Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

I
V
Z
V
V
V
1
2
1
1
1
1
90
=
=
+
=
+
=
+
∠
° −
−
S
S
S
S
R
j C
j C
j RC
V
C
RC
RC
w
w
w
w
w
w
S
(
)
(
tan
)
Thus, a Series RC Circuit adds a positive phase angle to the voltage phasor in transforming it into 
current phasor. The current in an RC circuit leads the voltage waveform by 90
°-
tan
-
1 
(
w
RC).
We calculate the impedance in the present instance by substituting the relevant numbers. 
Z
=
100 
- 
j318.3 
=
333.65 
∠-
72.54
°
Ω
.
\
I
1
=
325
∠-
90
°
÷
333.65 
∠-
72.54
°
=
0.975 
∠-
17.46
°
\
i
1
(t) 
=
0.975 cos(100
p
 t –17.46
°
) 
=
0.975 cos(100
p
 t –90
°
+ 
72.54
°
) 
=
0.975 sin(100
p
 t 
+ 
72.54
°
) A.
The circuit current leads the applied voltage by 72.54
°
.
Average power delivered to resistor 
=
(I
1rms
)
2
R 
=
(0.975/
√
2)
2
×
100 
=
47.53 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 capacitor. The first quantity is given by 
0.5V
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 
-
72.54
°
. Therefore average power delivered by the source is 0.5
×
325
×
0.975
×
cos(
-
72.54
°
) 
=
47.53 W. 
The voltage phasor across the capacitor 
=
-
j318.3 
×
0.975 
∠-
17.46
°
=
310.34 
∠-
107.46
°
\
Voltage across capacitor 
=
310.34 cos(100
p
 t –107.46
°
) 
=
310.34 sin(100
p
 t –17.46
°
) V.
\
The phase angle between capacitor voltage and current through it 
=
-
17.46
°
– 72.54
°
=
-
90
°
This is the expected value since the voltage across a capacitor is expected to lag behind its current 
under sinusoidal steady-state. Since cosine of –90
°
is zero, the average power delivered to the capacitor 
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 47.53 W.
(V)
v
S
(
t
)
i
1
(
t
)
t
300
200
100
–100
1
0.5
–0.5
–1
–1.5
(A)
t
in ms
t
+ 10
t
+ 20
t
+ 30
t
+ 40
(a)
–200
–300
(V)
v
C
(
t
)
i
1
(
t
)
t
300
200
100
–100
1
0.5
–0.5
–1
–1.5
(A)
t
in ms
t
+ 10
t
+ 20
t
+ 30
t
+ 40
(b)
–200
–300
Fig. 7.6-4 
(a) Applied voltage and current (b) Capacitor voltage 
and current in Example 7.6-2 

---

Sinusoidal Steady-State Response from Phasor Equivalent Circuit 
7.23
Figure 7.6-4 shows the applied voltage waveform lagging behind the circuit current in (a) and 
the capacitor voltage lagging behind the current by 90
°
in (b). Of course, a physical system can 
only delay the response with respect to input. Hence the phase lead that the current in a capacitive 
circuit exhibits under sinusoidal steady-state condition should not be understood as a time-advance. 
The apparent time-advance comes up only after the response undergoes a delay during the transient
period.
example: 7.6-3
Find the (i) source current, (ii) source power, (iii) output voltage and (iv) power delivered to 25
Ω
resistor in the circuit in Fig. 7.6-5.
v
S
v
O
v
S 
= 300 cos 100
t
V
i
x
i
y
+
–
+
–
+
–
+
–
1 
Ω
25 
Ω
0.25 
Ω
1 H
0.25 H
0.5
d
i
y
d
t
0.5
d
i
x
d
t
Fig. 7.6-5 
Circuit for Example 7.6-3 
Solution
The source voltage phasor is 300
∠
0
°
V.
The dependent sources involve first derivative of controlling currents. But differentiation in time-
domain is to be replaced by multiplication by j
w
in the phasor equivalent circuit. The value of 
w
in 
this example is 100 rad/s. The phasor equivalent circuit of the circuit is given in Fig. 7.6-6. The mesh 
equations are written as follows:
−
∠ ° + +
+
−
=
−
+
+
+
=
300 0
100
50
0
50
0 25
25
25
0
1
1
2
1
2
2
2