Prof. Antonio Gabaldón. ETS de Ing. Industrial de Cartagena (Spain)
2007
Chapter 3. Problem 2-6) By superposition, and some formal calculation (KVL and KCL
equations), find i2 in the circuit shown in the figure.
Data: w=20 rad/s; R1=2Ω; R2=3Ω; R3=4Ω; L=0.1H.
Solution:
We have two sources in the network (12V AC voltage source and a 3A DC current
source). The application of superposition theorem will be done in two networks: circuit (a) and
circuit (b) (no superposition scheme will be applied to the dependent source. There are two
reasons to do that: simplicity and the fact that a dependent source usually is a previously
transformed pasive device).
Network (a) (Source Eg 12V AC). We retain only the independent volage source, while
the independent current source is made zero (open circuit). Using phasor methods:
,
Figure 2.6.1. Circuit in AC (without current DC source, now an open circuit)
The partial response i’is given by loop and node equations:
Prof. Antonio Gabaldón. ETS de Ing. Industrial de Cartagena (Spain)
2007
)
)
)
)
)
)
)
i2 ' &i3 ; i3 ' 4i3 % i4Ω Y i4Ω ' &3i3
)
)
Eg ' (R1%R2)i3 & 3R3i3
Eg
)
i3 '
Eg
)
R1 % R2 & 3R3
Y i2 ' &
R1 % R2 & 3R3
Network (b) (source Ig in DC):next we retain the independent current source and make
the independent voltage source zero (short circuit, cc)
Figure 2.6.2. Circuit in DC (without sine source, now a short circuit)
In steady state the coil is a short circuit (see figure 6.5.2). Appliying KCL in nodes A and
B,
))
))
Node A) i2 % i3 ' Ig
))
))
Node B) iR3 % 4i3 ' i3
))
))
Y i3 ' I g & i2
and KVL in loop defined by resistors R1, R2 and R3:
))
))
R1i2 ' (R2 & 3R3)i3
thus, the second current i”2
))
i2 '
(R2 & 3R3)Ig
R1 % R2 & R3
and appliying superposition theorem, branch current i2, is the summatory of partial
response (currents in this case) obtained through network analysis in circuits (a) and (b):
Prof. Antonio Gabaldón. ETS de Ing. Industrial de Cartagena (Spain)
2007
)
))
i2 ' i2 % i2 '
(R2 & 3R3)Ig & Eg
R1 % R2 & R3