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1 | 5890-5893 | 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 Solution
(a) To find the impedance of the circuit, we first calculate XL and XC |
1 | 5891-5894 | 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 Solution
(a) To find the impedance of the circuit, we first calculate XL and XC XL = 2 pnL
= 2 × 3 |
1 | 5892-5895 | 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 Solution
(a) To find the impedance of the circuit, we first calculate XL and XC XL = 2 pnL
= 2 × 3 14 × 50 × 25 |
1 | 5893-5896 | Solution
(a) To find the impedance of the circuit, we first calculate XL and XC XL = 2 pnL
= 2 × 3 14 × 50 × 25 48 × 10–3 W = 8 W
21
XC
νC
=
π
6
1
4
2
3 |
1 | 5894-5897 | XL = 2 pnL
= 2 × 3 14 × 50 × 25 48 × 10–3 W = 8 W
21
XC
νC
=
π
6
1
4
2
3 14
50
796
10−
=
=
Ω
×
×
×
×
Therefore,
2
2
2
2
(
)
3
(8
4)
L
C
Z
R
X
X
=
+
−
=
+
−
= 5 W
(b) Phase difference, f = tan–1
C
L
X
X
R
−
=
−
= −
°
tan− |
1 | 5895-5898 | 14 × 50 × 25 48 × 10–3 W = 8 W
21
XC
νC
=
π
6
1
4
2
3 14
50
796
10−
=
=
Ω
×
×
×
×
Therefore,
2
2
2
2
(
)
3
(8
4)
L
C
Z
R
X
X
=
+
−
=
+
−
= 5 W
(b) Phase difference, f = tan–1
C
L
X
X
R
−
=
−
= −
°
tan− 1
4
38
53 1
EXAMPLE 7 |
1 | 5896-5899 | 48 × 10–3 W = 8 W
21
XC
νC
=
π
6
1
4
2
3 14
50
796
10−
=
=
Ω
×
×
×
×
Therefore,
2
2
2
2
(
)
3
(8
4)
L
C
Z
R
X
X
=
+
−
=
+
−
= 5 W
(b) Phase difference, f = tan–1
C
L
X
X
R
−
=
−
= −
°
tan− 1
4
38
53 1
EXAMPLE 7 8
FIGURE 7 |
1 | 5897-5900 | 14
50
796
10−
=
=
Ω
×
×
×
×
Therefore,
2
2
2
2
(
)
3
(8
4)
L
C
Z
R
X
X
=
+
−
=
+
−
= 5 W
(b) Phase difference, f = tan–1
C
L
X
X
R
−
=
−
= −
°
tan− 1
4
38
53 1
EXAMPLE 7 8
FIGURE 7 15
EXAMPLE 7 |
1 | 5898-5901 | 1
4
38
53 1
EXAMPLE 7 8
FIGURE 7 15
EXAMPLE 7 7
Rationalised 2023-24
193
Alternating Current
EXAMPLE 7 |
1 | 5899-5902 | 8
FIGURE 7 15
EXAMPLE 7 7
Rationalised 2023-24
193
Alternating Current
EXAMPLE 7 9
Since f is negative, the current in the circuit lags the voltage
across the source |
1 | 5900-5903 | 15
EXAMPLE 7 7
Rationalised 2023-24
193
Alternating Current
EXAMPLE 7 9
Since f is negative, the current in the circuit lags the voltage
across the source (c) The power dissipated in the circuit is
2
P
I R
=
Now, I
=im
=
=
2
1
2
283
5
40A
Therefore,
A
W
(40 )2
3
4800
P =
×
Ω =
(d) Power factor =
�
�
cos
cos –53 |
1 | 5901-5904 | 7
Rationalised 2023-24
193
Alternating Current
EXAMPLE 7 9
Since f is negative, the current in the circuit lags the voltage
across the source (c) The power dissipated in the circuit is
2
P
I R
=
Now, I
=im
=
=
2
1
2
283
5
40A
Therefore,
A
W
(40 )2
3
4800
P =
×
Ω =
(d) Power factor =
�
�
cos
cos –53 1
0 |
1 | 5902-5905 | 9
Since f is negative, the current in the circuit lags the voltage
across the source (c) The power dissipated in the circuit is
2
P
I R
=
Now, I
=im
=
=
2
1
2
283
5
40A
Therefore,
A
W
(40 )2
3
4800
P =
×
Ω =
(d) Power factor =
�
�
cos
cos –53 1
0 6
� �
� �
Example 7 |
1 | 5903-5906 | (c) The power dissipated in the circuit is
2
P
I R
=
Now, I
=im
=
=
2
1
2
283
5
40A
Therefore,
A
W
(40 )2
3
4800
P =
×
Ω =
(d) Power factor =
�
�
cos
cos –53 1
0 6
� �
� �
Example 7 9 Suppose the frequency of the source in the previous
example can be varied |
1 | 5904-5907 | 1
0 6
� �
� �
Example 7 9 Suppose the frequency of the source in the previous
example can be varied (a) What is the frequency of the source at
which resonance occurs |
1 | 5905-5908 | 6
� �
� �
Example 7 9 Suppose the frequency of the source in the previous
example can be varied (a) What is the frequency of the source at
which resonance occurs (b) Calculate the impedance, the current,
and the power dissipated at the resonant condition |
1 | 5906-5909 | 9 Suppose the frequency of the source in the previous
example can be varied (a) What is the frequency of the source at
which resonance occurs (b) Calculate the impedance, the current,
and the power dissipated at the resonant condition Solution
(a) The frequency at which the resonance occurs is
0
3
6
1
1
25 |
1 | 5907-5910 | (a) What is the frequency of the source at
which resonance occurs (b) Calculate the impedance, the current,
and the power dissipated at the resonant condition Solution
(a) The frequency at which the resonance occurs is
0
3
6
1
1
25 48
10
796
10
LC
ω
−
−
=
=
×
×
×
222 |
1 | 5908-5911 | (b) Calculate the impedance, the current,
and the power dissipated at the resonant condition Solution
(a) The frequency at which the resonance occurs is
0
3
6
1
1
25 48
10
796
10
LC
ω
−
−
=
=
×
×
×
222 1rad/s
=
0
221 |
1 | 5909-5912 | Solution
(a) The frequency at which the resonance occurs is
0
3
6
1
1
25 48
10
796
10
LC
ω
−
−
=
=
×
×
×
222 1rad/s
=
0
221 1 Hz
35 |
1 | 5910-5913 | 48
10
796
10
LC
ω
−
−
=
=
×
×
×
222 1rad/s
=
0
221 1 Hz
35 4Hz
2
2
3 |
1 | 5911-5914 | 1rad/s
=
0
221 1 Hz
35 4Hz
2
2
3 14
r
ν =ω
=
=
π
×
(b) The impedance Z at resonant condition is equal to the resistance:
3
Z
=R
=
Ω
The rms current at resonance is
=
=
=
=
ZV
RV
283
2
31
66 7 |
1 | 5912-5915 | 1 Hz
35 4Hz
2
2
3 14
r
ν =ω
=
=
π
×
(b) The impedance Z at resonant condition is equal to the resistance:
3
Z
=R
=
Ω
The rms current at resonance is
=
=
=
=
ZV
RV
283
2
31
66 7 A
The power dissipated at resonance is
2
(66 |
1 | 5913-5916 | 4Hz
2
2
3 14
r
ν =ω
=
=
π
×
(b) The impedance Z at resonant condition is equal to the resistance:
3
Z
=R
=
Ω
The rms current at resonance is
=
=
=
=
ZV
RV
283
2
31
66 7 A
The power dissipated at resonance is
2
(66 7)2
3
13 |
1 | 5914-5917 | 14
r
ν =ω
=
=
π
×
(b) The impedance Z at resonant condition is equal to the resistance:
3
Z
=R
=
Ω
The rms current at resonance is
=
=
=
=
ZV
RV
283
2
31
66 7 A
The power dissipated at resonance is
2
(66 7)2
3
13 35 kW
P
I
R
=
×
=
×
=
You can see that in the present case, power dissipated
at resonance is more than the power dissipated in Example 7 |
1 | 5915-5918 | A
The power dissipated at resonance is
2
(66 7)2
3
13 35 kW
P
I
R
=
×
=
×
=
You can see that in the present case, power dissipated
at resonance is more than the power dissipated in Example 7 8 |
1 | 5916-5919 | 7)2
3
13 35 kW
P
I
R
=
×
=
×
=
You can see that in the present case, power dissipated
at resonance is more than the power dissipated in Example 7 8 EXAMPLE 7 |
1 | 5917-5920 | 35 kW
P
I
R
=
×
=
×
=
You can see that in the present case, power dissipated
at resonance is more than the power dissipated in Example 7 8 EXAMPLE 7 8
EXAMPLE 7 |
1 | 5918-5921 | 8 EXAMPLE 7 8
EXAMPLE 7 10
Example 7 |
1 | 5919-5922 | EXAMPLE 7 8
EXAMPLE 7 10
Example 7 10 At an airport, a person is made to walk through the
doorway of a metal detector, for security reasons |
1 | 5920-5923 | 8
EXAMPLE 7 10
Example 7 10 At an airport, a person is made to walk through the
doorway of a metal detector, for security reasons If she/he is carrying
anything made of metal, the metal detector emits a sound |
1 | 5921-5924 | 10
Example 7 10 At an airport, a person is made to walk through the
doorway of a metal detector, for security reasons If she/he is carrying
anything made of metal, the metal detector emits a sound On what
principle does this detector work |
1 | 5922-5925 | 10 At an airport, a person is made to walk through the
doorway of a metal detector, for security reasons If she/he is carrying
anything made of metal, the metal detector emits a sound On what
principle does this detector work Solution The metal detector works on the principle of resonance in
ac circuits |
1 | 5923-5926 | If she/he is carrying
anything made of metal, the metal detector emits a sound On what
principle does this detector work Solution The metal detector works on the principle of resonance in
ac circuits When you walk through a metal detector, you are,
in fact, walking through a coil of many turns |
1 | 5924-5927 | On what
principle does this detector work Solution The metal detector works on the principle of resonance in
ac circuits When you walk through a metal detector, you are,
in fact, walking through a coil of many turns The coil is connected to
a capacitor tuned so that the circuit is in resonance |
1 | 5925-5928 | Solution The metal detector works on the principle of resonance in
ac circuits When you walk through a metal detector, you are,
in fact, walking through a coil of many turns The coil is connected to
a capacitor tuned so that the circuit is in resonance When
you walk through with metal in your pocket, the impedance of the
circuit changes – resulting in significant change in current in the
circuit |
1 | 5926-5929 | When you walk through a metal detector, you are,
in fact, walking through a coil of many turns The coil is connected to
a capacitor tuned so that the circuit is in resonance When
you walk through with metal in your pocket, the impedance of the
circuit changes – resulting in significant change in current in the
circuit This change in current is detected and the electronic circuitry
causes a sound to be emitted as an alarm |
1 | 5927-5930 | The coil is connected to
a capacitor tuned so that the circuit is in resonance When
you walk through with metal in your pocket, the impedance of the
circuit changes – resulting in significant change in current in the
circuit This change in current is detected and the electronic circuitry
causes a sound to be emitted as an alarm Rationalised 2023-24
Physics
194
7 |
1 | 5928-5931 | When
you walk through with metal in your pocket, the impedance of the
circuit changes – resulting in significant change in current in the
circuit This change in current is detected and the electronic circuitry
causes a sound to be emitted as an alarm Rationalised 2023-24
Physics
194
7 8 TRANSFORMERS
For many purposes, it is necessary to change (or transform) an alternating
voltage from one to another of greater or smaller value |
1 | 5929-5932 | This change in current is detected and the electronic circuitry
causes a sound to be emitted as an alarm Rationalised 2023-24
Physics
194
7 8 TRANSFORMERS
For many purposes, it is necessary to change (or transform) an alternating
voltage from one to another of greater or smaller value This is done with
a device called transformer using the principle of mutual induction |
1 | 5930-5933 | Rationalised 2023-24
Physics
194
7 8 TRANSFORMERS
For many purposes, it is necessary to change (or transform) an alternating
voltage from one to another of greater or smaller value This is done with
a device called transformer using the principle of mutual induction A transformer consists of two sets of coils, insulated from each other |
1 | 5931-5934 | 8 TRANSFORMERS
For many purposes, it is necessary to change (or transform) an alternating
voltage from one to another of greater or smaller value This is done with
a device called transformer using the principle of mutual induction A transformer consists of two sets of coils, insulated from each other They are wound on a soft-iron core, either one on top of the other as in
Fig |
1 | 5932-5935 | This is done with
a device called transformer using the principle of mutual induction A transformer consists of two sets of coils, insulated from each other They are wound on a soft-iron core, either one on top of the other as in
Fig 7 |
1 | 5933-5936 | A transformer consists of two sets of coils, insulated from each other They are wound on a soft-iron core, either one on top of the other as in
Fig 7 16(a) or on separate limbs of the core as in Fig |
1 | 5934-5937 | They are wound on a soft-iron core, either one on top of the other as in
Fig 7 16(a) or on separate limbs of the core as in Fig 7 |
1 | 5935-5938 | 7 16(a) or on separate limbs of the core as in Fig 7 16(b) |
1 | 5936-5939 | 16(a) or on separate limbs of the core as in Fig 7 16(b) One of the
coils called the primary coil has Np turns |
1 | 5937-5940 | 7 16(b) One of the
coils called the primary coil has Np turns The other coil is called the
secondary coil; it has Ns turns |
1 | 5938-5941 | 16(b) One of the
coils called the primary coil has Np turns The other coil is called the
secondary coil; it has Ns turns Often the primary coil is the input coil
and the secondary coil is the output coil of the transformer |
1 | 5939-5942 | One of the
coils called the primary coil has Np turns The other coil is called the
secondary coil; it has Ns turns Often the primary coil is the input coil
and the secondary coil is the output coil of the transformer FIGURE 7 |
1 | 5940-5943 | The other coil is called the
secondary coil; it has Ns turns Often the primary coil is the input coil
and the secondary coil is the output coil of the transformer FIGURE 7 16 Two arrangements for winding of primary and secondary coil in a transformer:
(a) two coils on top of each other, (b) two coils on separate limbs of the core |
1 | 5941-5944 | Often the primary coil is the input coil
and the secondary coil is the output coil of the transformer FIGURE 7 16 Two arrangements for winding of primary and secondary coil in a transformer:
(a) two coils on top of each other, (b) two coils on separate limbs of the core When an alternating voltage is applied to the primary, the resulting
current produces an alternating magnetic flux which links the secondary
and induces an emf in it |
1 | 5942-5945 | FIGURE 7 16 Two arrangements for winding of primary and secondary coil in a transformer:
(a) two coils on top of each other, (b) two coils on separate limbs of the core When an alternating voltage is applied to the primary, the resulting
current produces an alternating magnetic flux which links the secondary
and induces an emf in it The value of this emf depends on the number of
turns in the secondary |
1 | 5943-5946 | 16 Two arrangements for winding of primary and secondary coil in a transformer:
(a) two coils on top of each other, (b) two coils on separate limbs of the core When an alternating voltage is applied to the primary, the resulting
current produces an alternating magnetic flux which links the secondary
and induces an emf in it The value of this emf depends on the number of
turns in the secondary We consider an ideal transformer in which the
primary has negligible resistance and all the flux in the core links both
primary and secondary windings |
1 | 5944-5947 | When an alternating voltage is applied to the primary, the resulting
current produces an alternating magnetic flux which links the secondary
and induces an emf in it The value of this emf depends on the number of
turns in the secondary We consider an ideal transformer in which the
primary has negligible resistance and all the flux in the core links both
primary and secondary windings Let f be the flux in each turn in the core
at time t due to current in the primary when a voltage vp is applied to it |
1 | 5945-5948 | The value of this emf depends on the number of
turns in the secondary We consider an ideal transformer in which the
primary has negligible resistance and all the flux in the core links both
primary and secondary windings Let f be the flux in each turn in the core
at time t due to current in the primary when a voltage vp is applied to it Then the induced emf or voltage es, in the secondary with Ns turns is
d
d
s
Ns
φt
ε
= −
(7 |
1 | 5946-5949 | We consider an ideal transformer in which the
primary has negligible resistance and all the flux in the core links both
primary and secondary windings Let f be the flux in each turn in the core
at time t due to current in the primary when a voltage vp is applied to it Then the induced emf or voltage es, in the secondary with Ns turns is
d
d
s
Ns
φt
ε
= −
(7 31)
The alternating flux f also induces an emf, called back emf in the
primary |
1 | 5947-5950 | Let f be the flux in each turn in the core
at time t due to current in the primary when a voltage vp is applied to it Then the induced emf or voltage es, in the secondary with Ns turns is
d
d
s
Ns
φt
ε
= −
(7 31)
The alternating flux f also induces an emf, called back emf in the
primary This is
d
d
p
Np
φt
ε
= −
(7 |
1 | 5948-5951 | Then the induced emf or voltage es, in the secondary with Ns turns is
d
d
s
Ns
φt
ε
= −
(7 31)
The alternating flux f also induces an emf, called back emf in the
primary This is
d
d
p
Np
φt
ε
= −
(7 32)
But ep = vp |
1 | 5949-5952 | 31)
The alternating flux f also induces an emf, called back emf in the
primary This is
d
d
p
Np
φt
ε
= −
(7 32)
But ep = vp If this were not so, the primary current would be infinite
since the primary has zero resistance (as assumed) |
1 | 5950-5953 | This is
d
d
p
Np
φt
ε
= −
(7 32)
But ep = vp If this were not so, the primary current would be infinite
since the primary has zero resistance (as assumed) If the secondary is
an open circuit or the current taken from it is small, then to a good
approximation
es = vs
Rationalised 2023-24
195
Alternating Current
where vs is the voltage across the secondary |
1 | 5951-5954 | 32)
But ep = vp If this were not so, the primary current would be infinite
since the primary has zero resistance (as assumed) If the secondary is
an open circuit or the current taken from it is small, then to a good
approximation
es = vs
Rationalised 2023-24
195
Alternating Current
where vs is the voltage across the secondary Therefore, Eqs |
1 | 5952-5955 | If this were not so, the primary current would be infinite
since the primary has zero resistance (as assumed) If the secondary is
an open circuit or the current taken from it is small, then to a good
approximation
es = vs
Rationalised 2023-24
195
Alternating Current
where vs is the voltage across the secondary Therefore, Eqs (7 |
1 | 5953-5956 | If the secondary is
an open circuit or the current taken from it is small, then to a good
approximation
es = vs
Rationalised 2023-24
195
Alternating Current
where vs is the voltage across the secondary Therefore, Eqs (7 31) and
(7 |
1 | 5954-5957 | Therefore, Eqs (7 31) and
(7 32) can be written as
s
s
d
v
N
d t
φ
= −
[7 |
1 | 5955-5958 | (7 31) and
(7 32) can be written as
s
s
d
v
N
d t
φ
= −
[7 31(a)]
p
p
d
v
N
d t
φ
= −
[7 |
1 | 5956-5959 | 31) and
(7 32) can be written as
s
s
d
v
N
d t
φ
= −
[7 31(a)]
p
p
d
v
N
d t
φ
= −
[7 32(a)]
From Eqs |
1 | 5957-5960 | 32) can be written as
s
s
d
v
N
d t
φ
= −
[7 31(a)]
p
p
d
v
N
d t
φ
= −
[7 32(a)]
From Eqs [7 |
1 | 5958-5961 | 31(a)]
p
p
d
v
N
d t
φ
= −
[7 32(a)]
From Eqs [7 31 (a)] and [7 |
1 | 5959-5962 | 32(a)]
From Eqs [7 31 (a)] and [7 32 (a)], we have
s
s
p
p
v
N
v
=N
(7 |
1 | 5960-5963 | [7 31 (a)] and [7 32 (a)], we have
s
s
p
p
v
N
v
=N
(7 33)
Note that the above relation has been obtained using three
assumptions: (i) the primary resistance and current are small; (ii) the
same flux links both the primary and the secondary as very little flux
escapes from the core, and (iii) the secondary current is small |
1 | 5961-5964 | 31 (a)] and [7 32 (a)], we have
s
s
p
p
v
N
v
=N
(7 33)
Note that the above relation has been obtained using three
assumptions: (i) the primary resistance and current are small; (ii) the
same flux links both the primary and the secondary as very little flux
escapes from the core, and (iii) the secondary current is small If the transformer is assumed to be 100% efficient (no energy losses),
the power input is equal to the power output, and since p = i v,
ipvp = isvs
(7 |
1 | 5962-5965 | 32 (a)], we have
s
s
p
p
v
N
v
=N
(7 33)
Note that the above relation has been obtained using three
assumptions: (i) the primary resistance and current are small; (ii) the
same flux links both the primary and the secondary as very little flux
escapes from the core, and (iii) the secondary current is small If the transformer is assumed to be 100% efficient (no energy losses),
the power input is equal to the power output, and since p = i v,
ipvp = isvs
(7 34)
Although some energy is always lost, this is a good approximation,
since a well designed transformer may have an efficiency of more than
95% |
1 | 5963-5966 | 33)
Note that the above relation has been obtained using three
assumptions: (i) the primary resistance and current are small; (ii) the
same flux links both the primary and the secondary as very little flux
escapes from the core, and (iii) the secondary current is small If the transformer is assumed to be 100% efficient (no energy losses),
the power input is equal to the power output, and since p = i v,
ipvp = isvs
(7 34)
Although some energy is always lost, this is a good approximation,
since a well designed transformer may have an efficiency of more than
95% Combining Eqs |
1 | 5964-5967 | If the transformer is assumed to be 100% efficient (no energy losses),
the power input is equal to the power output, and since p = i v,
ipvp = isvs
(7 34)
Although some energy is always lost, this is a good approximation,
since a well designed transformer may have an efficiency of more than
95% Combining Eqs (7 |
1 | 5965-5968 | 34)
Although some energy is always lost, this is a good approximation,
since a well designed transformer may have an efficiency of more than
95% Combining Eqs (7 33) and (7 |
1 | 5966-5969 | Combining Eqs (7 33) and (7 34), we have
p
s
s
s
p
p
i
v
N
i
v
N
=
=
(7 |
1 | 5967-5970 | (7 33) and (7 34), we have
p
s
s
s
p
p
i
v
N
i
v
N
=
=
(7 35)
Since i and v both oscillate with the same frequency as the ac source,
Eq |
1 | 5968-5971 | 33) and (7 34), we have
p
s
s
s
p
p
i
v
N
i
v
N
=
=
(7 35)
Since i and v both oscillate with the same frequency as the ac source,
Eq (7 |
1 | 5969-5972 | 34), we have
p
s
s
s
p
p
i
v
N
i
v
N
=
=
(7 35)
Since i and v both oscillate with the same frequency as the ac source,
Eq (7 35) also gives the ratio of the amplitudes or rms values of
corresponding quantities |
1 | 5970-5973 | 35)
Since i and v both oscillate with the same frequency as the ac source,
Eq (7 35) also gives the ratio of the amplitudes or rms values of
corresponding quantities Now, we can see how a transformer affects the voltage and current |
1 | 5971-5974 | (7 35) also gives the ratio of the amplitudes or rms values of
corresponding quantities Now, we can see how a transformer affects the voltage and current We have:
V
NN
V
s
s
p
p
=
and I
N
N
I
s
p
s
p
=
(7 |
1 | 5972-5975 | 35) also gives the ratio of the amplitudes or rms values of
corresponding quantities Now, we can see how a transformer affects the voltage and current We have:
V
NN
V
s
s
p
p
=
and I
N
N
I
s
p
s
p
=
(7 36)
That is, if the secondary coil has a greater number of turns than the
primary (Ns > Np), the voltage is stepped up (Vs > Vp) |
1 | 5973-5976 | Now, we can see how a transformer affects the voltage and current We have:
V
NN
V
s
s
p
p
=
and I
N
N
I
s
p
s
p
=
(7 36)
That is, if the secondary coil has a greater number of turns than the
primary (Ns > Np), the voltage is stepped up (Vs > Vp) This type of
arrangement is called a step-up transformer |
1 | 5974-5977 | We have:
V
NN
V
s
s
p
p
=
and I
N
N
I
s
p
s
p
=
(7 36)
That is, if the secondary coil has a greater number of turns than the
primary (Ns > Np), the voltage is stepped up (Vs > Vp) This type of
arrangement is called a step-up transformer However, in this arrangement,
there is less current in the secondary than in the primary (Np/Ns < 1 and Is
< Ip) |
1 | 5975-5978 | 36)
That is, if the secondary coil has a greater number of turns than the
primary (Ns > Np), the voltage is stepped up (Vs > Vp) This type of
arrangement is called a step-up transformer However, in this arrangement,
there is less current in the secondary than in the primary (Np/Ns < 1 and Is
< Ip) For example, if the primary coil of a transformer has 100 turns and
the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2 |
1 | 5976-5979 | This type of
arrangement is called a step-up transformer However, in this arrangement,
there is less current in the secondary than in the primary (Np/Ns < 1 and Is
< Ip) For example, if the primary coil of a transformer has 100 turns and
the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2 Thus, a 220V
input at 10A will step-up to 440 V output at 5 |
1 | 5977-5980 | However, in this arrangement,
there is less current in the secondary than in the primary (Np/Ns < 1 and Is
< Ip) For example, if the primary coil of a transformer has 100 turns and
the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2 Thus, a 220V
input at 10A will step-up to 440 V output at 5 0 A |
1 | 5978-5981 | For example, if the primary coil of a transformer has 100 turns and
the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2 Thus, a 220V
input at 10A will step-up to 440 V output at 5 0 A If the secondary coil has less turns than the primary (Ns < Np),
we have a step-down transformer |
1 | 5979-5982 | Thus, a 220V
input at 10A will step-up to 440 V output at 5 0 A If the secondary coil has less turns than the primary (Ns < Np),
we have a step-down transformer In this case, Vs < Vp and Is > Ip |
1 | 5980-5983 | 0 A If the secondary coil has less turns than the primary (Ns < Np),
we have a step-down transformer In this case, Vs < Vp and Is > Ip That
is, the voltage is stepped down, or reduced, and the current
is increased |
1 | 5981-5984 | If the secondary coil has less turns than the primary (Ns < Np),
we have a step-down transformer In this case, Vs < Vp and Is > Ip That
is, the voltage is stepped down, or reduced, and the current
is increased The equations obtained above apply to ideal transformers (without
any energy losses) |
1 | 5982-5985 | In this case, Vs < Vp and Is > Ip That
is, the voltage is stepped down, or reduced, and the current
is increased The equations obtained above apply to ideal transformers (without
any energy losses) But in actual transformers, small energy losses do
occur due to the following reasons:
(i) Flux Leakage: There is always some flux leakage; that is, not all of
the flux due to primary passes through the secondary due to poor
Rationalised 2023-24
Physics
196
SUMMARY
1 |
1 | 5983-5986 | That
is, the voltage is stepped down, or reduced, and the current
is increased The equations obtained above apply to ideal transformers (without
any energy losses) But in actual transformers, small energy losses do
occur due to the following reasons:
(i) Flux Leakage: There is always some flux leakage; that is, not all of
the flux due to primary passes through the secondary due to poor
Rationalised 2023-24
Physics
196
SUMMARY
1 An alternating voltage
sin
=
ω
m
v
v
t applied to a resistor R drives a
current i = im sinwt in the resistor,
m
m
v
i
=R |
1 | 5984-5987 | The equations obtained above apply to ideal transformers (without
any energy losses) But in actual transformers, small energy losses do
occur due to the following reasons:
(i) Flux Leakage: There is always some flux leakage; that is, not all of
the flux due to primary passes through the secondary due to poor
Rationalised 2023-24
Physics
196
SUMMARY
1 An alternating voltage
sin
=
ω
m
v
v
t applied to a resistor R drives a
current i = im sinwt in the resistor,
m
m
v
i
=R The current is in phase with
the applied voltage |
1 | 5985-5988 | But in actual transformers, small energy losses do
occur due to the following reasons:
(i) Flux Leakage: There is always some flux leakage; that is, not all of
the flux due to primary passes through the secondary due to poor
Rationalised 2023-24
Physics
196
SUMMARY
1 An alternating voltage
sin
=
ω
m
v
v
t applied to a resistor R drives a
current i = im sinwt in the resistor,
m
m
v
i
=R The current is in phase with
the applied voltage 2 |
1 | 5986-5989 | An alternating voltage
sin
=
ω
m
v
v
t applied to a resistor R drives a
current i = im sinwt in the resistor,
m
m
v
i
=R The current is in phase with
the applied voltage 2 For an alternating current i = im sin wt passing through a resistor R, the
average power loss P (averaged over a cycle) due to joule heating is
( 1/2 )i 2
mR |
1 | 5987-5990 | The current is in phase with
the applied voltage 2 For an alternating current i = im sin wt passing through a resistor R, the
average power loss P (averaged over a cycle) due to joule heating is
( 1/2 )i 2
mR To express it in the same form as the dc power (P = I 2R), a
special value of current is used |
1 | 5988-5991 | 2 For an alternating current i = im sin wt passing through a resistor R, the
average power loss P (averaged over a cycle) due to joule heating is
( 1/2 )i 2
mR To express it in the same form as the dc power (P = I 2R), a
special value of current is used It is called root mean square (rms)
current and is donoted by I:
0 |
1 | 5989-5992 | For an alternating current i = im sin wt passing through a resistor R, the
average power loss P (averaged over a cycle) due to joule heating is
( 1/2 )i 2
mR To express it in the same form as the dc power (P = I 2R), a
special value of current is used It is called root mean square (rms)
current and is donoted by I:
0 707
2
m
m
i
I
i
=
=
Similarly, the rms voltage is defined by
0 |
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