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1 | 5790-5793 | 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 00 mH, C =
1 |
1 | 5791-5794 | Figure 7 16 shows the variation of im with w
in a RLC series circuit with L = 1 00 mH, C =
1 00 nF for two values of R: (i) R = 100 W
and (ii) R = 200 W |
1 | 5792-5795 | 16 shows the variation of im with w
in a RLC series circuit with L = 1 00 mH, C =
1 00 nF for two values of R: (i) R = 100 W
and (ii) R = 200 W For the source applied vm =
100 V |
1 | 5793-5796 | 00 mH, C =
1 00 nF for two values of R: (i) R = 100 W
and (ii) R = 200 W For the source applied vm =
100 V w0 for this case is
1
LC
= 1 |
1 | 5794-5797 | 00 nF for two values of R: (i) R = 100 W
and (ii) R = 200 W For the source applied vm =
100 V w0 for this case is
1
LC
= 1 00×106
rad/s |
1 | 5795-5798 | For the source applied vm =
100 V w0 for this case is
1
LC
= 1 00×106
rad/s We see that the current amplitude is
maximum at the resonant frequency |
1 | 5796-5799 | w0 for this case is
1
LC
= 1 00×106
rad/s We see that the current amplitude is
maximum at the resonant frequency Since im =
vm / R at resonance, the current amplitude for
case (i) is twice to that for case (ii) |
1 | 5797-5800 | 00×106
rad/s We see that the current amplitude is
maximum at the resonant frequency Since im =
vm / R at resonance, the current amplitude for
case (i) is twice to that for case (ii) Resonant circuits have a variety of applications, for example, in the
tuning mechanism of a radio or a TV set |
1 | 5798-5801 | We see that the current amplitude is
maximum at the resonant frequency Since im =
vm / R at resonance, the current amplitude for
case (i) is twice to that for case (ii) Resonant circuits have a variety of applications, for example, in the
tuning mechanism of a radio or a TV set The antenna of a radio accepts
signals from many broadcasting stations |
1 | 5799-5802 | Since im =
vm / R at resonance, the current amplitude for
case (i) is twice to that for case (ii) Resonant circuits have a variety of applications, for example, in the
tuning mechanism of a radio or a TV set The antenna of a radio accepts
signals from many broadcasting stations The signals picked up in the
antenna acts as a source in the tuning circuit of the radio, so the circuit
can be driven at many frequencies |
1 | 5800-5803 | Resonant circuits have a variety of applications, for example, in the
tuning mechanism of a radio or a TV set The antenna of a radio accepts
signals from many broadcasting stations The signals picked up in the
antenna acts as a source in the tuning circuit of the radio, so the circuit
can be driven at many frequencies But to hear one particular radio
station, we tune the radio |
1 | 5801-5804 | The antenna of a radio accepts
signals from many broadcasting stations The signals picked up in the
antenna acts as a source in the tuning circuit of the radio, so the circuit
can be driven at many frequencies But to hear one particular radio
station, we tune the radio In tuning, we vary the capacitance of a
capacitor in the tuning circuit such that the resonant frequency of the
circuit becomes nearly equal to the frequency of the radio signal received |
1 | 5802-5805 | The signals picked up in the
antenna acts as a source in the tuning circuit of the radio, so the circuit
can be driven at many frequencies But to hear one particular radio
station, we tune the radio In tuning, we vary the capacitance of a
capacitor in the tuning circuit such that the resonant frequency of the
circuit becomes nearly equal to the frequency of the radio signal received When this happens, the amplitude of the current with the frequency of
the signal of the particular radio station in the circuit is maximum |
1 | 5803-5806 | But to hear one particular radio
station, we tune the radio In tuning, we vary the capacitance of a
capacitor in the tuning circuit such that the resonant frequency of the
circuit becomes nearly equal to the frequency of the radio signal received When this happens, the amplitude of the current with the frequency of
the signal of the particular radio station in the circuit is maximum It is important to note that resonance phenomenon is exhibited by a
circuit only if both L and C are present in the circuit |
1 | 5804-5807 | In tuning, we vary the capacitance of a
capacitor in the tuning circuit such that the resonant frequency of the
circuit becomes nearly equal to the frequency of the radio signal received When this happens, the amplitude of the current with the frequency of
the signal of the particular radio station in the circuit is maximum It is important to note that resonance phenomenon is exhibited by a
circuit only if both L and C are present in the circuit Only then do the
voltages across L and C cancel each other (both being out of phase)
and the current amplitude is vm/R, the total source voltage appearing
across R |
1 | 5805-5808 | When this happens, the amplitude of the current with the frequency of
the signal of the particular radio station in the circuit is maximum It is important to note that resonance phenomenon is exhibited by a
circuit only if both L and C are present in the circuit Only then do the
voltages across L and C cancel each other (both being out of phase)
and the current amplitude is vm/R, the total source voltage appearing
across R This means that we cannot have resonance in a RL or
RC circuit |
1 | 5806-5809 | It is important to note that resonance phenomenon is exhibited by a
circuit only if both L and C are present in the circuit Only then do the
voltages across L and C cancel each other (both being out of phase)
and the current amplitude is vm/R, the total source voltage appearing
across R This means that we cannot have resonance in a RL or
RC circuit FIGURE 7 |
1 | 5807-5810 | Only then do the
voltages across L and C cancel each other (both being out of phase)
and the current amplitude is vm/R, the total source voltage appearing
across R This means that we cannot have resonance in a RL or
RC circuit FIGURE 7 14 Variation of im with w for two
cases: (i) R = 100 W, (ii) R = 200 W,
L = 1 |
1 | 5808-5811 | This means that we cannot have resonance in a RL or
RC circuit FIGURE 7 14 Variation of im with w for two
cases: (i) R = 100 W, (ii) R = 200 W,
L = 1 00 mH |
1 | 5809-5812 | FIGURE 7 14 Variation of im with w for two
cases: (i) R = 100 W, (ii) R = 200 W,
L = 1 00 mH Rationalised 2023-24
Physics
190
Example 7 |
1 | 5810-5813 | 14 Variation of im with w for two
cases: (i) R = 100 W, (ii) R = 200 W,
L = 1 00 mH Rationalised 2023-24
Physics
190
Example 7 6 A resistor of 200 W and a capacitor of 15 |
1 | 5811-5814 | 00 mH Rationalised 2023-24
Physics
190
Example 7 6 A resistor of 200 W and a capacitor of 15 0 mF are
connected in series to a 220 V, 50 Hz ac source |
1 | 5812-5815 | Rationalised 2023-24
Physics
190
Example 7 6 A resistor of 200 W and a capacitor of 15 0 mF are
connected in series to a 220 V, 50 Hz ac source (a) Calculate the
current in the circuit; (b) Calculate the voltage (rms) across the
resistor and the capacitor |
1 | 5813-5816 | 6 A resistor of 200 W and a capacitor of 15 0 mF are
connected in series to a 220 V, 50 Hz ac source (a) Calculate the
current in the circuit; (b) Calculate the voltage (rms) across the
resistor and the capacitor Is the algebraic sum of these voltages
more than the source voltage |
1 | 5814-5817 | 0 mF are
connected in series to a 220 V, 50 Hz ac source (a) Calculate the
current in the circuit; (b) Calculate the voltage (rms) across the
resistor and the capacitor Is the algebraic sum of these voltages
more than the source voltage If yes, resolve the paradox |
1 | 5815-5818 | (a) Calculate the
current in the circuit; (b) Calculate the voltage (rms) across the
resistor and the capacitor Is the algebraic sum of these voltages
more than the source voltage If yes, resolve the paradox Solution
Given
F
6
200
,
15 |
1 | 5816-5819 | Is the algebraic sum of these voltages
more than the source voltage If yes, resolve the paradox Solution
Given
F
6
200
,
15 0
15 |
1 | 5817-5820 | If yes, resolve the paradox Solution
Given
F
6
200
,
15 0
15 0
10
F
R
C
−
=
Ω
=
µ
=
×
220 V,
50Hz
V
ν
=
=
(a)
In order to calculate the current, we need the impedance of
the circuit |
1 | 5818-5821 | Solution
Given
F
6
200
,
15 0
15 0
10
F
R
C
−
=
Ω
=
µ
=
×
220 V,
50Hz
V
ν
=
=
(a)
In order to calculate the current, we need the impedance of
the circuit It is
2
2
2
2
(2
)
C
Z
R
X
R
π νC
−
=
+
=
+
F
2
6
2
(200
)
(2
3 |
1 | 5819-5822 | 0
15 0
10
F
R
C
−
=
Ω
=
µ
=
×
220 V,
50Hz
V
ν
=
=
(a)
In order to calculate the current, we need the impedance of
the circuit It is
2
2
2
2
(2
)
C
Z
R
X
R
π νC
−
=
+
=
+
F
2
6
2
(200
)
(2
3 14
50
15 |
1 | 5820-5823 | 0
10
F
R
C
−
=
Ω
=
µ
=
×
220 V,
50Hz
V
ν
=
=
(a)
In order to calculate the current, we need the impedance of
the circuit It is
2
2
2
2
(2
)
C
Z
R
X
R
π νC
−
=
+
=
+
F
2
6
2
(200
)
(2
3 14
50
15 0
10
)
−
−
=
Ω
+
×
×
×
×
2
2
(200
)
(212 |
1 | 5821-5824 | It is
2
2
2
2
(2
)
C
Z
R
X
R
π νC
−
=
+
=
+
F
2
6
2
(200
)
(2
3 14
50
15 0
10
)
−
−
=
Ω
+
×
×
×
×
2
2
(200
)
(212 3
)
=
Ω
+
Ω
=291 |
1 | 5822-5825 | 14
50
15 0
10
)
−
−
=
Ω
+
×
×
×
×
2
2
(200
)
(212 3
)
=
Ω
+
Ω
=291 67
Ω
Therefore, the current in the circuit is
220V
0 |
1 | 5823-5826 | 0
10
)
−
−
=
Ω
+
×
×
×
×
2
2
(200
)
(212 3
)
=
Ω
+
Ω
=291 67
Ω
Therefore, the current in the circuit is
220V
0 755 A
291 |
1 | 5824-5827 | 3
)
=
Ω
+
Ω
=291 67
Ω
Therefore, the current in the circuit is
220V
0 755 A
291 5
V
I
=Z
=
=
Ω
(b)
Since the current is the same throughout the circuit, we have
(0 |
1 | 5825-5828 | 67
Ω
Therefore, the current in the circuit is
220V
0 755 A
291 5
V
I
=Z
=
=
Ω
(b)
Since the current is the same throughout the circuit, we have
(0 755 A)(200
)
151V
VR
=I R
=
Ω =
(0 |
1 | 5826-5829 | 755 A
291 5
V
I
=Z
=
=
Ω
(b)
Since the current is the same throughout the circuit, we have
(0 755 A)(200
)
151V
VR
=I R
=
Ω =
(0 755 A)(212 |
1 | 5827-5830 | 5
V
I
=Z
=
=
Ω
(b)
Since the current is the same throughout the circuit, we have
(0 755 A)(200
)
151V
VR
=I R
=
Ω =
(0 755 A)(212 3
)
160 |
1 | 5828-5831 | 755 A)(200
)
151V
VR
=I R
=
Ω =
(0 755 A)(212 3
)
160 3 V
C
C
V
=I X
=
Ω =
The algebraic sum of the two voltages, VR and VC is 311 |
1 | 5829-5832 | 755 A)(212 3
)
160 3 V
C
C
V
=I X
=
Ω =
The algebraic sum of the two voltages, VR and VC is 311 3 V which
is more than the source voltage of 220 V |
1 | 5830-5833 | 3
)
160 3 V
C
C
V
=I X
=
Ω =
The algebraic sum of the two voltages, VR and VC is 311 3 V which
is more than the source voltage of 220 V How to resolve this
paradox |
1 | 5831-5834 | 3 V
C
C
V
=I X
=
Ω =
The algebraic sum of the two voltages, VR and VC is 311 3 V which
is more than the source voltage of 220 V How to resolve this
paradox As you have learnt in the text, the two voltages are not
in the same phase |
1 | 5832-5835 | 3 V which
is more than the source voltage of 220 V How to resolve this
paradox As you have learnt in the text, the two voltages are not
in the same phase Therefore, they cannot be added like ordinary
numbers |
1 | 5833-5836 | How to resolve this
paradox As you have learnt in the text, the two voltages are not
in the same phase Therefore, they cannot be added like ordinary
numbers The two voltages are out of phase by ninety degrees |
1 | 5834-5837 | As you have learnt in the text, the two voltages are not
in the same phase Therefore, they cannot be added like ordinary
numbers The two voltages are out of phase by ninety degrees Therefore, the total of these voltages must be obtained using the
Pythagorean theorem:
2
2
R C
R
C
V
V
V
+
=
+
= 220 V
Thus, if the phase difference between two voltages is properly taken
into account, the total voltage across the resistor and the capacitor
is equal to the voltage of the source |
1 | 5835-5838 | Therefore, they cannot be added like ordinary
numbers The two voltages are out of phase by ninety degrees Therefore, the total of these voltages must be obtained using the
Pythagorean theorem:
2
2
R C
R
C
V
V
V
+
=
+
= 220 V
Thus, if the phase difference between two voltages is properly taken
into account, the total voltage across the resistor and the capacitor
is equal to the voltage of the source EXAMPLE 7 |
1 | 5836-5839 | The two voltages are out of phase by ninety degrees Therefore, the total of these voltages must be obtained using the
Pythagorean theorem:
2
2
R C
R
C
V
V
V
+
=
+
= 220 V
Thus, if the phase difference between two voltages is properly taken
into account, the total voltage across the resistor and the capacitor
is equal to the voltage of the source EXAMPLE 7 6
7 |
1 | 5837-5840 | Therefore, the total of these voltages must be obtained using the
Pythagorean theorem:
2
2
R C
R
C
V
V
V
+
=
+
= 220 V
Thus, if the phase difference between two voltages is properly taken
into account, the total voltage across the resistor and the capacitor
is equal to the voltage of the source EXAMPLE 7 6
7 7 POWER IN AC CIRCUIT: THE POWER FACTOR
We have seen that a voltage v = vm sinwt applied to a series RLC circuit
drives a current in the circuit given by i = im sin(wt + f) where
m
m
v
i
=Z
and φ =
−
−
tan 1
X
X
R
C
L
Therefore, the instantaneous power p supplied by the source is
Rationalised 2023-24
191
Alternating Current
(
) [
]
sin
sin(
)
m
m
p
v i
v
t
i
t
ω
ω
φ
=
=
×
+
[
]
cos
cos(2
)
2
m m
v i
t
φ
ω
φ
=
−
+
(7 |
1 | 5838-5841 | EXAMPLE 7 6
7 7 POWER IN AC CIRCUIT: THE POWER FACTOR
We have seen that a voltage v = vm sinwt applied to a series RLC circuit
drives a current in the circuit given by i = im sin(wt + f) where
m
m
v
i
=Z
and φ =
−
−
tan 1
X
X
R
C
L
Therefore, the instantaneous power p supplied by the source is
Rationalised 2023-24
191
Alternating Current
(
) [
]
sin
sin(
)
m
m
p
v i
v
t
i
t
ω
ω
φ
=
=
×
+
[
]
cos
cos(2
)
2
m m
v i
t
φ
ω
φ
=
−
+
(7 29)
The average power over a cycle is given by the average of the two terms in
R |
1 | 5839-5842 | 6
7 7 POWER IN AC CIRCUIT: THE POWER FACTOR
We have seen that a voltage v = vm sinwt applied to a series RLC circuit
drives a current in the circuit given by i = im sin(wt + f) where
m
m
v
i
=Z
and φ =
−
−
tan 1
X
X
R
C
L
Therefore, the instantaneous power p supplied by the source is
Rationalised 2023-24
191
Alternating Current
(
) [
]
sin
sin(
)
m
m
p
v i
v
t
i
t
ω
ω
φ
=
=
×
+
[
]
cos
cos(2
)
2
m m
v i
t
φ
ω
φ
=
−
+
(7 29)
The average power over a cycle is given by the average of the two terms in
R H |
1 | 5840-5843 | 7 POWER IN AC CIRCUIT: THE POWER FACTOR
We have seen that a voltage v = vm sinwt applied to a series RLC circuit
drives a current in the circuit given by i = im sin(wt + f) where
m
m
v
i
=Z
and φ =
−
−
tan 1
X
X
R
C
L
Therefore, the instantaneous power p supplied by the source is
Rationalised 2023-24
191
Alternating Current
(
) [
]
sin
sin(
)
m
m
p
v i
v
t
i
t
ω
ω
φ
=
=
×
+
[
]
cos
cos(2
)
2
m m
v i
t
φ
ω
φ
=
−
+
(7 29)
The average power over a cycle is given by the average of the two terms in
R H S |
1 | 5841-5844 | 29)
The average power over a cycle is given by the average of the two terms in
R H S of Eq |
1 | 5842-5845 | H S of Eq (7 |
1 | 5843-5846 | S of Eq (7 37) |
1 | 5844-5847 | of Eq (7 37) It is only the second term which is time-dependent |
1 | 5845-5848 | (7 37) It is only the second term which is time-dependent Its average is zero (the positive half of the cosine cancels the negative
half) |
1 | 5846-5849 | 37) It is only the second term which is time-dependent Its average is zero (the positive half of the cosine cancels the negative
half) Therefore,
cos
2
m m
v i
P
φ
=
cos
2
2
m
m
v
i
φ
=
V Icos
φ
=
[7 |
1 | 5847-5850 | It is only the second term which is time-dependent Its average is zero (the positive half of the cosine cancels the negative
half) Therefore,
cos
2
m m
v i
P
φ
=
cos
2
2
m
m
v
i
φ
=
V Icos
φ
=
[7 30(a)]
This can also be written as,
2
cos
P
I Z
φ
=
[7 |
1 | 5848-5851 | Its average is zero (the positive half of the cosine cancels the negative
half) Therefore,
cos
2
m m
v i
P
φ
=
cos
2
2
m
m
v
i
φ
=
V Icos
φ
=
[7 30(a)]
This can also be written as,
2
cos
P
I Z
φ
=
[7 30(b)]
So, the average power dissipated depends not only on the voltage and
current but also on the cosine of the phase angle f between them |
1 | 5849-5852 | Therefore,
cos
2
m m
v i
P
φ
=
cos
2
2
m
m
v
i
φ
=
V Icos
φ
=
[7 30(a)]
This can also be written as,
2
cos
P
I Z
φ
=
[7 30(b)]
So, the average power dissipated depends not only on the voltage and
current but also on the cosine of the phase angle f between them The
quantity cosf is called the power factor |
1 | 5850-5853 | 30(a)]
This can also be written as,
2
cos
P
I Z
φ
=
[7 30(b)]
So, the average power dissipated depends not only on the voltage and
current but also on the cosine of the phase angle f between them The
quantity cosf is called the power factor Let us discuss the following
cases:
Case (i) Resistive circuit: If the circuit contains only pure R, it is called
resistive |
1 | 5851-5854 | 30(b)]
So, the average power dissipated depends not only on the voltage and
current but also on the cosine of the phase angle f between them The
quantity cosf is called the power factor Let us discuss the following
cases:
Case (i) Resistive circuit: If the circuit contains only pure R, it is called
resistive In that case f = 0, cos f = 1 |
1 | 5852-5855 | The
quantity cosf is called the power factor Let us discuss the following
cases:
Case (i) Resistive circuit: If the circuit contains only pure R, it is called
resistive In that case f = 0, cos f = 1 There is maximum power dissipation |
1 | 5853-5856 | Let us discuss the following
cases:
Case (i) Resistive circuit: If the circuit contains only pure R, it is called
resistive In that case f = 0, cos f = 1 There is maximum power dissipation Case (ii) Purely inductive or capacitive circuit: If the circuit contains
only an inductor or capacitor, we know that the phase difference between
voltage and current is p/2 |
1 | 5854-5857 | In that case f = 0, cos f = 1 There is maximum power dissipation Case (ii) Purely inductive or capacitive circuit: If the circuit contains
only an inductor or capacitor, we know that the phase difference between
voltage and current is p/2 Therefore, cos f = 0, and no power is dissipated
even though a current is flowing in the circuit |
1 | 5855-5858 | There is maximum power dissipation Case (ii) Purely inductive or capacitive circuit: If the circuit contains
only an inductor or capacitor, we know that the phase difference between
voltage and current is p/2 Therefore, cos f = 0, and no power is dissipated
even though a current is flowing in the circuit This current is sometimes
referred to as wattless current |
1 | 5856-5859 | Case (ii) Purely inductive or capacitive circuit: If the circuit contains
only an inductor or capacitor, we know that the phase difference between
voltage and current is p/2 Therefore, cos f = 0, and no power is dissipated
even though a current is flowing in the circuit This current is sometimes
referred to as wattless current Case (iii) LCR series circuit: In an LCR series circuit, power dissipated is
given by Eq |
1 | 5857-5860 | Therefore, cos f = 0, and no power is dissipated
even though a current is flowing in the circuit This current is sometimes
referred to as wattless current Case (iii) LCR series circuit: In an LCR series circuit, power dissipated is
given by Eq (7 |
1 | 5858-5861 | This current is sometimes
referred to as wattless current Case (iii) LCR series circuit: In an LCR series circuit, power dissipated is
given by Eq (7 30) where f = tan–1 (Xc – XL )/ R |
1 | 5859-5862 | Case (iii) LCR series circuit: In an LCR series circuit, power dissipated is
given by Eq (7 30) where f = tan–1 (Xc – XL )/ R So, f may be non-zero in
a RL or RC or RCL circuit |
1 | 5860-5863 | (7 30) where f = tan–1 (Xc – XL )/ R So, f may be non-zero in
a RL or RC or RCL circuit Even in such cases, power is dissipated only in
the resistor |
1 | 5861-5864 | 30) where f = tan–1 (Xc – XL )/ R So, f may be non-zero in
a RL or RC or RCL circuit Even in such cases, power is dissipated only in
the resistor Case (iv) Power dissipated at resonance in LCR circuit: At resonance
Xc – XL= 0, and f = 0 |
1 | 5862-5865 | So, f may be non-zero in
a RL or RC or RCL circuit Even in such cases, power is dissipated only in
the resistor Case (iv) Power dissipated at resonance in LCR circuit: At resonance
Xc – XL= 0, and f = 0 Therefore, cosf = 1 and P = I 2Z = I 2 R |
1 | 5863-5866 | Even in such cases, power is dissipated only in
the resistor Case (iv) Power dissipated at resonance in LCR circuit: At resonance
Xc – XL= 0, and f = 0 Therefore, cosf = 1 and P = I 2Z = I 2 R That is,
maximum power is dissipated in a circuit (through R) at resonance |
1 | 5864-5867 | Case (iv) Power dissipated at resonance in LCR circuit: At resonance
Xc – XL= 0, and f = 0 Therefore, cosf = 1 and P = I 2Z = I 2 R That is,
maximum power is dissipated in a circuit (through R) at resonance Example 7 |
1 | 5865-5868 | Therefore, cosf = 1 and P = I 2Z = I 2 R That is,
maximum power is dissipated in a circuit (through R) at resonance Example 7 7 (a) For circuits used for transporting electric power, a
low power factor implies large power loss in transmission |
1 | 5866-5869 | That is,
maximum power is dissipated in a circuit (through R) at resonance Example 7 7 (a) For circuits used for transporting electric power, a
low power factor implies large power loss in transmission Explain |
1 | 5867-5870 | Example 7 7 (a) For circuits used for transporting electric power, a
low power factor implies large power loss in transmission Explain (b) Power factor can often be improved by the use of a capacitor of
appropriate capacitance in the circuit |
1 | 5868-5871 | 7 (a) For circuits used for transporting electric power, a
low power factor implies large power loss in transmission Explain (b) Power factor can often be improved by the use of a capacitor of
appropriate capacitance in the circuit Explain |
1 | 5869-5872 | Explain (b) Power factor can often be improved by the use of a capacitor of
appropriate capacitance in the circuit Explain Solution (a) We know that P = I V cosf where cosf is the power factor |
1 | 5870-5873 | (b) Power factor can often be improved by the use of a capacitor of
appropriate capacitance in the circuit Explain Solution (a) We know that P = I V cosf where cosf is the power factor To supply a given power at a given voltage, if cosf is small, we have to
increase current accordingly |
1 | 5871-5874 | Explain Solution (a) We know that P = I V cosf where cosf is the power factor To supply a given power at a given voltage, if cosf is small, we have to
increase current accordingly But this will lead to large power loss
(I2R) in transmission |
1 | 5872-5875 | Solution (a) We know that P = I V cosf where cosf is the power factor To supply a given power at a given voltage, if cosf is small, we have to
increase current accordingly But this will lead to large power loss
(I2R) in transmission (b)Suppose in a circuit, current I lags the voltage by an angle f |
1 | 5873-5876 | To supply a given power at a given voltage, if cosf is small, we have to
increase current accordingly But this will lead to large power loss
(I2R) in transmission (b)Suppose in a circuit, current I lags the voltage by an angle f Then
power factor cosf =R/Z |
1 | 5874-5877 | But this will lead to large power loss
(I2R) in transmission (b)Suppose in a circuit, current I lags the voltage by an angle f Then
power factor cosf =R/Z We can improve the power factor (tending to 1) by making Z tend to
R |
1 | 5875-5878 | (b)Suppose in a circuit, current I lags the voltage by an angle f Then
power factor cosf =R/Z We can improve the power factor (tending to 1) by making Z tend to
R Let us understand, with the help of a phasor diagram (Fig |
1 | 5876-5879 | Then
power factor cosf =R/Z We can improve the power factor (tending to 1) by making Z tend to
R Let us understand, with the help of a phasor diagram (Fig 7 |
1 | 5877-5880 | We can improve the power factor (tending to 1) by making Z tend to
R Let us understand, with the help of a phasor diagram (Fig 7 15)
EXAMPLE 7 |
1 | 5878-5881 | Let us understand, with the help of a phasor diagram (Fig 7 15)
EXAMPLE 7 7
Rationalised 2023-24
Physics
192
how this can be achieved |
1 | 5879-5882 | 7 15)
EXAMPLE 7 7
Rationalised 2023-24
Physics
192
how this can be achieved Let us resolve I into two components |
1 | 5880-5883 | 15)
EXAMPLE 7 7
Rationalised 2023-24
Physics
192
how this can be achieved Let us resolve I into two components Ip
along the applied voltage V and Iq perpendicular to the applied
voltage |
1 | 5881-5884 | 7
Rationalised 2023-24
Physics
192
how this can be achieved Let us resolve I into two components Ip
along the applied voltage V and Iq perpendicular to the applied
voltage Iq as you have learnt in Section 7 |
1 | 5882-5885 | Let us resolve I into two components Ip
along the applied voltage V and Iq perpendicular to the applied
voltage Iq as you have learnt in Section 7 7, is called the wattless
component since corresponding to this component of current, there
is no power loss |
1 | 5883-5886 | Ip
along the applied voltage V and Iq perpendicular to the applied
voltage Iq as you have learnt in Section 7 7, is called the wattless
component since corresponding to this component of current, there
is no power loss IP is known as the power component because it is
in phase with the voltage and corresponds to power loss in the circuit |
1 | 5884-5887 | Iq as you have learnt in Section 7 7, is called the wattless
component since corresponding to this component of current, there
is no power loss IP is known as the power component because it is
in phase with the voltage and corresponds to power loss in the circuit It’s clear from this analysis that if we want to improve power factor,
we must completely neutralize the lagging wattless current Iq by an
equal leading wattless current I¢q |
1 | 5885-5888 | 7, is called the wattless
component since corresponding to this component of current, there
is no power loss IP is known as the power component because it is
in phase with the voltage and corresponds to power loss in the circuit It’s clear from this analysis that if we want to improve power factor,
we must completely neutralize the lagging wattless current Iq by an
equal leading wattless current I¢q This can be done by connecting
a capacitor of appropriate value in parallel so that Iq and I¢q cancel
each other and P is effectively Ip V |
1 | 5886-5889 | IP is known as the power component because it is
in phase with the voltage and corresponds to power loss in the circuit It’s clear from this analysis that if we want to improve power factor,
we must completely neutralize the lagging wattless current Iq by an
equal leading wattless current I¢q This can be done by connecting
a capacitor of appropriate value in parallel so that Iq and I¢q cancel
each other and P is effectively Ip V Example 7 |
1 | 5887-5890 | It’s clear from this analysis that if we want to improve power factor,
we must completely neutralize the lagging wattless current Iq by an
equal leading wattless current I¢q This can be done by connecting
a capacitor of appropriate value in parallel so that Iq and I¢q cancel
each other and P is effectively Ip V Example 7 8 A sinusoidal voltage of peak value 283 V and frequency
50 Hz is applied to a series LCR circuit in which
R = 3 W, L = 25 |
1 | 5888-5891 | This can be done by connecting
a capacitor of appropriate value in parallel so that Iq and I¢q cancel
each other and P is effectively Ip V Example 7 8 A sinusoidal voltage of peak value 283 V and frequency
50 Hz is applied to a series LCR circuit in which
R = 3 W, L = 25 48 mH, and C = 796 mF |
1 | 5889-5892 | Example 7 8 A sinusoidal voltage of peak value 283 V and frequency
50 Hz is applied to a series LCR circuit in which
R = 3 W, L = 25 48 mH, and C = 796 mF Find (a) the impedance of the
circuit; (b) the phase difference between the voltage across the source
and the current; (c) the power dissipated in the circuit; and (d) the
power factor |
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