v
C
(
t
)
IC
=
V
O
I
S
+
–
2 k
Ω
2 k
Ω
2
u
(
t
) +
(
t
)
δ
Fig. 11.7-5
13. A series RC circuit with zero initial current is driven by v
S
(t)
=
d
(t)
-
d
(t
-
1). Its time constant
is 1 s. (i) Starting from impulse response, find the voltage across the resistor in the circuit when
driven by the input v
S
(t). (ii) Using the result derive an expression for voltage across the resistor
when the circuit is driven by a rectangular pulse of unit amplitude and 1 s duration.
14. Derive expressions for maximum voltage across resistor in an initially relaxed series RC circuit
when it is driven by e
a
t
u(t) A with
a
≠
-
1/
t
, where
t
is the time constant of the circuit. Also find
an expression for the time instant at which this maximum voltage occurs.
Problems
11.31
15. An input of k
d
(t)
+
2e
-
2 t /
t
V is applied to an initially relaxed series RC circuit with time constant
of
t
s. The output across capacitor for t
≤
0
+
is observed to contain only e
-
2 t /
t
waveshape. What
is the value of k?
16. The desired voltage across a parallel RC with initial condition as shown in the Fig. 11.7-6 is given
by v(t)
=
2t for t
≤
0 and 0 for t < 0. Find the i
S
(t) to be applied to the circuit if the initial condition
is 5 V. Sketch the required i
S
(t).
+
v
C
(
t
)
i
S
(
t
)
–
10 k
Ω
100
µ
F
Fig. 11.7-6
17. The switches in Fig. 11.7-7 are ideal. (i) Find the voltage across C and plot it. (ii) Find the current
through C and plot it. (iii) Find the voltage across the first resistor and plot it.
+
+
t
= 0.12 s
t
= 0
10 V
22
µ
F
5 k
Ω
5 k
Ω
S
2
S
1
–
–
Fig. 11.7-7
18. Initial voltage at t
=
-∞
in the capacitor in the circuit in Fig. 11.7-8 was zero. Find the voltage
across capacitor and current through it for t
≤
0
+
and plot them.
+ 600
µ
F
2
u
(–
t
) A
3
u
(
t
) A
3
Ω
6
Ω
–
Fig. 11.7-8
19. The switch in the circuit in Fig. 11.7-9 was closed for a long time and is opened at t
=
0. Find and
plot the current in C and voltage across C as functions of time.
+
t
= 0
10 V
10 mA
1 mF
–
+
–
1 k
Ω
1 k
Ω
Fig. 11.7-9
20. Find the impulse response of the voltage variable v(t) in the circuit in Fig. 11.7-10.
+
10
µ
F
–
+
–
+
–
20 k
Ω
20 k
Ω
20 k
Ω
30 k
Ω
0.6
(
t
)
δ
v
Fig. 11.7-10
11.32
First-Order
RC
Circuits
21. v
C
(t) at 0
-
in the circuit in Fig. 11.7-11 is
-
V. The switch S is closed at t
=
0. Show that v
C
(t) will
cross zero at t
=
0.69RC s.
+
v
C
(
t
)
S
R C
V
–
+
–
Fig. 11.7-11
22. A pulse of height V V and width T s is applied to a series RC circuit with zero initial condition
as in Fig. 11.7-12 . (i) Plot the output for (a) RC
=
T (b) RC
=
10T and (c) RC
=
0.1T (ii) Find the
relation between RC and T if the area of output pulse after T s is to be less than 10% of area of
output pulse from 0 to T s.
+
V
R
v
O
(
t
)
C
T
–
Fig. 11.7-12
23. What must be the value of k in the circuit in Fig. 11.7-13, if v(t)
=
0 for t
≤
0
+
?
+
+
0.1 mF
3
u
(
t
)
k
(
t
)
δ
5 k
Ω
5 k
Ω
v
5 k
Ω
–
–
+
–
+
–
Fig. 11.7-13
24. A single pulse defined as v
S
(t)
=
10t for 0
≤
t
≤
1 and 0 otherwise is applied to a series RC circuit
with a time constant of 0.3 s. Find and plot the capacitor voltage.
25. Find the time constants for each circuit in Fig. 11.7-14 for voltage excitation and current excitation.
All resistors are 1k
W
and all capacitors are 1
m
F.
Fig. 11.7-14
26. The Inverter in the circuit in Fig. 11.7-15 is a digital electronic gate circuit and its input and output
behaviour is as shown in the waveforms. Each inverter gate has 15pF input capacitance. The
gate circuit will draw only zero current from
+
V supply if its input is held at V V or 0V steadily.
Problems
11.33
The output of one such gate is connected as input to 4 such gates. Calculate (i) the power
dissipation in the driving gate and (ii) the average power supply current drawn by the driving gate
when the input to driving gate is a square wave varying between
+
V and 0 with a frequency f for
(a) V
=
5V, f
=
100kHz (b) V
=
5V f
=
10MHz (c) V
=
15V, f
=
100kHz (d) V
=
15V, f
=
10MHz.
Input
Output
V
+
V
V
T
Inverter
Inverter
Inverter
Inverter
Inverter
Fig. 11.7-15
27. A symmetric square wave of
±
V amplitude and period of T s is applied to a high pass RC circuit
as shown in Fig. 11.7-16. After a few cycles of initial transient the output waveform settles to a
periodic steady-state as shown where V
1
′
and V
2
′
will be equal to V
1
and V
2
, respectively, under
periodic steady-state. The slanting portion of output wave will be exponential with time constant
=
RC. However, if RC >> T (equivalently, if the cut-off frequency of the circuit is much less than
the square wave frequency) we can approximate the exponential by straight-line segments. Use
this approximation and find expressions for V
1
and V
2
. Also find an expression for the so-called
“percentage tilt” defined as 100
×
(V
1
-
V
2
)/V
+
–
+
–
V
1
V
1
'
V
2
'
V
2
V
–V
v
O
(
t
)
v
O
(
t
)
v
S
(
t
)
v
S
(
t
)
V
C
R
T
t
t
–V
Fig. 11.7-16
28. Let V
s
(t) be an arbitrary time varying periodic voltage source with a cycle average value of V
dc
.
This means that V
s
(t) can be written as V
dc
+
V
ac
(t), where V
ac
(t) is a time varying periodic
component with equal positive half cycle and negative half cycle areas. Let that area be A V-s.
11.34
First-Order
RC
Circuits
100 V
–100 V
100 V
100 V
1 ms
1 ms
sine wave
2 ms
2
8
1
1
0.8
t
in ms
t
in ms
t
in ms
Fig. 11.7-17
This waveform V
s
(t) is applied to a series RC circuit and output voltage is taken across C. Assume
that RC >> T where T is the period of V
s
(t) . Show that under periodic steady-state the (i) average
value of output voltage is V
dc
(ii) the peak-to-peak ripple in output voltage
≈
A/
t
V, where
t
=
RC.
(iii) Calculate the quantities in (i) and (ii) for the three inputs given in Fig. 11.7-17, if
t
is 20ms.
(The Series RC Circuit can be used to extract average value of the input. The basic issue involved
in this application is the tradeoff between ripple in the average value Vs response time)
29. The switch S
1
in Fig. 11.7-18 is closed at t
=
0 with zero initial condition in the capacitor. The
switch S
2
is kept open for T
1
s and closed for (T
-
T
1
) s periodically. Obtain the voltage across
capacitor and plot it assuming the following data. V
=
12V, R
=
12 k
W
, C
=
1
m
F, T
1
=
10ms,
T
=
11ms and Resistance of S
2
when it is ON
=
100
W
.
+
+
+
1
2
On
Off
T
1
T
2
T
v
12
(
t
) =
v
O
(
t
)
v
O
(
t
)
S
1
S
2
S
2
S
2
R
V
C
t
–
–
–
Fig. 11.7-18
30. The source-free response in an RC circuit takes the capacitor voltage from 10 V to 3V in 20ms.
What is the bandwidth of a low-pass RC filter made using the same components?
31. The steady-state voltage across resistor (v
R
) in a series RC circuit has an amplitude of 7.07V
when the circuit is driven by an AC voltage of amplitude 10V and angular frequency
w
rad/s.
(i) Find the phase angle of v
R
with respect to the input sinusoid. (ii) If another sinusoidal voltage
of 15V amplitude and 3
w
rad/s frequency is applied to the circuit, find the amplitude and phase
of v
R
under steady-state condition.
Problems
11.35
32. The bandwidth of a low-pass RC filter is 2kHz. If a 10V DC source is applied to this series RC
circuit (capacitor is uncharged before applying this source) at t
=
107ms, find t at which the
voltage across resistor will be 3V.
33. A Series RC circuit is excited by a sinusoidal voltage source at 3kHz. The voltage across the
resistor is seen to lead the input voltage by 30
°
. If this RC circuit undergoes source-free response,
how much time will it take to dissipate 50% of its initial stored energy in the resistance?
34. A Series RC circuit is excited by a sinusoidal voltage source at 1kHz. The voltage across the
capacitor is seen to lag the input voltage by 30
°
. What is the cyclic frequency of input sinusoid
such that the voltage across the resistor will lead the input by 30
°
?
35. A Series RC circuit is excited by a sinusoidal voltage source at 10kHz. The voltage across the
capacitor is seen to be 50% of the input voltage in amplitude. What is the cyclic frequency of input
sinusoid such that the voltage across the resistor will be 50% of the input voltage in amplitude?
36. A Series RC circuit is excited by a sinusoidal voltage source at 100kHz. The voltage across the
capacitor is seen to be 30% of the input voltage in amplitude. What is the cyclic frequency of input
sinusoid such that the voltage across the resistor will lead the input voltage by 60
°
?
37. An AC voltage source
=
V sin
w
t is applied to a series RC circuit from t
=
0. The circuit current is
found to be
=
0.7sin (
w
t
-
p
/3 )A for t
≤
0
+
. Was there any initial voltage across the capacitor? If
so, what is its magnitude and relative polarity?
38. The fall time of a series RC circuit is 6.6 ms. If the same components are used to make a parallel
RC circuit excited by a sinusoidal current source i
S
(t)
=
0.5 sin 300
p
t A, find the steady-state
current in the resistor in the circuit.
39. A parallel RC circuit has C
=
10
m
F. It is excited by an AC current source of 100mA amplitude and
5kHz frequency. The sinusoidal voltage across the combination is seen to lag the applied input by
60
°
. Find the value of R and the amplitude of voltage across the combination.
40. A current signal i
S
(t)
=
(2 sin
w
t – 1.2 sin 3
w
t
+
0.8 sin 5
w
t )u(t) mA is applied to a parallel RC
circuit of bandwidth 0.8
w
. Find and plot the steady-state waveform of voltage across the circuit.
This page is intentionally left blank.
12.1
S e r i e s a n d Pa r a l l e l
RLC
C i r c u i t s
CHAPTEROBJECTIVES
• Source-free Response of Series RLC Circuit
• The Series LC Circuit – A Special Case
• The Series LC Circuit with Small Damping – Another Special Case
• Standard Formats for Second-Order Circuit Zero-Input Response
• Sinusoidal Forced-Response of RLC Circuits from Differential Equation
• Frequency Response of RLC Circuits from Phasor Equivalent Circuit
• Qualitative Discussion on Frequency Response of Series RLC Circuit
• A More Detailed Look at the Band-pass Output of Series RLC Circuit – Series Resonance
• Quality Factor of Practical Inductors and Capacitors
• Zero-Input Response and Zero-State Response of Parallel RLC Circuit
• Sinusoidal Steady-State Frequency Response of Parallel RLC Circuit – Parallel Resonance
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