Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 5690-5693 | Hence,
the inductance of the coil increases Consequently, the inductive
reactance of the coil increases As a result, a larger fraction of the
applied ac voltage appears across the inductor, leaving less voltage
across the bulb Therefore, the glow of the light bulb decreases |
1 | 5691-5694 | Consequently, the inductive
reactance of the coil increases As a result, a larger fraction of the
applied ac voltage appears across the inductor, leaving less voltage
across the bulb Therefore, the glow of the light bulb decreases 7 |
1 | 5692-5695 | As a result, a larger fraction of the
applied ac voltage appears across the inductor, leaving less voltage
across the bulb Therefore, the glow of the light bulb decreases 7 6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT
Figure 7 |
1 | 5693-5696 | Therefore, the glow of the light bulb decreases 7 6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT
Figure 7 10 shows a series LCR circuit connected to an ac source e |
1 | 5694-5697 | 7 6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT
Figure 7 10 shows a series LCR circuit connected to an ac source e As
usual, we take the voltage of the source to be v = vm sin wt |
1 | 5695-5698 | 6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT
Figure 7 10 shows a series LCR circuit connected to an ac source e As
usual, we take the voltage of the source to be v = vm sin wt If q is the charge on the capacitor and i the
current, at time t, we have, from Kirchhoff’s loop
rule:
dd
i
q
L
i R
v
t
C
+
+
=
(7 |
1 | 5696-5699 | 10 shows a series LCR circuit connected to an ac source e As
usual, we take the voltage of the source to be v = vm sin wt If q is the charge on the capacitor and i the
current, at time t, we have, from Kirchhoff’s loop
rule:
dd
i
q
L
i R
v
t
C
+
+
=
(7 20)
We want to determine the instantaneous
current i and its phase relationship to the applied
alternating voltage v |
1 | 5697-5700 | As
usual, we take the voltage of the source to be v = vm sin wt If q is the charge on the capacitor and i the
current, at time t, we have, from Kirchhoff’s loop
rule:
dd
i
q
L
i R
v
t
C
+
+
=
(7 20)
We want to determine the instantaneous
current i and its phase relationship to the applied
alternating voltage v We shall solve this problem
by two methods |
1 | 5698-5701 | If q is the charge on the capacitor and i the
current, at time t, we have, from Kirchhoff’s loop
rule:
dd
i
q
L
i R
v
t
C
+
+
=
(7 20)
We want to determine the instantaneous
current i and its phase relationship to the applied
alternating voltage v We shall solve this problem
by two methods First, we use the technique of
phasors and in the second method, we solve
Eq |
1 | 5699-5702 | 20)
We want to determine the instantaneous
current i and its phase relationship to the applied
alternating voltage v We shall solve this problem
by two methods First, we use the technique of
phasors and in the second method, we solve
Eq (7 |
1 | 5700-5703 | We shall solve this problem
by two methods First, we use the technique of
phasors and in the second method, we solve
Eq (7 20) analytically to obtain the time–
dependence of i |
1 | 5701-5704 | First, we use the technique of
phasors and in the second method, we solve
Eq (7 20) analytically to obtain the time–
dependence of i FIGURE 7 |
1 | 5702-5705 | (7 20) analytically to obtain the time–
dependence of i FIGURE 7 10 A series LCR circuit
connected to an ac source |
1 | 5703-5706 | 20) analytically to obtain the time–
dependence of i FIGURE 7 10 A series LCR circuit
connected to an ac source Rationalised 2023-24
187
Alternating Current
7 |
1 | 5704-5707 | FIGURE 7 10 A series LCR circuit
connected to an ac source Rationalised 2023-24
187
Alternating Current
7 6 |
1 | 5705-5708 | 10 A series LCR circuit
connected to an ac source Rationalised 2023-24
187
Alternating Current
7 6 1 Phasor-diagram solution
From the circuit shown in Fig |
1 | 5706-5709 | Rationalised 2023-24
187
Alternating Current
7 6 1 Phasor-diagram solution
From the circuit shown in Fig 7 |
1 | 5707-5710 | 6 1 Phasor-diagram solution
From the circuit shown in Fig 7 10, we see that the resistor, inductor
and capacitor are in series |
1 | 5708-5711 | 1 Phasor-diagram solution
From the circuit shown in Fig 7 10, we see that the resistor, inductor
and capacitor are in series Therefore, the ac current in each element is
the same at any time, having the same amplitude and phase |
1 | 5709-5712 | 7 10, we see that the resistor, inductor
and capacitor are in series Therefore, the ac current in each element is
the same at any time, having the same amplitude and phase Let it be
i = im sin(wt+f)
(7 |
1 | 5710-5713 | 10, we see that the resistor, inductor
and capacitor are in series Therefore, the ac current in each element is
the same at any time, having the same amplitude and phase Let it be
i = im sin(wt+f)
(7 21)
where f is the phase difference between the voltage across the source and
the current in the circuit |
1 | 5711-5714 | Therefore, the ac current in each element is
the same at any time, having the same amplitude and phase Let it be
i = im sin(wt+f)
(7 21)
where f is the phase difference between the voltage across the source and
the current in the circuit On the basis of what we have learnt in the previous
sections, we shall construct a phasor diagram for the present case |
1 | 5712-5715 | Let it be
i = im sin(wt+f)
(7 21)
where f is the phase difference between the voltage across the source and
the current in the circuit On the basis of what we have learnt in the previous
sections, we shall construct a phasor diagram for the present case Let I be the phasor representing the current in the circuit as given by
Eq |
1 | 5713-5716 | 21)
where f is the phase difference between the voltage across the source and
the current in the circuit On the basis of what we have learnt in the previous
sections, we shall construct a phasor diagram for the present case Let I be the phasor representing the current in the circuit as given by
Eq (7 |
1 | 5714-5717 | On the basis of what we have learnt in the previous
sections, we shall construct a phasor diagram for the present case Let I be the phasor representing the current in the circuit as given by
Eq (7 21) |
1 | 5715-5718 | Let I be the phasor representing the current in the circuit as given by
Eq (7 21) Further, let VL, VR, VC, and V represent the voltage across the
inductor, resistor, capacitor and the source, respectively |
1 | 5716-5719 | (7 21) Further, let VL, VR, VC, and V represent the voltage across the
inductor, resistor, capacitor and the source, respectively From previous
section, we know that VR is parallel to I, VC is p/2
behind I and VL is p/2 ahead of I |
1 | 5717-5720 | 21) Further, let VL, VR, VC, and V represent the voltage across the
inductor, resistor, capacitor and the source, respectively From previous
section, we know that VR is parallel to I, VC is p/2
behind I and VL is p/2 ahead of I VL, VR, VC and I
are shown in Fig |
1 | 5718-5721 | Further, let VL, VR, VC, and V represent the voltage across the
inductor, resistor, capacitor and the source, respectively From previous
section, we know that VR is parallel to I, VC is p/2
behind I and VL is p/2 ahead of I VL, VR, VC and I
are shown in Fig 7 |
1 | 5719-5722 | From previous
section, we know that VR is parallel to I, VC is p/2
behind I and VL is p/2 ahead of I VL, VR, VC and I
are shown in Fig 7 11(a) with apppropriate phase-
relations |
1 | 5720-5723 | VL, VR, VC and I
are shown in Fig 7 11(a) with apppropriate phase-
relations The length of these phasors or the amplitude
of VR, VC and VL are:
vRm = im R, vCm = im XC, vLm = im XL
(7 |
1 | 5721-5724 | 7 11(a) with apppropriate phase-
relations The length of these phasors or the amplitude
of VR, VC and VL are:
vRm = im R, vCm = im XC, vLm = im XL
(7 22)
The voltage Equation (7 |
1 | 5722-5725 | 11(a) with apppropriate phase-
relations The length of these phasors or the amplitude
of VR, VC and VL are:
vRm = im R, vCm = im XC, vLm = im XL
(7 22)
The voltage Equation (7 20) for the circuit can
be written as
vL + vR + vC = v
(7 |
1 | 5723-5726 | The length of these phasors or the amplitude
of VR, VC and VL are:
vRm = im R, vCm = im XC, vLm = im XL
(7 22)
The voltage Equation (7 20) for the circuit can
be written as
vL + vR + vC = v
(7 23)
The phasor relation whose vertical component
gives the above equation is
VL + VR + VC = V
(7 |
1 | 5724-5727 | 22)
The voltage Equation (7 20) for the circuit can
be written as
vL + vR + vC = v
(7 23)
The phasor relation whose vertical component
gives the above equation is
VL + VR + VC = V
(7 24)
This relation is represented in Fig |
1 | 5725-5728 | 20) for the circuit can
be written as
vL + vR + vC = v
(7 23)
The phasor relation whose vertical component
gives the above equation is
VL + VR + VC = V
(7 24)
This relation is represented in Fig 7 |
1 | 5726-5729 | 23)
The phasor relation whose vertical component
gives the above equation is
VL + VR + VC = V
(7 24)
This relation is represented in Fig 7 11(b) |
1 | 5727-5730 | 24)
This relation is represented in Fig 7 11(b) Since
VC and VL are always along the same line and in
opposite directions, they can be combined into a single phasor (VC + VL)
which has a magnitude ½vCm – vLm½ |
1 | 5728-5731 | 7 11(b) Since
VC and VL are always along the same line and in
opposite directions, they can be combined into a single phasor (VC + VL)
which has a magnitude ½vCm – vLm½ Since V is represented as the
hypotenuse of a right-triangle whose sides are VR and (VC + VL), the
pythagorean theorem gives:
(
)
2
2
2
m
Rm
Cm
Lm
v
v
v
v
=
+
−
Substituting the values of vRm, vCm, and vLm from Eq |
1 | 5729-5732 | 11(b) Since
VC and VL are always along the same line and in
opposite directions, they can be combined into a single phasor (VC + VL)
which has a magnitude ½vCm – vLm½ Since V is represented as the
hypotenuse of a right-triangle whose sides are VR and (VC + VL), the
pythagorean theorem gives:
(
)
2
2
2
m
Rm
Cm
Lm
v
v
v
v
=
+
−
Substituting the values of vRm, vCm, and vLm from Eq (7 |
1 | 5730-5733 | Since
VC and VL are always along the same line and in
opposite directions, they can be combined into a single phasor (VC + VL)
which has a magnitude ½vCm – vLm½ Since V is represented as the
hypotenuse of a right-triangle whose sides are VR and (VC + VL), the
pythagorean theorem gives:
(
)
2
2
2
m
Rm
Cm
Lm
v
v
v
v
=
+
−
Substituting the values of vRm, vCm, and vLm from Eq (7 22) into the above
equation, we have
2
2
2
(
)
(
)
m
m
m
C
m
L
v
i R
i X
i X
=
+
−
=
+
−
i
R
X
X
m
C
L
2
2
2
(
)
or,
2
2
(
)
m
m
C
L
v
i
R
X
X
=
+
−
[7 |
1 | 5731-5734 | Since V is represented as the
hypotenuse of a right-triangle whose sides are VR and (VC + VL), the
pythagorean theorem gives:
(
)
2
2
2
m
Rm
Cm
Lm
v
v
v
v
=
+
−
Substituting the values of vRm, vCm, and vLm from Eq (7 22) into the above
equation, we have
2
2
2
(
)
(
)
m
m
m
C
m
L
v
i R
i X
i X
=
+
−
=
+
−
i
R
X
X
m
C
L
2
2
2
(
)
or,
2
2
(
)
m
m
C
L
v
i
R
X
X
=
+
−
[7 25(a)]
By analogy to the resistance in a circuit, we introduce the impedance Z
in an ac circuit:
m
m
v
i
=Z
[7 |
1 | 5732-5735 | (7 22) into the above
equation, we have
2
2
2
(
)
(
)
m
m
m
C
m
L
v
i R
i X
i X
=
+
−
=
+
−
i
R
X
X
m
C
L
2
2
2
(
)
or,
2
2
(
)
m
m
C
L
v
i
R
X
X
=
+
−
[7 25(a)]
By analogy to the resistance in a circuit, we introduce the impedance Z
in an ac circuit:
m
m
v
i
=Z
[7 25(b)]
where
2
2
(
)
C
L
Z
R
X
X
=
+
−
(7 |
1 | 5733-5736 | 22) into the above
equation, we have
2
2
2
(
)
(
)
m
m
m
C
m
L
v
i R
i X
i X
=
+
−
=
+
−
i
R
X
X
m
C
L
2
2
2
(
)
or,
2
2
(
)
m
m
C
L
v
i
R
X
X
=
+
−
[7 25(a)]
By analogy to the resistance in a circuit, we introduce the impedance Z
in an ac circuit:
m
m
v
i
=Z
[7 25(b)]
where
2
2
(
)
C
L
Z
R
X
X
=
+
−
(7 26)
FIGURE 7 |
1 | 5734-5737 | 25(a)]
By analogy to the resistance in a circuit, we introduce the impedance Z
in an ac circuit:
m
m
v
i
=Z
[7 25(b)]
where
2
2
(
)
C
L
Z
R
X
X
=
+
−
(7 26)
FIGURE 7 11 (a) Relation between the
phasors VL, VR, VC, and I, (b) Relation
between the phasors VL, VR, and (VL + VC)
for the circuit in Fig |
1 | 5735-5738 | 25(b)]
where
2
2
(
)
C
L
Z
R
X
X
=
+
−
(7 26)
FIGURE 7 11 (a) Relation between the
phasors VL, VR, VC, and I, (b) Relation
between the phasors VL, VR, and (VL + VC)
for the circuit in Fig 7 |
1 | 5736-5739 | 26)
FIGURE 7 11 (a) Relation between the
phasors VL, VR, VC, and I, (b) Relation
between the phasors VL, VR, and (VL + VC)
for the circuit in Fig 7 10 |
1 | 5737-5740 | 11 (a) Relation between the
phasors VL, VR, VC, and I, (b) Relation
between the phasors VL, VR, and (VL + VC)
for the circuit in Fig 7 10 Rationalised 2023-24
Physics
188
Since phasor I is always parallel to phasor VR, the phase angle f
is the angle between VR and V and can be determined from
Fig |
1 | 5738-5741 | 7 10 Rationalised 2023-24
Physics
188
Since phasor I is always parallel to phasor VR, the phase angle f
is the angle between VR and V and can be determined from
Fig 7 |
1 | 5739-5742 | 10 Rationalised 2023-24
Physics
188
Since phasor I is always parallel to phasor VR, the phase angle f
is the angle between VR and V and can be determined from
Fig 7 12:
tan
Cm
Lm
Rm
v
v
v
φ
−
=
Using Eq |
1 | 5740-5743 | Rationalised 2023-24
Physics
188
Since phasor I is always parallel to phasor VR, the phase angle f
is the angle between VR and V and can be determined from
Fig 7 12:
tan
Cm
Lm
Rm
v
v
v
φ
−
=
Using Eq (7 |
1 | 5741-5744 | 7 12:
tan
Cm
Lm
Rm
v
v
v
φ
−
=
Using Eq (7 22), we have
tan
C
L
X
X
R
φ
−
=
(7 |
1 | 5742-5745 | 12:
tan
Cm
Lm
Rm
v
v
v
φ
−
=
Using Eq (7 22), we have
tan
C
L
X
X
R
φ
−
=
(7 27)
Equations (7 |
1 | 5743-5746 | (7 22), we have
tan
C
L
X
X
R
φ
−
=
(7 27)
Equations (7 26) and (7 |
1 | 5744-5747 | 22), we have
tan
C
L
X
X
R
φ
−
=
(7 27)
Equations (7 26) and (7 27) are graphically shown in Fig |
1 | 5745-5748 | 27)
Equations (7 26) and (7 27) are graphically shown in Fig (7 |
1 | 5746-5749 | 26) and (7 27) are graphically shown in Fig (7 12) |
1 | 5747-5750 | 27) are graphically shown in Fig (7 12) This is called Impedance diagram which is a right-triangle with
Z as its hypotenuse |
1 | 5748-5751 | (7 12) This is called Impedance diagram which is a right-triangle with
Z as its hypotenuse Equation 7 |
1 | 5749-5752 | 12) This is called Impedance diagram which is a right-triangle with
Z as its hypotenuse Equation 7 25(a) gives the amplitude of the current and Eq |
1 | 5750-5753 | This is called Impedance diagram which is a right-triangle with
Z as its hypotenuse Equation 7 25(a) gives the amplitude of the current and Eq (7 |
1 | 5751-5754 | Equation 7 25(a) gives the amplitude of the current and Eq (7 27)
gives the phase angle |
1 | 5752-5755 | 25(a) gives the amplitude of the current and Eq (7 27)
gives the phase angle With these, Eq |
1 | 5753-5756 | (7 27)
gives the phase angle With these, Eq (7 |
1 | 5754-5757 | 27)
gives the phase angle With these, Eq (7 21) is completely specified |
1 | 5755-5758 | With these, Eq (7 21) is completely specified If XC > XL, f is positive and the circuit is predominantly capacitive |
1 | 5756-5759 | (7 21) is completely specified If XC > XL, f is positive and the circuit is predominantly capacitive Consequently, the current in the circuit leads the source voltage |
1 | 5757-5760 | 21) is completely specified If XC > XL, f is positive and the circuit is predominantly capacitive Consequently, the current in the circuit leads the source voltage If
XC < XL, f is negative and the circuit is predominantly inductive |
1 | 5758-5761 | If XC > XL, f is positive and the circuit is predominantly capacitive Consequently, the current in the circuit leads the source voltage If
XC < XL, f is negative and the circuit is predominantly inductive Consequently, the current in the circuit lags the source voltage |
1 | 5759-5762 | Consequently, the current in the circuit leads the source voltage If
XC < XL, f is negative and the circuit is predominantly inductive Consequently, the current in the circuit lags the source voltage Figure 7 |
1 | 5760-5763 | If
XC < XL, f is negative and the circuit is predominantly inductive Consequently, the current in the circuit lags the source voltage Figure 7 13 shows the phasor diagram and variation of v and i with w t
for the case XC > XL |
1 | 5761-5764 | Consequently, the current in the circuit lags the source voltage Figure 7 13 shows the phasor diagram and variation of v and i with w t
for the case XC > XL Thus, we have obtained the amplitude
and phase of current for an LCR series circuit
using the technique of phasors |
1 | 5762-5765 | Figure 7 13 shows the phasor diagram and variation of v and i with w t
for the case XC > XL Thus, we have obtained the amplitude
and phase of current for an LCR series circuit
using the technique of phasors But this
method of analysing ac circuits suffers from
certain disadvantages |
1 | 5763-5766 | 13 shows the phasor diagram and variation of v and i with w t
for the case XC > XL Thus, we have obtained the amplitude
and phase of current for an LCR series circuit
using the technique of phasors But this
method of analysing ac circuits suffers from
certain disadvantages First, the phasor
diagram say nothing about the initial
condition |
1 | 5764-5767 | Thus, we have obtained the amplitude
and phase of current for an LCR series circuit
using the technique of phasors But this
method of analysing ac circuits suffers from
certain disadvantages First, the phasor
diagram say nothing about the initial
condition One can take any arbitrary value
of t (say, t1, as done throughout this chapter)
and draw different phasors which show the
relative angle between different phasors |
1 | 5765-5768 | But this
method of analysing ac circuits suffers from
certain disadvantages First, the phasor
diagram say nothing about the initial
condition One can take any arbitrary value
of t (say, t1, as done throughout this chapter)
and draw different phasors which show the
relative angle between different phasors The solution so obtained is called the
steady-state solution |
1 | 5766-5769 | First, the phasor
diagram say nothing about the initial
condition One can take any arbitrary value
of t (say, t1, as done throughout this chapter)
and draw different phasors which show the
relative angle between different phasors The solution so obtained is called the
steady-state solution This is not a general
solution |
1 | 5767-5770 | One can take any arbitrary value
of t (say, t1, as done throughout this chapter)
and draw different phasors which show the
relative angle between different phasors The solution so obtained is called the
steady-state solution This is not a general
solution Additionally, we do have a
transient solution which exists even for
v = 0 |
1 | 5768-5771 | The solution so obtained is called the
steady-state solution This is not a general
solution Additionally, we do have a
transient solution which exists even for
v = 0 The general solution is the sum of the
transient solution and the steady-state
solution |
1 | 5769-5772 | This is not a general
solution Additionally, we do have a
transient solution which exists even for
v = 0 The general solution is the sum of the
transient solution and the steady-state
solution After a sufficiently long time, the effects of the transient solution
die out and the behaviour of the circuit is described by the steady-state
solution |
1 | 5770-5773 | Additionally, we do have a
transient solution which exists even for
v = 0 The general solution is the sum of the
transient solution and the steady-state
solution After a sufficiently long time, the effects of the transient solution
die out and the behaviour of the circuit is described by the steady-state
solution 7 |
1 | 5771-5774 | The general solution is the sum of the
transient solution and the steady-state
solution After a sufficiently long time, the effects of the transient solution
die out and the behaviour of the circuit is described by the steady-state
solution 7 6 |
1 | 5772-5775 | After a sufficiently long time, the effects of the transient solution
die out and the behaviour of the circuit is described by the steady-state
solution 7 6 2 Resonance
An interesting characteristic of the series RLC circuit is the phenomenon
of resonance |
1 | 5773-5776 | 7 6 2 Resonance
An interesting characteristic of the series RLC circuit is the phenomenon
of resonance The phenomenon of resonance is common among systems
that have a tendency to oscillate at a particular frequency |
1 | 5774-5777 | 6 2 Resonance
An interesting characteristic of the series RLC circuit is the phenomenon
of resonance The phenomenon of resonance is common among systems
that have a tendency to oscillate at a particular frequency This frequency
is called the system’s natural frequency |
1 | 5775-5778 | 2 Resonance
An interesting characteristic of the series RLC circuit is the phenomenon
of resonance The phenomenon of resonance is common among systems
that have a tendency to oscillate at a particular frequency This frequency
is called the system’s natural frequency If such a system is driven by an
energy source at a frequency that is near the natural frequency, the
amplitude of oscillation is found to be large |
1 | 5776-5779 | The phenomenon of resonance is common among systems
that have a tendency to oscillate at a particular frequency This frequency
is called the system’s natural frequency If such a system is driven by an
energy source at a frequency that is near the natural frequency, the
amplitude of oscillation is found to be large A familiar example of this
phenomenon is a child on a swing |
1 | 5777-5780 | This frequency
is called the system’s natural frequency If such a system is driven by an
energy source at a frequency that is near the natural frequency, the
amplitude of oscillation is found to be large A familiar example of this
phenomenon is a child on a swing The swing has a natural frequency
for swinging back and forth like a pendulum |
1 | 5778-5781 | If such a system is driven by an
energy source at a frequency that is near the natural frequency, the
amplitude of oscillation is found to be large A familiar example of this
phenomenon is a child on a swing The swing has a natural frequency
for swinging back and forth like a pendulum If the child pulls on the
FIGURE 7 |
1 | 5779-5782 | A familiar example of this
phenomenon is a child on a swing The swing has a natural frequency
for swinging back and forth like a pendulum If the child pulls on the
FIGURE 7 12 Impedance
diagram |
1 | 5780-5783 | The swing has a natural frequency
for swinging back and forth like a pendulum If the child pulls on the
FIGURE 7 12 Impedance
diagram FIGURE 7 |
1 | 5781-5784 | If the child pulls on the
FIGURE 7 12 Impedance
diagram FIGURE 7 13 (a) Phasor diagram of V and I |
1 | 5782-5785 | 12 Impedance
diagram FIGURE 7 13 (a) Phasor diagram of V and I (b) Graphs of v and i versus w t for a series LCR
circuit where XC > XL |
1 | 5783-5786 | FIGURE 7 13 (a) Phasor diagram of V and I (b) Graphs of v and i versus w t for a series LCR
circuit where XC > XL Rationalised 2023-24
189
Alternating Current
rope at regular intervals and the frequency of the pulls is almost the
same as the frequency of swinging, the amplitude of the swinging will be
large (Chapter 13, Class XI) |
1 | 5784-5787 | 13 (a) Phasor diagram of V and I (b) Graphs of v and i versus w t for a series LCR
circuit where XC > XL Rationalised 2023-24
189
Alternating Current
rope at regular intervals and the frequency of the pulls is almost the
same as the frequency of swinging, the amplitude of the swinging will be
large (Chapter 13, Class XI) For an RLC circuit driven with voltage of amplitude vm and frequency
w, we found that the current amplitude is given by
2
2
(
)
m
m
m
C
L
v
v
i
Z
R
X
X
=
=
+
−
with Xc = 1/wC and XL = w L |
1 | 5785-5788 | (b) Graphs of v and i versus w t for a series LCR
circuit where XC > XL Rationalised 2023-24
189
Alternating Current
rope at regular intervals and the frequency of the pulls is almost the
same as the frequency of swinging, the amplitude of the swinging will be
large (Chapter 13, Class XI) For an RLC circuit driven with voltage of amplitude vm and frequency
w, we found that the current amplitude is given by
2
2
(
)
m
m
m
C
L
v
v
i
Z
R
X
X
=
=
+
−
with Xc = 1/wC and XL = w L So if w is varied, then at a particular frequency
w0, Xc = XL, and the impedance is minimum (
)
2
02
Z
R
R
=
+
= |
1 | 5786-5789 | Rationalised 2023-24
189
Alternating Current
rope at regular intervals and the frequency of the pulls is almost the
same as the frequency of swinging, the amplitude of the swinging will be
large (Chapter 13, Class XI) For an RLC circuit driven with voltage of amplitude vm and frequency
w, we found that the current amplitude is given by
2
2
(
)
m
m
m
C
L
v
v
i
Z
R
X
X
=
=
+
−
with Xc = 1/wC and XL = w L So if w is varied, then at a particular frequency
w0, Xc = XL, and the impedance is minimum (
)
2
02
Z
R
R
=
+
= This
frequency is called the resonant frequency:
0
0
1
or
c
L
X
X
L
C
ω
ω
=
=
or
0
1
LC
ω
=
(7 |
1 | 5787-5790 | For an RLC circuit driven with voltage of amplitude vm and frequency
w, we found that the current amplitude is given by
2
2
(
)
m
m
m
C
L
v
v
i
Z
R
X
X
=
=
+
−
with Xc = 1/wC and XL = w L So if w is varied, then at a particular frequency
w0, Xc = XL, and the impedance is minimum (
)
2
02
Z
R
R
=
+
= This
frequency is called the resonant frequency:
0
0
1
or
c
L
X
X
L
C
ω
ω
=
=
or
0
1
LC
ω
=
(7 28)
At resonant frequency, the current amplitude
is maximum; im = vm/R |
1 | 5788-5791 | So if w is varied, then at a particular frequency
w0, Xc = XL, and the impedance is minimum (
)
2
02
Z
R
R
=
+
= This
frequency is called the resonant frequency:
0
0
1
or
c
L
X
X
L
C
ω
ω
=
=
or
0
1
LC
ω
=
(7 28)
At resonant frequency, the current amplitude
is maximum; im = vm/R Figure 7 |
1 | 5789-5792 | This
frequency is called the resonant frequency:
0
0
1
or
c
L
X
X
L
C
ω
ω
=
=
or
0
1
LC
ω
=
(7 28)
At resonant frequency, the current amplitude
is maximum; im = vm/R Figure 7 16 shows the variation of im with w
in a RLC series circuit with L = 1 |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.