
 Electrical energy input.
?
 Magnetic energy stored.
?
 Mechanical work done.
?
Electrical energy input
•
Consider a simple magnetic system of toroid, excited by a 
single source.
•
The instantaneous voltage equation for the electric circuit can 
written by applying kirchhoff’s law as: -
                                   V
t 
= I.r + e  ----------Eqn(1)
Page 4


Singly Excited Magnetic 
System
Singly Excited Magnetic System
In this article we will derive the expression for:-
?
 Electrical energy input.
?
 Magnetic energy stored.
?
 Mechanical work done.
?
Electrical energy input
•
Consider a simple magnetic system of toroid, excited by a 
single source.
•
The instantaneous voltage equation for the electric circuit can 
written by applying kirchhoff’s law as: -
                                   V
t 
= I.r + e  ----------Eqn(1)
•
Where ‘e’ in the Eqn(1) is the reaction Emf taken as voltage drop in the 
direction of the current ‘I’.
 Therefore   
                                                -----Eqn(2)
Combining Eqn(1) & Eqn(2) we have:- 
     
                                                         
                                                         ------Eqn(3)
Here ‘?’ is the instantaneous flux linkage with the circuit.
Multiplying Eqn(3) by ‘Idt’ on both side we have:-
dt
d
e
?
=
Equation 3.0
? + = d I rdt I Idt V
t
. . .
2
dt
d
r I V
t
?
+ = .
Page 5


Singly Excited Magnetic 
System
Singly Excited Magnetic System
In this article we will derive the expression for:-
?
 Electrical energy input.
?
 Magnetic energy stored.
?
 Mechanical work done.
?
Electrical energy input
•
Consider a simple magnetic system of toroid, excited by a 
single source.
•
The instantaneous voltage equation for the electric circuit can 
written by applying kirchhoff’s law as: -
                                   V
t 
= I.r + e  ----------Eqn(1)
•
Where ‘e’ in the Eqn(1) is the reaction Emf taken as voltage drop in the 
direction of the current ‘I’.
 Therefore   
                                                -----Eqn(2)
Combining Eqn(1) & Eqn(2) we have:- 
     
                                                         
                                                         ------Eqn(3)
Here ‘?’ is the instantaneous flux linkage with the circuit.
Multiplying Eqn(3) by ‘Idt’ on both side we have:-
dt
d
e
?
=
Equation 3.0
? + = d I rdt I Idt V
t
. . .
2
dt
d
r I V
t
?
+ = .
?
As we know that—
                  
                                                         ----Eqn(4)
Assuming that ‘F’ links all the N-turns of the coil, the flux linkage ‘?’ are 
equal to ‘N.F’ Wb turns.
Therefore from Eqn(4)
                                                                              ------Eqn(5)
? = ?
? = - ?
? = - d I Idt e
d I Idt Ir V
d I rdt I Idt V
t
t
. .
. ) (
. . .
2
? = =
=
Id Idt e dW
dW Idt e
elec
elec
.
.
F = F = ? = d F Nd I d I dW
elec
. . .
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