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1 | 5590-5593 | unsw edu au//jw/AC html
Rationalised 2023-24
183
Alternating Current
EXAMPLE 7 |
1 | 5591-5594 | edu au//jw/AC html
Rationalised 2023-24
183
Alternating Current
EXAMPLE 7 2
We see that the current reaches its maximum value later than the
voltage by one-fourth of a period
4T
2
=
π/
ω |
1 | 5592-5595 | au//jw/AC html
Rationalised 2023-24
183
Alternating Current
EXAMPLE 7 2
We see that the current reaches its maximum value later than the
voltage by one-fourth of a period
4T
2
=
π/
ω You have seen that an
inductor has reactance that limits current similar to resistance in a
dc circuit |
1 | 5593-5596 | html
Rationalised 2023-24
183
Alternating Current
EXAMPLE 7 2
We see that the current reaches its maximum value later than the
voltage by one-fourth of a period
4T
2
=
π/
ω You have seen that an
inductor has reactance that limits current similar to resistance in a
dc circuit Does it also consume power like a resistance |
1 | 5594-5597 | 2
We see that the current reaches its maximum value later than the
voltage by one-fourth of a period
4T
2
=
π/
ω You have seen that an
inductor has reactance that limits current similar to resistance in a
dc circuit Does it also consume power like a resistance Let us try to
find out |
1 | 5595-5598 | You have seen that an
inductor has reactance that limits current similar to resistance in a
dc circuit Does it also consume power like a resistance Let us try to
find out The instantaneous power supplied to the inductor is
p
i v
i
t
v
t
L
m
m
=
=
−
(
)
sin
sin
ω
ω
π
2 ×
(
)
(
)
cos
sin
m
i vm
t
t
ω
ω
= −
(
)
sin 2
2
m
i vm
ωt
= −
So, the average power over a complete cycle is
(
)
L
sin 2
2
m
i vm
P
ωt
=
−
(
)
sin 2
2
m
i vm
ωt
= −
= 0,
since the average of sin (2wt) over a complete cycle is zero |
1 | 5596-5599 | Does it also consume power like a resistance Let us try to
find out The instantaneous power supplied to the inductor is
p
i v
i
t
v
t
L
m
m
=
=
−
(
)
sin
sin
ω
ω
π
2 ×
(
)
(
)
cos
sin
m
i vm
t
t
ω
ω
= −
(
)
sin 2
2
m
i vm
ωt
= −
So, the average power over a complete cycle is
(
)
L
sin 2
2
m
i vm
P
ωt
=
−
(
)
sin 2
2
m
i vm
ωt
= −
= 0,
since the average of sin (2wt) over a complete cycle is zero Thus, the average power supplied to an inductor over one complete
cycle is zero |
1 | 5597-5600 | Let us try to
find out The instantaneous power supplied to the inductor is
p
i v
i
t
v
t
L
m
m
=
=
−
(
)
sin
sin
ω
ω
π
2 ×
(
)
(
)
cos
sin
m
i vm
t
t
ω
ω
= −
(
)
sin 2
2
m
i vm
ωt
= −
So, the average power over a complete cycle is
(
)
L
sin 2
2
m
i vm
P
ωt
=
−
(
)
sin 2
2
m
i vm
ωt
= −
= 0,
since the average of sin (2wt) over a complete cycle is zero Thus, the average power supplied to an inductor over one complete
cycle is zero Example 7 |
1 | 5598-5601 | The instantaneous power supplied to the inductor is
p
i v
i
t
v
t
L
m
m
=
=
−
(
)
sin
sin
ω
ω
π
2 ×
(
)
(
)
cos
sin
m
i vm
t
t
ω
ω
= −
(
)
sin 2
2
m
i vm
ωt
= −
So, the average power over a complete cycle is
(
)
L
sin 2
2
m
i vm
P
ωt
=
−
(
)
sin 2
2
m
i vm
ωt
= −
= 0,
since the average of sin (2wt) over a complete cycle is zero Thus, the average power supplied to an inductor over one complete
cycle is zero Example 7 2 A pure inductor of 25 |
1 | 5599-5602 | Thus, the average power supplied to an inductor over one complete
cycle is zero Example 7 2 A pure inductor of 25 0 mH is connected to a source of
220 V |
1 | 5600-5603 | Example 7 2 A pure inductor of 25 0 mH is connected to a source of
220 V Find the inductive reactance and rms current in the circuit if
the frequency of the source is 50 Hz |
1 | 5601-5604 | 2 A pure inductor of 25 0 mH is connected to a source of
220 V Find the inductive reactance and rms current in the circuit if
the frequency of the source is 50 Hz Solution The inductive reactance,
–
= |
1 | 5602-5605 | 0 mH is connected to a source of
220 V Find the inductive reactance and rms current in the circuit if
the frequency of the source is 50 Hz Solution The inductive reactance,
–
= 3
2
2
3 14
50
25
10
πν
×
×
×
×
Ω
XL
L =
= 7 |
1 | 5603-5606 | Find the inductive reactance and rms current in the circuit if
the frequency of the source is 50 Hz Solution The inductive reactance,
–
= 3
2
2
3 14
50
25
10
πν
×
×
×
×
Ω
XL
L =
= 7 85W
The rms current in the circuit is
V
A
220
28
7 |
1 | 5604-5607 | Solution The inductive reactance,
–
= 3
2
2
3 14
50
25
10
πν
×
×
×
×
Ω
XL
L =
= 7 85W
The rms current in the circuit is
V
A
220
28
7 85
L
V
I
=X
=
=
Ω
FIGURE 7 |
1 | 5605-5608 | 3
2
2
3 14
50
25
10
πν
×
×
×
×
Ω
XL
L =
= 7 85W
The rms current in the circuit is
V
A
220
28
7 85
L
V
I
=X
=
=
Ω
FIGURE 7 6 (a) A Phasor diagram for the circuit in Fig |
1 | 5606-5609 | 85W
The rms current in the circuit is
V
A
220
28
7 85
L
V
I
=X
=
=
Ω
FIGURE 7 6 (a) A Phasor diagram for the circuit in Fig 7 |
1 | 5607-5610 | 85
L
V
I
=X
=
=
Ω
FIGURE 7 6 (a) A Phasor diagram for the circuit in Fig 7 5 |
1 | 5608-5611 | 6 (a) A Phasor diagram for the circuit in Fig 7 5 (b) Graph of v and i versus wt |
1 | 5609-5612 | 7 5 (b) Graph of v and i versus wt Rationalised 2023-24
Physics
184
7 |
1 | 5610-5613 | 5 (b) Graph of v and i versus wt Rationalised 2023-24
Physics
184
7 5 AC VOLTAGE APPLIED TO A CAPACITOR
Figure 7 |
1 | 5611-5614 | (b) Graph of v and i versus wt Rationalised 2023-24
Physics
184
7 5 AC VOLTAGE APPLIED TO A CAPACITOR
Figure 7 7 shows an ac source e generating ac voltage v = vm sin wt
connected to a capacitor only, a purely capacitive ac circuit |
1 | 5612-5615 | Rationalised 2023-24
Physics
184
7 5 AC VOLTAGE APPLIED TO A CAPACITOR
Figure 7 7 shows an ac source e generating ac voltage v = vm sin wt
connected to a capacitor only, a purely capacitive ac circuit When a capacitor is connected to a voltage source
in a dc circuit, current will flow for the short time
required to charge the capacitor |
1 | 5613-5616 | 5 AC VOLTAGE APPLIED TO A CAPACITOR
Figure 7 7 shows an ac source e generating ac voltage v = vm sin wt
connected to a capacitor only, a purely capacitive ac circuit When a capacitor is connected to a voltage source
in a dc circuit, current will flow for the short time
required to charge the capacitor As charge
accumulates on the capacitor plates, the voltage
across them increases, opposing the current |
1 | 5614-5617 | 7 shows an ac source e generating ac voltage v = vm sin wt
connected to a capacitor only, a purely capacitive ac circuit When a capacitor is connected to a voltage source
in a dc circuit, current will flow for the short time
required to charge the capacitor As charge
accumulates on the capacitor plates, the voltage
across them increases, opposing the current That is,
a capacitor in a dc circuit will limit or oppose the
current as it charges |
1 | 5615-5618 | When a capacitor is connected to a voltage source
in a dc circuit, current will flow for the short time
required to charge the capacitor As charge
accumulates on the capacitor plates, the voltage
across them increases, opposing the current That is,
a capacitor in a dc circuit will limit or oppose the
current as it charges When the capacitor is fully
charged, the current in the circuit falls to zero |
1 | 5616-5619 | As charge
accumulates on the capacitor plates, the voltage
across them increases, opposing the current That is,
a capacitor in a dc circuit will limit or oppose the
current as it charges When the capacitor is fully
charged, the current in the circuit falls to zero When the capacitor is connected to an ac source,
as in Fig |
1 | 5617-5620 | That is,
a capacitor in a dc circuit will limit or oppose the
current as it charges When the capacitor is fully
charged, the current in the circuit falls to zero When the capacitor is connected to an ac source,
as in Fig 7 |
1 | 5618-5621 | When the capacitor is fully
charged, the current in the circuit falls to zero When the capacitor is connected to an ac source,
as in Fig 7 7, it limits or regulates the current, but
does not completely prevent the flow of charge |
1 | 5619-5622 | When the capacitor is connected to an ac source,
as in Fig 7 7, it limits or regulates the current, but
does not completely prevent the flow of charge The
capacitor is alternately charged and discharged as
the current reverses each half cycle |
1 | 5620-5623 | 7 7, it limits or regulates the current, but
does not completely prevent the flow of charge The
capacitor is alternately charged and discharged as
the current reverses each half cycle Let q be the
charge on the capacitor at any time t |
1 | 5621-5624 | 7, it limits or regulates the current, but
does not completely prevent the flow of charge The
capacitor is alternately charged and discharged as
the current reverses each half cycle Let q be the
charge on the capacitor at any time t The instantaneous voltage v across
the capacitor is
q
v
=C
(7 |
1 | 5622-5625 | The
capacitor is alternately charged and discharged as
the current reverses each half cycle Let q be the
charge on the capacitor at any time t The instantaneous voltage v across
the capacitor is
q
v
=C
(7 15)
From the Kirchhoff’s loop rule, the voltage across the source and the
capacitor are equal,
sin
m
q
v
t
C
ω
=
To find the current, we use the relation
dd
q
i
t
=
(
)
dd
sin
cos(
)
m
m
i
v C
t
C v
t
t
ω
ω
ω
=
=
Using the relation, cos(
)
sin
ω
ω
t
t
=
+
π
2
, we have
i
i
t
=m
+
sin ω
2π
(7 |
1 | 5623-5626 | Let q be the
charge on the capacitor at any time t The instantaneous voltage v across
the capacitor is
q
v
=C
(7 15)
From the Kirchhoff’s loop rule, the voltage across the source and the
capacitor are equal,
sin
m
q
v
t
C
ω
=
To find the current, we use the relation
dd
q
i
t
=
(
)
dd
sin
cos(
)
m
m
i
v C
t
C v
t
t
ω
ω
ω
=
=
Using the relation, cos(
)
sin
ω
ω
t
t
=
+
π
2
, we have
i
i
t
=m
+
sin ω
2π
(7 16)
where the amplitude of the oscillating current is im = w Cvm |
1 | 5624-5627 | The instantaneous voltage v across
the capacitor is
q
v
=C
(7 15)
From the Kirchhoff’s loop rule, the voltage across the source and the
capacitor are equal,
sin
m
q
v
t
C
ω
=
To find the current, we use the relation
dd
q
i
t
=
(
)
dd
sin
cos(
)
m
m
i
v C
t
C v
t
t
ω
ω
ω
=
=
Using the relation, cos(
)
sin
ω
ω
t
t
=
+
π
2
, we have
i
i
t
=m
+
sin ω
2π
(7 16)
where the amplitude of the oscillating current is im = w Cvm We can rewrite
it as
(1/
)
m
m
v
i
ωC
=
Comparing it to im= vm/R for a purely resistive circuit, we find that
(1/wC) plays the role of resistance |
1 | 5625-5628 | 15)
From the Kirchhoff’s loop rule, the voltage across the source and the
capacitor are equal,
sin
m
q
v
t
C
ω
=
To find the current, we use the relation
dd
q
i
t
=
(
)
dd
sin
cos(
)
m
m
i
v C
t
C v
t
t
ω
ω
ω
=
=
Using the relation, cos(
)
sin
ω
ω
t
t
=
+
π
2
, we have
i
i
t
=m
+
sin ω
2π
(7 16)
where the amplitude of the oscillating current is im = w Cvm We can rewrite
it as
(1/
)
m
m
v
i
ωC
=
Comparing it to im= vm/R for a purely resistive circuit, we find that
(1/wC) plays the role of resistance It is called capacitive reactance and
is denoted by Xc,
Xc= 1/wC
(7 |
1 | 5626-5629 | 16)
where the amplitude of the oscillating current is im = w Cvm We can rewrite
it as
(1/
)
m
m
v
i
ωC
=
Comparing it to im= vm/R for a purely resistive circuit, we find that
(1/wC) plays the role of resistance It is called capacitive reactance and
is denoted by Xc,
Xc= 1/wC
(7 17)
so that the amplitude of the current is
m
m
C
v
i
=X
(7 |
1 | 5627-5630 | We can rewrite
it as
(1/
)
m
m
v
i
ωC
=
Comparing it to im= vm/R for a purely resistive circuit, we find that
(1/wC) plays the role of resistance It is called capacitive reactance and
is denoted by Xc,
Xc= 1/wC
(7 17)
so that the amplitude of the current is
m
m
C
v
i
=X
(7 18)
FIGURE 7 |
1 | 5628-5631 | It is called capacitive reactance and
is denoted by Xc,
Xc= 1/wC
(7 17)
so that the amplitude of the current is
m
m
C
v
i
=X
(7 18)
FIGURE 7 7 An ac source
connected to a capacitor |
1 | 5629-5632 | 17)
so that the amplitude of the current is
m
m
C
v
i
=X
(7 18)
FIGURE 7 7 An ac source
connected to a capacitor Rationalised 2023-24
185
Alternating Current
FIGURE 7 |
1 | 5630-5633 | 18)
FIGURE 7 7 An ac source
connected to a capacitor Rationalised 2023-24
185
Alternating Current
FIGURE 7 8 (a) A Phasor diagram for the circuit
in Fig |
1 | 5631-5634 | 7 An ac source
connected to a capacitor Rationalised 2023-24
185
Alternating Current
FIGURE 7 8 (a) A Phasor diagram for the circuit
in Fig 7 |
1 | 5632-5635 | Rationalised 2023-24
185
Alternating Current
FIGURE 7 8 (a) A Phasor diagram for the circuit
in Fig 7 8 |
1 | 5633-5636 | 8 (a) A Phasor diagram for the circuit
in Fig 7 8 (b) Graph of v and i versus wt |
1 | 5634-5637 | 7 8 (b) Graph of v and i versus wt The dimension of capacitive reactance is the
same as that of resistance and its SI unit is
ohm (W) |
1 | 5635-5638 | 8 (b) Graph of v and i versus wt The dimension of capacitive reactance is the
same as that of resistance and its SI unit is
ohm (W) The capacitive reactance limits the
amplitude of the current in a purely capacitive
circuit in the same way as the resistance limits
the current in a purely resistive circuit |
1 | 5636-5639 | (b) Graph of v and i versus wt The dimension of capacitive reactance is the
same as that of resistance and its SI unit is
ohm (W) The capacitive reactance limits the
amplitude of the current in a purely capacitive
circuit in the same way as the resistance limits
the current in a purely resistive circuit But it
is inversely proportional to the frequency and
the capacitance |
1 | 5637-5640 | The dimension of capacitive reactance is the
same as that of resistance and its SI unit is
ohm (W) The capacitive reactance limits the
amplitude of the current in a purely capacitive
circuit in the same way as the resistance limits
the current in a purely resistive circuit But it
is inversely proportional to the frequency and
the capacitance A comparison of Eq |
1 | 5638-5641 | The capacitive reactance limits the
amplitude of the current in a purely capacitive
circuit in the same way as the resistance limits
the current in a purely resistive circuit But it
is inversely proportional to the frequency and
the capacitance A comparison of Eq (7 |
1 | 5639-5642 | But it
is inversely proportional to the frequency and
the capacitance A comparison of Eq (7 16) with the
equation of source voltage, Eq |
1 | 5640-5643 | A comparison of Eq (7 16) with the
equation of source voltage, Eq (7 |
1 | 5641-5644 | (7 16) with the
equation of source voltage, Eq (7 1) shows that
the current is p/2 ahead of voltage |
1 | 5642-5645 | 16) with the
equation of source voltage, Eq (7 1) shows that
the current is p/2 ahead of voltage Figure 7 |
1 | 5643-5646 | (7 1) shows that
the current is p/2 ahead of voltage Figure 7 8(a) shows the phasor diagram at an instant t1 |
1 | 5644-5647 | 1) shows that
the current is p/2 ahead of voltage Figure 7 8(a) shows the phasor diagram at an instant t1 Here the current
phasor I is p/2 ahead of the voltage phasor V as they rotate
counterclockwise |
1 | 5645-5648 | Figure 7 8(a) shows the phasor diagram at an instant t1 Here the current
phasor I is p/2 ahead of the voltage phasor V as they rotate
counterclockwise Figure 7 |
1 | 5646-5649 | 8(a) shows the phasor diagram at an instant t1 Here the current
phasor I is p/2 ahead of the voltage phasor V as they rotate
counterclockwise Figure 7 8(b) shows the variation of voltage and current
with time |
1 | 5647-5650 | Here the current
phasor I is p/2 ahead of the voltage phasor V as they rotate
counterclockwise Figure 7 8(b) shows the variation of voltage and current
with time We see that the current reaches its maximum value earlier than
the voltage by one-fourth of a period |
1 | 5648-5651 | Figure 7 8(b) shows the variation of voltage and current
with time We see that the current reaches its maximum value earlier than
the voltage by one-fourth of a period The instantaneous power supplied to the capacitor is
pc = i v = im cos(wt)vm sin(wt)
= imvm cos(wt) sin(wt)
sin(2
)
2
m
i vm
ωt
=
(7 |
1 | 5649-5652 | 8(b) shows the variation of voltage and current
with time We see that the current reaches its maximum value earlier than
the voltage by one-fourth of a period The instantaneous power supplied to the capacitor is
pc = i v = im cos(wt)vm sin(wt)
= imvm cos(wt) sin(wt)
sin(2
)
2
m
i vm
ωt
=
(7 19)
So, as in the case of an inductor, the average power
sin(2
)
sin(2
)
0
2
2
m
m
m
m
C
i v
i v
P
t
t
ω
ω
=
=
=
since <sin (2wt)> = 0 over a complete cycle |
1 | 5650-5653 | We see that the current reaches its maximum value earlier than
the voltage by one-fourth of a period The instantaneous power supplied to the capacitor is
pc = i v = im cos(wt)vm sin(wt)
= imvm cos(wt) sin(wt)
sin(2
)
2
m
i vm
ωt
=
(7 19)
So, as in the case of an inductor, the average power
sin(2
)
sin(2
)
0
2
2
m
m
m
m
C
i v
i v
P
t
t
ω
ω
=
=
=
since <sin (2wt)> = 0 over a complete cycle Thus, we see that in the case of an inductor, the current lags the voltage
by p/2 and in the case of a capacitor, the current leads the voltage by p/2 |
1 | 5651-5654 | The instantaneous power supplied to the capacitor is
pc = i v = im cos(wt)vm sin(wt)
= imvm cos(wt) sin(wt)
sin(2
)
2
m
i vm
ωt
=
(7 19)
So, as in the case of an inductor, the average power
sin(2
)
sin(2
)
0
2
2
m
m
m
m
C
i v
i v
P
t
t
ω
ω
=
=
=
since <sin (2wt)> = 0 over a complete cycle Thus, we see that in the case of an inductor, the current lags the voltage
by p/2 and in the case of a capacitor, the current leads the voltage by p/2 Example 7 |
1 | 5652-5655 | 19)
So, as in the case of an inductor, the average power
sin(2
)
sin(2
)
0
2
2
m
m
m
m
C
i v
i v
P
t
t
ω
ω
=
=
=
since <sin (2wt)> = 0 over a complete cycle Thus, we see that in the case of an inductor, the current lags the voltage
by p/2 and in the case of a capacitor, the current leads the voltage by p/2 Example 7 3 A lamp is connected in series with a capacitor |
1 | 5653-5656 | Thus, we see that in the case of an inductor, the current lags the voltage
by p/2 and in the case of a capacitor, the current leads the voltage by p/2 Example 7 3 A lamp is connected in series with a capacitor Predict
your observations for dc and ac connections |
1 | 5654-5657 | Example 7 3 A lamp is connected in series with a capacitor Predict
your observations for dc and ac connections What happens in each
case if the capacitance of the capacitor is reduced |
1 | 5655-5658 | 3 A lamp is connected in series with a capacitor Predict
your observations for dc and ac connections What happens in each
case if the capacitance of the capacitor is reduced Solution When a dc source is connected to a capacitor, the capacitor
gets charged and after charging no current flows in the circuit and
the lamp will not glow |
1 | 5656-5659 | Predict
your observations for dc and ac connections What happens in each
case if the capacitance of the capacitor is reduced Solution When a dc source is connected to a capacitor, the capacitor
gets charged and after charging no current flows in the circuit and
the lamp will not glow There will be no change even if C is reduced |
1 | 5657-5660 | What happens in each
case if the capacitance of the capacitor is reduced Solution When a dc source is connected to a capacitor, the capacitor
gets charged and after charging no current flows in the circuit and
the lamp will not glow There will be no change even if C is reduced With ac source, the capacitor offers capacitative reactance (1/wC )
and the current flows in the circuit |
1 | 5658-5661 | Solution When a dc source is connected to a capacitor, the capacitor
gets charged and after charging no current flows in the circuit and
the lamp will not glow There will be no change even if C is reduced With ac source, the capacitor offers capacitative reactance (1/wC )
and the current flows in the circuit Consequently, the lamp will shine |
1 | 5659-5662 | There will be no change even if C is reduced With ac source, the capacitor offers capacitative reactance (1/wC )
and the current flows in the circuit Consequently, the lamp will shine Reducing C will increase reactance and the lamp will shine less brightly
than before |
1 | 5660-5663 | With ac source, the capacitor offers capacitative reactance (1/wC )
and the current flows in the circuit Consequently, the lamp will shine Reducing C will increase reactance and the lamp will shine less brightly
than before Example 7 |
1 | 5661-5664 | Consequently, the lamp will shine Reducing C will increase reactance and the lamp will shine less brightly
than before Example 7 4 A 15 |
1 | 5662-5665 | Reducing C will increase reactance and the lamp will shine less brightly
than before Example 7 4 A 15 0 mF capacitor is connected to a 220 V, 50 Hz source |
1 | 5663-5666 | Example 7 4 A 15 0 mF capacitor is connected to a 220 V, 50 Hz source Find the capacitive reactance and the current (rms and peak) in the
circuit |
1 | 5664-5667 | 4 A 15 0 mF capacitor is connected to a 220 V, 50 Hz source Find the capacitive reactance and the current (rms and peak) in the
circuit If the frequency is doubled, what happens to the capacitive
reactance and the current |
1 | 5665-5668 | 0 mF capacitor is connected to a 220 V, 50 Hz source Find the capacitive reactance and the current (rms and peak) in the
circuit If the frequency is doubled, what happens to the capacitive
reactance and the current Solution The capacitive reactance is
6F
1
1
212
2
2 (50Hz)(15 |
1 | 5666-5669 | Find the capacitive reactance and the current (rms and peak) in the
circuit If the frequency is doubled, what happens to the capacitive
reactance and the current Solution The capacitive reactance is
6F
1
1
212
2
2 (50Hz)(15 0
10
)
XC
νC
−
=
=
=
Ω
π
π
×
The rms current is
EXAMPLE 7 |
1 | 5667-5670 | If the frequency is doubled, what happens to the capacitive
reactance and the current Solution The capacitive reactance is
6F
1
1
212
2
2 (50Hz)(15 0
10
)
XC
νC
−
=
=
=
Ω
π
π
×
The rms current is
EXAMPLE 7 3
EXAMPLE 7 |
1 | 5668-5671 | Solution The capacitive reactance is
6F
1
1
212
2
2 (50Hz)(15 0
10
)
XC
νC
−
=
=
=
Ω
π
π
×
The rms current is
EXAMPLE 7 3
EXAMPLE 7 4
Rationalised 2023-24
Physics
186
EXAMPLE 7 |
1 | 5669-5672 | 0
10
)
XC
νC
−
=
=
=
Ω
π
π
×
The rms current is
EXAMPLE 7 3
EXAMPLE 7 4
Rationalised 2023-24
Physics
186
EXAMPLE 7 5
EXAMPLE 7 |
1 | 5670-5673 | 3
EXAMPLE 7 4
Rationalised 2023-24
Physics
186
EXAMPLE 7 5
EXAMPLE 7 4
V
A
220
1 |
1 | 5671-5674 | 4
Rationalised 2023-24
Physics
186
EXAMPLE 7 5
EXAMPLE 7 4
V
A
220
1 04
212
C
V
I
=X
=
=
Ω
The peak current is
2
(1 |
1 | 5672-5675 | 5
EXAMPLE 7 4
V
A
220
1 04
212
C
V
I
=X
=
=
Ω
The peak current is
2
(1 41)(1 |
1 | 5673-5676 | 4
V
A
220
1 04
212
C
V
I
=X
=
=
Ω
The peak current is
2
(1 41)(1 04
)
1 |
1 | 5674-5677 | 04
212
C
V
I
=X
=
=
Ω
The peak current is
2
(1 41)(1 04
)
1 47
im
I
A
A
=
=
=
This current oscillates between +1 |
1 | 5675-5678 | 41)(1 04
)
1 47
im
I
A
A
=
=
=
This current oscillates between +1 47A and –1 |
1 | 5676-5679 | 04
)
1 47
im
I
A
A
=
=
=
This current oscillates between +1 47A and –1 47 A, and is ahead of
the voltage by p/2 |
1 | 5677-5680 | 47
im
I
A
A
=
=
=
This current oscillates between +1 47A and –1 47 A, and is ahead of
the voltage by p/2 If the frequency is doubled, the capacitive reactance is halved and
consequently, the current is doubled |
1 | 5678-5681 | 47A and –1 47 A, and is ahead of
the voltage by p/2 If the frequency is doubled, the capacitive reactance is halved and
consequently, the current is doubled Example 7 |
1 | 5679-5682 | 47 A, and is ahead of
the voltage by p/2 If the frequency is doubled, the capacitive reactance is halved and
consequently, the current is doubled Example 7 5 A light bulb and an open coil inductor are connected to
an ac source through a key as shown in Fig |
1 | 5680-5683 | If the frequency is doubled, the capacitive reactance is halved and
consequently, the current is doubled Example 7 5 A light bulb and an open coil inductor are connected to
an ac source through a key as shown in Fig 7 |
1 | 5681-5684 | Example 7 5 A light bulb and an open coil inductor are connected to
an ac source through a key as shown in Fig 7 9 |
1 | 5682-5685 | 5 A light bulb and an open coil inductor are connected to
an ac source through a key as shown in Fig 7 9 FIGURE 7 |
1 | 5683-5686 | 7 9 FIGURE 7 9
The switch is closed and after sometime, an iron rod is inserted into
the interior of the inductor |
1 | 5684-5687 | 9 FIGURE 7 9
The switch is closed and after sometime, an iron rod is inserted into
the interior of the inductor The glow of the light bulb (a) increases; (b)
decreases; (c) is unchanged, as the iron rod is inserted |
1 | 5685-5688 | FIGURE 7 9
The switch is closed and after sometime, an iron rod is inserted into
the interior of the inductor The glow of the light bulb (a) increases; (b)
decreases; (c) is unchanged, as the iron rod is inserted Give your
answer with reasons |
1 | 5686-5689 | 9
The switch is closed and after sometime, an iron rod is inserted into
the interior of the inductor The glow of the light bulb (a) increases; (b)
decreases; (c) is unchanged, as the iron rod is inserted Give your
answer with reasons Solution As the iron rod is inserted, the magnetic field inside the coil
magnetizes the iron increasing the magnetic field inside it |
1 | 5687-5690 | The glow of the light bulb (a) increases; (b)
decreases; (c) is unchanged, as the iron rod is inserted Give your
answer with reasons Solution As the iron rod is inserted, the magnetic field inside the coil
magnetizes the iron increasing the magnetic field inside it Hence,
the inductance of the coil increases |
1 | 5688-5691 | Give your
answer with reasons Solution As the iron rod is inserted, the magnetic field inside the coil
magnetizes the iron increasing the magnetic field inside it Hence,
the inductance of the coil increases Consequently, the inductive
reactance of the coil increases |
1 | 5689-5692 | Solution As the iron rod is inserted, the magnetic field inside the coil
magnetizes the iron increasing the magnetic field inside it 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 |
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