Ideal Transformer — Example Circuit 1
Figure 1: Example circuit including an ideal transformer.
Using Eq. (9.26) on the webpage http://qucs.sourceforge.net/tech/node48.html and superposing the node admittance matrix for the conductances results in

   
−I01
V1
G1
0
0
−G1
−1

  
 0
G2
−G2
0
T
  V2   0 


  
 0
−G2 G2 + G3
0
−T
 · V3  = −I02 

−G1
0
0
G1 + G4
1  V4  +I01 
0
It
1
−T
T
−1
0

,
(1)
where V1 , V2 , V3 , and V4 are the node voltages, G1 , G2 , G3 , and G4 are ohmic conductances,
and I01 , I02 are independent current sources.
Ideal Transformer — Example Circuit 2
Figure 2: Example circuit including an ideal transformer.
Firstly, applying KCL to node 1 yields
node 1:
I0 −
IS
I0 − GS · US
I0 − G S · V 1
GS · V 1
+
+
+
−
It
It
It
It
=
=
=
=
0
0
0
I0
(2)
where I0 is an independent current source, may it be dc or ac. The corresponding branch
current It is introduced as a new variable.
Then, applying KCL to node 2 yields
node 2:
IL + T · It = 0
GL · UL + T · It = 0
G L · V 2 + T · It = 0
(3)
.
Finally, establishing the branch equation of the ideal transformer.
ideal transformer:
All three Equations (1), (2), and (3)

GS 0
 0 GL
1 −T
V1 − T · Ut = 0
V1 − T · V2 = 0
are cast into matrix form:
    
−1
V1
I0
T  ·  V2  =  0 
.
0
It
0
.
(4)
(5)
Discussion
• All elements as shown in Fig. 1:
The rank of the matrix is 3. That is, there is a unique solution for this problem.
Current I0 causes current It which itself causes a load current IL = T·It . Current IL
causes a voltage UL = V2 which in turn causes a voltage Ut and therefore a voltage
V1 = US .
• I0 = 0:
Since there is no more source in the circuit, all branch currents and branch voltages
are zero. If one may assume an initial condition nonequal to zero for any branch
current or branch voltage, the system of equations yields no solution. Hence, the
only solution is that all currents and voltages are zero.
• GS = 0:
In this case, the current source is an ideal current source. Current I0 equals current
It , and there is a unique solution for this circuit.
• GL = 0:
In this case, the output of the ideal transformer sees an open circuit. It is expected
that IL = 0 and therefore It = 0. There is a unique solution as long as GS is not
zero. Then, IS = I0 and V1 = IS /GS . Accordingly, V2 = V1 /T = UL .
• GS = 0 and GL = 0:
In this case, the circuit is over-idealized, and there is no solution. There would be a
unique solution if a voltage source U0 were connected instead of current source I0 .