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1 | 5490-5493 | 6)
In terms of I, the average power, denoted by P is
2
2
21
m
p
P
i R
I R
=
=
=
(7 7)
Similarly, we define the rms voltage or effective voltage by
V =
2
m
v
= 0 707 vm
(7 8)
From Eq |
1 | 5491-5494 | 7)
Similarly, we define the rms voltage or effective voltage by
V =
2
m
v
= 0 707 vm
(7 8)
From Eq (7 |
1 | 5492-5495 | 707 vm
(7 8)
From Eq (7 3), we have
vm = imR
or,
2
2
m
m
v
i
R
=
or, V = IR
(7 |
1 | 5493-5496 | 8)
From Eq (7 3), we have
vm = imR
or,
2
2
m
m
v
i
R
=
or, V = IR
(7 9)
Equation (7 |
1 | 5494-5497 | (7 3), we have
vm = imR
or,
2
2
m
m
v
i
R
=
or, V = IR
(7 9)
Equation (7 9) gives the relation between ac current and ac voltage
and is similar to that in the dc case |
1 | 5495-5498 | 3), we have
vm = imR
or,
2
2
m
m
v
i
R
=
or, V = IR
(7 9)
Equation (7 9) gives the relation between ac current and ac voltage
and is similar to that in the dc case This shows the advantage of
introducing the concept of rms values |
1 | 5496-5499 | 9)
Equation (7 9) gives the relation between ac current and ac voltage
and is similar to that in the dc case This shows the advantage of
introducing the concept of rms values In terms of rms values, the equation
for power [Eq |
1 | 5497-5500 | 9) gives the relation between ac current and ac voltage
and is similar to that in the dc case This shows the advantage of
introducing the concept of rms values In terms of rms values, the equation
for power [Eq (7 |
1 | 5498-5501 | This shows the advantage of
introducing the concept of rms values In terms of rms values, the equation
for power [Eq (7 7)] and relation between current and voltage in ac circuits
are essentially the same as those for the dc case |
1 | 5499-5502 | In terms of rms values, the equation
for power [Eq (7 7)] and relation between current and voltage in ac circuits
are essentially the same as those for the dc case It is customary to measure and specify rms values for ac quantities |
1 | 5500-5503 | (7 7)] and relation between current and voltage in ac circuits
are essentially the same as those for the dc case It is customary to measure and specify rms values for ac quantities For
example, the household line voltage of 220 V is an rms value with a peak
voltage of
vm = 2 V = (1 |
1 | 5501-5504 | 7)] and relation between current and voltage in ac circuits
are essentially the same as those for the dc case It is customary to measure and specify rms values for ac quantities For
example, the household line voltage of 220 V is an rms value with a peak
voltage of
vm = 2 V = (1 414)(220 V) = 311 V
In fact, the I or rms current is the equivalent dc current that would
produce the same average power loss as the alternating current |
1 | 5502-5505 | It is customary to measure and specify rms values for ac quantities For
example, the household line voltage of 220 V is an rms value with a peak
voltage of
vm = 2 V = (1 414)(220 V) = 311 V
In fact, the I or rms current is the equivalent dc current that would
produce the same average power loss as the alternating current Equation
(7 |
1 | 5503-5506 | For
example, the household line voltage of 220 V is an rms value with a peak
voltage of
vm = 2 V = (1 414)(220 V) = 311 V
In fact, the I or rms current is the equivalent dc current that would
produce the same average power loss as the alternating current Equation
(7 7) can also be written as
P = V2 / R = I V (since V = I R)
Example 7 |
1 | 5504-5507 | 414)(220 V) = 311 V
In fact, the I or rms current is the equivalent dc current that would
produce the same average power loss as the alternating current Equation
(7 7) can also be written as
P = V2 / R = I V (since V = I R)
Example 7 1 A light bulb is rated at 100W for a 220 V supply |
1 | 5505-5508 | Equation
(7 7) can also be written as
P = V2 / R = I V (since V = I R)
Example 7 1 A light bulb is rated at 100W for a 220 V supply Find
(a) the resistance of the bulb; (b) the peak voltage of the source; and
(c) the rms current through the bulb |
1 | 5506-5509 | 7) can also be written as
P = V2 / R = I V (since V = I R)
Example 7 1 A light bulb is rated at 100W for a 220 V supply Find
(a) the resistance of the bulb; (b) the peak voltage of the source; and
(c) the rms current through the bulb Solution
(a) We are given P = 100 W and V = 220 V |
1 | 5507-5510 | 1 A light bulb is rated at 100W for a 220 V supply Find
(a) the resistance of the bulb; (b) the peak voltage of the source; and
(c) the rms current through the bulb Solution
(a) We are given P = 100 W and V = 220 V The resistance of the
bulb is
(
)
2
2
220 V
484
100 W
V
R
=P
=
=
Ω
(b) The peak voltage of the source is
V
2
311
vm
V
=
=
(c) Since, P = I V
100 W
0 |
1 | 5508-5511 | Find
(a) the resistance of the bulb; (b) the peak voltage of the source; and
(c) the rms current through the bulb Solution
(a) We are given P = 100 W and V = 220 V The resistance of the
bulb is
(
)
2
2
220 V
484
100 W
V
R
=P
=
=
Ω
(b) The peak voltage of the source is
V
2
311
vm
V
=
=
(c) Since, P = I V
100 W
0 454A
220 V
�
�
�
P
I
V
EXAMPLE 7 |
1 | 5509-5512 | Solution
(a) We are given P = 100 W and V = 220 V The resistance of the
bulb is
(
)
2
2
220 V
484
100 W
V
R
=P
=
=
Ω
(b) The peak voltage of the source is
V
2
311
vm
V
=
=
(c) Since, P = I V
100 W
0 454A
220 V
�
�
�
P
I
V
EXAMPLE 7 1
Rationalised 2023-24
181
Alternating Current
7 |
1 | 5510-5513 | The resistance of the
bulb is
(
)
2
2
220 V
484
100 W
V
R
=P
=
=
Ω
(b) The peak voltage of the source is
V
2
311
vm
V
=
=
(c) Since, P = I V
100 W
0 454A
220 V
�
�
�
P
I
V
EXAMPLE 7 1
Rationalised 2023-24
181
Alternating Current
7 3 REPRESENTATION OF AC CURRENT AND VOLTAGE
BY ROTATING VECTORS — PHASORS
In the previous section, we learnt that the current through a resistor is
in phase with the ac voltage |
1 | 5511-5514 | 454A
220 V
�
�
�
P
I
V
EXAMPLE 7 1
Rationalised 2023-24
181
Alternating Current
7 3 REPRESENTATION OF AC CURRENT AND VOLTAGE
BY ROTATING VECTORS — PHASORS
In the previous section, we learnt that the current through a resistor is
in phase with the ac voltage But this is not so in the case of an inductor,
a capacitor or a combination of these circuit elements |
1 | 5512-5515 | 1
Rationalised 2023-24
181
Alternating Current
7 3 REPRESENTATION OF AC CURRENT AND VOLTAGE
BY ROTATING VECTORS — PHASORS
In the previous section, we learnt that the current through a resistor is
in phase with the ac voltage But this is not so in the case of an inductor,
a capacitor or a combination of these circuit elements In order to show
phase relationship between voltage and current
in an ac circuit, we use the notion of phasors |
1 | 5513-5516 | 3 REPRESENTATION OF AC CURRENT AND VOLTAGE
BY ROTATING VECTORS — PHASORS
In the previous section, we learnt that the current through a resistor is
in phase with the ac voltage But this is not so in the case of an inductor,
a capacitor or a combination of these circuit elements In order to show
phase relationship between voltage and current
in an ac circuit, we use the notion of phasors The analysis of an ac circuit is facilitated by the
use of a phasor diagram |
1 | 5514-5517 | But this is not so in the case of an inductor,
a capacitor or a combination of these circuit elements In order to show
phase relationship between voltage and current
in an ac circuit, we use the notion of phasors The analysis of an ac circuit is facilitated by the
use of a phasor diagram A phasor* is a vector
which rotates about the origin with angular
speed w, as shown in Fig |
1 | 5515-5518 | In order to show
phase relationship between voltage and current
in an ac circuit, we use the notion of phasors The analysis of an ac circuit is facilitated by the
use of a phasor diagram A phasor* is a vector
which rotates about the origin with angular
speed w, as shown in Fig 7 |
1 | 5516-5519 | The analysis of an ac circuit is facilitated by the
use of a phasor diagram A phasor* is a vector
which rotates about the origin with angular
speed w, as shown in Fig 7 4 |
1 | 5517-5520 | A phasor* is a vector
which rotates about the origin with angular
speed w, as shown in Fig 7 4 The vertical
components of phasors V and I represent the
sinusoidally varying quantities v and i |
1 | 5518-5521 | 7 4 The vertical
components of phasors V and I represent the
sinusoidally varying quantities v and i The
magnitudes of phasors V and I represent the
amplitudes or the peak values vm and im of these
oscillating quantities |
1 | 5519-5522 | 4 The vertical
components of phasors V and I represent the
sinusoidally varying quantities v and i The
magnitudes of phasors V and I represent the
amplitudes or the peak values vm and im of these
oscillating quantities Figure 7 |
1 | 5520-5523 | The vertical
components of phasors V and I represent the
sinusoidally varying quantities v and i The
magnitudes of phasors V and I represent the
amplitudes or the peak values vm and im of these
oscillating quantities Figure 7 4(a) shows the
voltage and current phasors and their
relationship at time t1 for the case of an ac source
connected to a resistor i |
1 | 5521-5524 | The
magnitudes of phasors V and I represent the
amplitudes or the peak values vm and im of these
oscillating quantities Figure 7 4(a) shows the
voltage and current phasors and their
relationship at time t1 for the case of an ac source
connected to a resistor i e |
1 | 5522-5525 | Figure 7 4(a) shows the
voltage and current phasors and their
relationship at time t1 for the case of an ac source
connected to a resistor i e , corresponding to the
circuit shown in Fig |
1 | 5523-5526 | 4(a) shows the
voltage and current phasors and their
relationship at time t1 for the case of an ac source
connected to a resistor i e , corresponding to the
circuit shown in Fig 7 |
1 | 5524-5527 | e , corresponding to the
circuit shown in Fig 7 1 |
1 | 5525-5528 | , corresponding to the
circuit shown in Fig 7 1 The projection of
voltage and current phasors on vertical axis, i |
1 | 5526-5529 | 7 1 The projection of
voltage and current phasors on vertical axis, i e |
1 | 5527-5530 | 1 The projection of
voltage and current phasors on vertical axis, i e , vm sinw t and im sinw t,
respectively represent the value of voltage and current at that instant |
1 | 5528-5531 | The projection of
voltage and current phasors on vertical axis, i e , vm sinw t and im sinw t,
respectively represent the value of voltage and current at that instant As
they rotate with frequency w, curves in Fig |
1 | 5529-5532 | e , vm sinw t and im sinw t,
respectively represent the value of voltage and current at that instant As
they rotate with frequency w, curves in Fig 7 |
1 | 5530-5533 | , vm sinw t and im sinw t,
respectively represent the value of voltage and current at that instant As
they rotate with frequency w, curves in Fig 7 4(b) are generated |
1 | 5531-5534 | As
they rotate with frequency w, curves in Fig 7 4(b) are generated From Fig |
1 | 5532-5535 | 7 4(b) are generated From Fig 7 |
1 | 5533-5536 | 4(b) are generated From Fig 7 4(a) we see that phasors V and I for the case of a resistor are
in the same direction |
1 | 5534-5537 | From Fig 7 4(a) we see that phasors V and I for the case of a resistor are
in the same direction This is so for all times |
1 | 5535-5538 | 7 4(a) we see that phasors V and I for the case of a resistor are
in the same direction This is so for all times This means that the phase
angle between the voltage and the current is zero |
1 | 5536-5539 | 4(a) we see that phasors V and I for the case of a resistor are
in the same direction This is so for all times This means that the phase
angle between the voltage and the current is zero 7 |
1 | 5537-5540 | This is so for all times This means that the phase
angle between the voltage and the current is zero 7 4 AC VOLTAGE APPLIED TO AN INDUCTOR
Figure 7 |
1 | 5538-5541 | This means that the phase
angle between the voltage and the current is zero 7 4 AC VOLTAGE APPLIED TO AN INDUCTOR
Figure 7 5 shows an ac source connected to an inductor |
1 | 5539-5542 | 7 4 AC VOLTAGE APPLIED TO AN INDUCTOR
Figure 7 5 shows an ac source connected to an inductor Usually,
inductors have appreciable resistance in their windings, but we shall
assume that this inductor has negligible resistance |
1 | 5540-5543 | 4 AC VOLTAGE APPLIED TO AN INDUCTOR
Figure 7 5 shows an ac source connected to an inductor Usually,
inductors have appreciable resistance in their windings, but we shall
assume that this inductor has negligible resistance Thus, the circuit is a purely inductive ac circuit |
1 | 5541-5544 | 5 shows an ac source connected to an inductor Usually,
inductors have appreciable resistance in their windings, but we shall
assume that this inductor has negligible resistance Thus, the circuit is a purely inductive ac circuit Let
the voltage across the source be v = vm sinw t |
1 | 5542-5545 | Usually,
inductors have appreciable resistance in their windings, but we shall
assume that this inductor has negligible resistance Thus, the circuit is a purely inductive ac circuit Let
the voltage across the source be v = vm sinw t Using
the Kirchhoff’s loop rule,
∑ε ( )t =
0 , and since there
is no resistor in the circuit,
d
0
d
i
v
L
t
−
=
(7 |
1 | 5543-5546 | Thus, the circuit is a purely inductive ac circuit Let
the voltage across the source be v = vm sinw t Using
the Kirchhoff’s loop rule,
∑ε ( )t =
0 , and since there
is no resistor in the circuit,
d
0
d
i
v
L
t
−
=
(7 10)
where the second term is the self-induced Faraday
emf in the inductor; and L is the self-inductance of
FIGURE 7 |
1 | 5544-5547 | Let
the voltage across the source be v = vm sinw t Using
the Kirchhoff’s loop rule,
∑ε ( )t =
0 , and since there
is no resistor in the circuit,
d
0
d
i
v
L
t
−
=
(7 10)
where the second term is the self-induced Faraday
emf in the inductor; and L is the self-inductance of
FIGURE 7 4 (a) A phasor diagram for the
circuit in Fig 7 |
1 | 5545-5548 | Using
the Kirchhoff’s loop rule,
∑ε ( )t =
0 , and since there
is no resistor in the circuit,
d
0
d
i
v
L
t
−
=
(7 10)
where the second term is the self-induced Faraday
emf in the inductor; and L is the self-inductance of
FIGURE 7 4 (a) A phasor diagram for the
circuit in Fig 7 1 |
1 | 5546-5549 | 10)
where the second term is the self-induced Faraday
emf in the inductor; and L is the self-inductance of
FIGURE 7 4 (a) A phasor diagram for the
circuit in Fig 7 1 (b) Graph of v and
i versus wt |
1 | 5547-5550 | 4 (a) A phasor diagram for the
circuit in Fig 7 1 (b) Graph of v and
i versus wt FIGURE 7 |
1 | 5548-5551 | 1 (b) Graph of v and
i versus wt FIGURE 7 5 An ac source
connected to an inductor |
1 | 5549-5552 | (b) Graph of v and
i versus wt FIGURE 7 5 An ac source
connected to an inductor *
Though voltage and current in ac circuit are represented by phasors – rotating
vectors, they are not vectors themselves |
1 | 5550-5553 | FIGURE 7 5 An ac source
connected to an inductor *
Though voltage and current in ac circuit are represented by phasors – rotating
vectors, they are not vectors themselves They are scalar quantities |
1 | 5551-5554 | 5 An ac source
connected to an inductor *
Though voltage and current in ac circuit are represented by phasors – rotating
vectors, they are not vectors themselves They are scalar quantities It so happens
that the amplitudes and phases of harmonically varying scalars combine
mathematically in the same way as do the projections of rotating vectors of
corresponding magnitudes and directions |
1 | 5552-5555 | *
Though voltage and current in ac circuit are represented by phasors – rotating
vectors, they are not vectors themselves They are scalar quantities It so happens
that the amplitudes and phases of harmonically varying scalars combine
mathematically in the same way as do the projections of rotating vectors of
corresponding magnitudes and directions The rotating vectors that represent
harmonically varying scalar quantities are introduced only to provide us with a
simple way of adding these quantities using a rule that we already know |
1 | 5553-5556 | They are scalar quantities It so happens
that the amplitudes and phases of harmonically varying scalars combine
mathematically in the same way as do the projections of rotating vectors of
corresponding magnitudes and directions The rotating vectors that represent
harmonically varying scalar quantities are introduced only to provide us with a
simple way of adding these quantities using a rule that we already know Rationalised 2023-24
Physics
182
the inductor |
1 | 5554-5557 | It so happens
that the amplitudes and phases of harmonically varying scalars combine
mathematically in the same way as do the projections of rotating vectors of
corresponding magnitudes and directions The rotating vectors that represent
harmonically varying scalar quantities are introduced only to provide us with a
simple way of adding these quantities using a rule that we already know Rationalised 2023-24
Physics
182
the inductor The negative sign follows from Lenz’s law (Chapter 6) |
1 | 5555-5558 | The rotating vectors that represent
harmonically varying scalar quantities are introduced only to provide us with a
simple way of adding these quantities using a rule that we already know Rationalised 2023-24
Physics
182
the inductor The negative sign follows from Lenz’s law (Chapter 6) Combining Eqs |
1 | 5556-5559 | Rationalised 2023-24
Physics
182
the inductor The negative sign follows from Lenz’s law (Chapter 6) Combining Eqs (7 |
1 | 5557-5560 | The negative sign follows from Lenz’s law (Chapter 6) Combining Eqs (7 1) and (7 |
1 | 5558-5561 | Combining Eqs (7 1) and (7 10), we have
d
sin
d
vm
i
v
t
t
L
L
ω
=
=
(7 |
1 | 5559-5562 | (7 1) and (7 10), we have
d
sin
d
vm
i
v
t
t
L
L
ω
=
=
(7 11)
Equation (7 |
1 | 5560-5563 | 1) and (7 10), we have
d
sin
d
vm
i
v
t
t
L
L
ω
=
=
(7 11)
Equation (7 11) implies that the equation for i(t), the current as a
function of time, must be such that its slope di/dt is a sinusoidally varying
quantity, with the same phase as the source voltage and an amplitude
given by vm/L |
1 | 5561-5564 | 10), we have
d
sin
d
vm
i
v
t
t
L
L
ω
=
=
(7 11)
Equation (7 11) implies that the equation for i(t), the current as a
function of time, must be such that its slope di/dt is a sinusoidally varying
quantity, with the same phase as the source voltage and an amplitude
given by vm/L To obtain the current, we integrate di/dt with respect to
time:
d
d d
d
ti
t
v
L
t
t
m
∫
∫
=
sin(
)
ω
and get,
cos(
)
constant
vm
i
t
L
= −
ω
+
ω
The integration constant has the dimension of current and is time-
independent |
1 | 5562-5565 | 11)
Equation (7 11) implies that the equation for i(t), the current as a
function of time, must be such that its slope di/dt is a sinusoidally varying
quantity, with the same phase as the source voltage and an amplitude
given by vm/L To obtain the current, we integrate di/dt with respect to
time:
d
d d
d
ti
t
v
L
t
t
m
∫
∫
=
sin(
)
ω
and get,
cos(
)
constant
vm
i
t
L
= −
ω
+
ω
The integration constant has the dimension of current and is time-
independent Since the source has an emf which oscillates symmetrically
about zero, the current it sustains also oscillates symmetrically about
zero, so that no constant or time-independent component of the current
exists |
1 | 5563-5566 | 11) implies that the equation for i(t), the current as a
function of time, must be such that its slope di/dt is a sinusoidally varying
quantity, with the same phase as the source voltage and an amplitude
given by vm/L To obtain the current, we integrate di/dt with respect to
time:
d
d d
d
ti
t
v
L
t
t
m
∫
∫
=
sin(
)
ω
and get,
cos(
)
constant
vm
i
t
L
= −
ω
+
ω
The integration constant has the dimension of current and is time-
independent Since the source has an emf which oscillates symmetrically
about zero, the current it sustains also oscillates symmetrically about
zero, so that no constant or time-independent component of the current
exists Therefore, the integration constant is zero |
1 | 5564-5567 | To obtain the current, we integrate di/dt with respect to
time:
d
d d
d
ti
t
v
L
t
t
m
∫
∫
=
sin(
)
ω
and get,
cos(
)
constant
vm
i
t
L
= −
ω
+
ω
The integration constant has the dimension of current and is time-
independent Since the source has an emf which oscillates symmetrically
about zero, the current it sustains also oscillates symmetrically about
zero, so that no constant or time-independent component of the current
exists Therefore, the integration constant is zero Using
−
=
−
cos(
)
sin
ω
ω
t
t
2π
, we have
i
i
t
=m
−
sin ω
2π
(7 |
1 | 5565-5568 | Since the source has an emf which oscillates symmetrically
about zero, the current it sustains also oscillates symmetrically about
zero, so that no constant or time-independent component of the current
exists Therefore, the integration constant is zero Using
−
=
−
cos(
)
sin
ω
ω
t
t
2π
, we have
i
i
t
=m
−
sin ω
2π
(7 12)
where
m
m
v
i
= ωL
is the amplitude of the current |
1 | 5566-5569 | Therefore, the integration constant is zero Using
−
=
−
cos(
)
sin
ω
ω
t
t
2π
, we have
i
i
t
=m
−
sin ω
2π
(7 12)
where
m
m
v
i
= ωL
is the amplitude of the current The quantity w L is
analogous to the resistance and is called inductive reactance, denoted
by XL:
XL = w L
(7 |
1 | 5567-5570 | Using
−
=
−
cos(
)
sin
ω
ω
t
t
2π
, we have
i
i
t
=m
−
sin ω
2π
(7 12)
where
m
m
v
i
= ωL
is the amplitude of the current The quantity w L is
analogous to the resistance and is called inductive reactance, denoted
by XL:
XL = w L
(7 13)
The amplitude of the current is, then
m
m
L
v
i
=X
(7 |
1 | 5568-5571 | 12)
where
m
m
v
i
= ωL
is the amplitude of the current The quantity w L is
analogous to the resistance and is called inductive reactance, denoted
by XL:
XL = w L
(7 13)
The amplitude of the current is, then
m
m
L
v
i
=X
(7 14)
The dimension of inductive reactance is the same as that of resistance
and its SI unit is ohm (W) |
1 | 5569-5572 | The quantity w L is
analogous to the resistance and is called inductive reactance, denoted
by XL:
XL = w L
(7 13)
The amplitude of the current is, then
m
m
L
v
i
=X
(7 14)
The dimension of inductive reactance is the same as that of resistance
and its SI unit is ohm (W) The inductive reactance limits the current in a
purely inductive circuit in the same way as the resistance limits the
current in a purely resistive circuit |
1 | 5570-5573 | 13)
The amplitude of the current is, then
m
m
L
v
i
=X
(7 14)
The dimension of inductive reactance is the same as that of resistance
and its SI unit is ohm (W) The inductive reactance limits the current in a
purely inductive circuit in the same way as the resistance limits the
current in a purely resistive circuit The inductive reactance is directly
proportional to the inductance and to the frequency of the current |
1 | 5571-5574 | 14)
The dimension of inductive reactance is the same as that of resistance
and its SI unit is ohm (W) The inductive reactance limits the current in a
purely inductive circuit in the same way as the resistance limits the
current in a purely resistive circuit The inductive reactance is directly
proportional to the inductance and to the frequency of the current A comparison of Eqs |
1 | 5572-5575 | The inductive reactance limits the current in a
purely inductive circuit in the same way as the resistance limits the
current in a purely resistive circuit The inductive reactance is directly
proportional to the inductance and to the frequency of the current A comparison of Eqs (7 |
1 | 5573-5576 | The inductive reactance is directly
proportional to the inductance and to the frequency of the current A comparison of Eqs (7 1) and (7 |
1 | 5574-5577 | A comparison of Eqs (7 1) and (7 12) for the source voltage and the
current in an inductor shows that the current lags the voltage by p/2 or
one-quarter (1/4) cycle |
1 | 5575-5578 | (7 1) and (7 12) for the source voltage and the
current in an inductor shows that the current lags the voltage by p/2 or
one-quarter (1/4) cycle Figure 7 |
1 | 5576-5579 | 1) and (7 12) for the source voltage and the
current in an inductor shows that the current lags the voltage by p/2 or
one-quarter (1/4) cycle Figure 7 6 (a) shows the voltage and the current
phasors in the present case at instant t1 |
1 | 5577-5580 | 12) for the source voltage and the
current in an inductor shows that the current lags the voltage by p/2 or
one-quarter (1/4) cycle Figure 7 6 (a) shows the voltage and the current
phasors in the present case at instant t1 The current phasor I is p/2
behind the voltage phasor V |
1 | 5578-5581 | Figure 7 6 (a) shows the voltage and the current
phasors in the present case at instant t1 The current phasor I is p/2
behind the voltage phasor V When rotated with frequency w counter-
clockwise, they generate the voltage and current given by Eqs |
1 | 5579-5582 | 6 (a) shows the voltage and the current
phasors in the present case at instant t1 The current phasor I is p/2
behind the voltage phasor V When rotated with frequency w counter-
clockwise, they generate the voltage and current given by Eqs (7 |
1 | 5580-5583 | The current phasor I is p/2
behind the voltage phasor V When rotated with frequency w counter-
clockwise, they generate the voltage and current given by Eqs (7 1) and
(7 |
1 | 5581-5584 | When rotated with frequency w counter-
clockwise, they generate the voltage and current given by Eqs (7 1) and
(7 12), respectively and as shown in Fig |
1 | 5582-5585 | (7 1) and
(7 12), respectively and as shown in Fig 7 |
1 | 5583-5586 | 1) and
(7 12), respectively and as shown in Fig 7 6(b) |
1 | 5584-5587 | 12), respectively and as shown in Fig 7 6(b) Interactive animation on Phasor diagrams of ac circuits containing, R, L, C and RLC series circuits:
http://www |
1 | 5585-5588 | 7 6(b) Interactive animation on Phasor diagrams of ac circuits containing, R, L, C and RLC series circuits:
http://www animations |
1 | 5586-5589 | 6(b) Interactive animation on Phasor diagrams of ac circuits containing, R, L, C and RLC series circuits:
http://www animations physics |
1 | 5587-5590 | Interactive animation on Phasor diagrams of ac circuits containing, R, L, C and RLC series circuits:
http://www animations physics unsw |
1 | 5588-5591 | animations physics unsw edu |
1 | 5589-5592 | physics unsw edu au//jw/AC |
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