7.20
The Sinusoidal Steady-State Response
i
1
+
1 H
(a)
–
100
Ω
325 sin100
t
V
π
I
1
+
(b)
–
j
314.15
Ω
325
∠
–90°
100
Ω
Fig. 7.6-1
(a) The circuit in time-domain and (b) The phasor equivalent
circuit for Example 7.6-1
This is a single mesh circuit and the mesh current
I
1
is identified in the phasor equivalent circuit in
Fig. 7.6-1 (b). The
mesh equation is obtained as
(100
+
j314.15)
I
1
= 325
∠-
90
°
Solving for
I
1
, we get,
I
1
325
90
100
314 15
325
90
329 7 72 34
0 986
162 3
=
∠ −
+
=
∠ −
∠
=
∠ −
°
°
°
(
. )
.
.
.
.
j
44
°
A
Going back to time-domain by
inverse phasor transformation, we get,
i t
t
t
1
0 986
100
162 34
0 986
100
90
72 34
( )
.
cos(
.
)
.
cos(
.
)
=
−
°
=
− ° −
°
p
p
A
A
A
=
−
°
0 986
100
72 34
.
sin(
.
)
p
t
The source voltage and circuit current waveforms are shown in Fig. 7.6-2 (a) and (b).
The current waveform as drawn in (a) is wrong. Remember that we have obtained the sinusoidal
steady-state solution only and not the complete circuit solution for all
t > 0. Sinusoidal steady-state
gets established only in the long run. The time taken for that will depend on circuit parameters. We
will learn how to estimate the time required for a given circuit to reach
sinusoidal steady-state in
later chapters. We may accept the fact that it takes about 5
L/R seconds (
i.e., about 50 ms in this
circuit) for an
R
-
L circuit to reach steady-state. Therefore, strictly speaking, the sinusoidal steady-
state waveforms should be marked in time-axis as shown in Fig. 7.6-2 (b). It is understood that the
t
used in the axis marking in (b) can have any value greater than 50 ms or so.
400
(V)
v
S
(
t
)
i
1
(
t
)
300
200
100
–100
1.5
1
0.5
–0.5
–1
–1.5
(A)
t
in ms
10
20
30
40
(a)
–200
–300
400
(V)
v
S
(
t
)
i
1
(
t
)
300
200
100
–100
1.5
1
0.5
–0.5
–1
–1.5
(A)
t
in ms
t
+ 10
t
+ 20
t
+ 30
t
+ 40
t
(b)
–200
–300
Fig. 7.6-2
Source voltage and circuit current waveforms in Example 7.6-1 with
(a) Misleading time-axis marking (b) Correct time-axis marking
Sinusoidal Steady-State Response from Phasor Equivalent Circuit
7.21
The waveform as shown in (a) is wrong from another point of view too. We remember that the
voltage applied to the circuit was zero prior to
t
=
0. According to (a), the current suddenly changed
from zero to a –ve value at
t
=
0. It is true that this value of current will exist in the circuit whenever
voltage goes through a positive-going zero-crossing once the circuit has reached steady-state. But the
current cannot do that at the first zero-crossing of voltage itself since it will be the violation of law
of causality then. How did the circuit know while it was at
t
=
0 that the zero voltage that it is being
subjected to at that instant is somehow different from the zero voltage that it was subjected to at the
prior instants? Could it have anticipated that the voltage is going to rise and could it have raised its
current instantaneously as per its anticipation about what the voltage waveform is going to do in future
after
t
=
0 while it was at
t
=
0? No physical system can do that sort of a thing.
All physical systems
are non-anticipatory. The last sentence is yet another form of law of causality.
Hence the current
waveform as shown in (a) violates law of causality.
We note from this example that (i) the impedance of an
R
-
L circuit has positive angle which is
tan
-
1
(
w
L/
R) in general (ii) the current in an
R
-
L circuit lags the voltage waveform under steady-state
conditions
by tan
-
1
(
w
L/
R) in general.
Average
power delivered to resistor
=
(
I
1rm
s
)
2
R
=
(0.986/
√
2)
2
×
100
=
48.6 W
Average power delivered to the resistor can also be calculated by calculating the power delivered
by the voltage source minus the average power delivered
to the inductor. The first quantity is given by
0.5
V
m
I
1m
cos
q
where
q
is the phase angle by which the voltage phasor
leads the current phasor. The
angle in this case is
+
72.34
°
. Therefore average power delivered by the source is 0.5
×
325
×
0.986
×
cos(72.34
°
)
=
48.6 W.
The voltage phasor across the inductor
=
j314.15
×
0.986
∠-
162.34
°
=
309.75
∠-
72.34
°
V.
\
Voltage across inductor
=
309.75 cos(100
p
t –72.34
°
)
=
310.34 sin(100
p
t
+
17.66
°
) V.
\
The phase angle between inductor
voltage and current
=
+
17.66
°
– (
-
72.34
°
)
=
+
90
°
This is the expected value since the voltage across an inductor is expected to
lead ahead of its
current under sinusoidal steady-state. Since cosine of 90
°
is zero, the average power delivered to the
inductor under sinusoidal steady-state condition is zero. Therefore, the average power delivered to the
resistor is the same as the average power delivered by the voltage source and is equal to 48.6 W.
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