DC Pandey Solutions: Alternating Current

# DC Pandey Solutions: Alternating Current | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

Introductory Exercise 25.1
1.    R
V
I
= = =
DC
100
10
10W
Z
V
I
= = =
AC
150
10
15W
X Z R
L
= -
2 2
= - ( ) ( ) 15 10
2 2
=5 5W
L
X X
f
L L
= =
w p 2
=
´ ´
5 5
2 50 3.14
»0.036 H
\ V IX
L L
= = 50 5 V
=111.8 V
2. For phase an gle to be zero,
X X
L C
=
Þ       w
w
L
C
=
1

Þ L
C f C
= =
1 1
2
2 2
w p ( )
=
´
-
1
360 10
2 6
( )
»7.7 H
As   X X
L C
=
\      Z R =
I
V
Z
= = =
120
20
6 A
In tro duc tory Ex er cise 25.2
1. Res o nat ing fre quency,
w
r
L C
= =
´ ´
-
1 1
2 1 0
6
0 . 0 3
=
1 0
6
4

f
r
r
= =
´ ´
w
p 2
10
2 6
4
3.14
» 1105 Hz
Phase angle at resonance is always 0°.
2. Re sis tance of arc lamp,
R
V
I
=
DC
= =
40
10
4W
Impedance of series combination,
Z
V
I
= = =
AC
200
10
20W
Power factor = f = = cos
R
Z
4
20
=
1
5

Alternating Current
25
Page 2

Introductory Exercise 25.1
1.    R
V
I
= = =
DC
100
10
10W
Z
V
I
= = =
AC
150
10
15W
X Z R
L
= -
2 2
= - ( ) ( ) 15 10
2 2
=5 5W
L
X X
f
L L
= =
w p 2
=
´ ´
5 5
2 50 3.14
»0.036 H
\ V IX
L L
= = 50 5 V
=111.8 V
2. For phase an gle to be zero,
X X
L C
=
Þ       w
w
L
C
=
1

Þ L
C f C
= =
1 1
2
2 2
w p ( )
=
´
-
1
360 10
2 6
( )
»7.7 H
As   X X
L C
=
\      Z R =
I
V
Z
= = =
120
20
6 A
In tro duc tory Ex er cise 25.2
1. Res o nat ing fre quency,
w
r
L C
= =
´ ´
-
1 1
2 1 0
6
0 . 0 3
=
1 0
6
4

f
r
r
= =
´ ´
w
p 2
10
2 6
4
3.14
» 1105 Hz
Phase angle at resonance is always 0°.
2. Re sis tance of arc lamp,
R
V
I
=
DC
= =
40
10
4W
Impedance of series combination,
Z
V
I
= = =
AC
200
10
20W
Power factor = f = = cos
R
Z
4
20
=
1
5

Alternating Current
25
AIEEE Corner
Sub jec ti ve Ques tions (Level-1)
1. (a) X L fL
L
= = w p 2
= ´ ´ ´ 2 50 2 3.14
=628W
(b) X L
L
=w Þ L
X
L
=
w
= =
´ ´
X
f
L
2
2
2 50 p 3.14
= 6.37 mH
(c) X
C fC
C
= =
1 1
2 w p
=
´ ´ ´ ´
-
1
2 50 2 10
6
3.14

= = 1592 1.59 k W W
(d) X
C
C
=
1
w
Þ C
X
C
=
1
w
=
´ ´ ´
=
1
2 50 2 3.14
1.59mF
2. (a) Z R X X
L C
= + -
2 2
( )
= + -
æ
è
ç
ç
ö
ø
÷
÷
R L
C
2
2
1
w
w
= + ´ -
´ ´
æ
è
ç
ç
ö
ø
÷
÷ -
( ) 300 400
1
400 8 10
2
6
2
0.25
=367.6W
I
V
Z
0
0
120
= = =
367.6
0.326 A
(b) f =
-
-
tan
1
X X
R
L C
= » - °
-
tan
1
300
212.5
35.3
As X X
C L
> voltage will lag behind
current by 35.3°.
(c) V I R
R
= = ´
0
300 0.326 = 97.8 V,
V I X
L L
= =
0
32.6V
V I X
C C
= = ´
0
0.326 312.5
= » 101.875 V 120 V
3. (a) Power fac tor at res o nance is al ways 1,
as Z R = , Power factor = f = = cos
R
Z
1.
(b) P
I E E
R
=
f
=
0 0 0
2
2 2
cos
=
´
=
( ) 150
3 150
75
2
W
(c) Because resonance is still maintained,
average power consumed will remain
same, i.e., 75 W.
4. (a) As voltage is lag behind current,
inductor should be added to the circuit
to raise the power factor.
(b) Power factor = f = cos
R
Z
Þ Z
R
=
f
=
cos
60
0.720
=
250
3
W
X Z R
C
= -
2 2
=
æ
è
ç
ö
ø
÷
-
250
3
60
2
2
( )
=58W
C
X
C
=
1
w
=
1
2pf X
C
=
´ ´ ´
1
2 50 58 3.14
=54mF
For resonance,
w
r
LC
=
1

Þ    L
C
r
=
1
2
w
=
1
2
2
( ) pf C
Þ      L =
´ ´ ´ ´
-
1
2 50 54 10
2 6
( ) 3.14
=0.185 H
131
Page 3

Introductory Exercise 25.1
1.    R
V
I
= = =
DC
100
10
10W
Z
V
I
= = =
AC
150
10
15W
X Z R
L
= -
2 2
= - ( ) ( ) 15 10
2 2
=5 5W
L
X X
f
L L
= =
w p 2
=
´ ´
5 5
2 50 3.14
»0.036 H
\ V IX
L L
= = 50 5 V
=111.8 V
2. For phase an gle to be zero,
X X
L C
=
Þ       w
w
L
C
=
1

Þ L
C f C
= =
1 1
2
2 2
w p ( )
=
´
-
1
360 10
2 6
( )
»7.7 H
As   X X
L C
=
\      Z R =
I
V
Z
= = =
120
20
6 A
In tro duc tory Ex er cise 25.2
1. Res o nat ing fre quency,
w
r
L C
= =
´ ´
-
1 1
2 1 0
6
0 . 0 3
=
1 0
6
4

f
r
r
= =
´ ´
w
p 2
10
2 6
4
3.14
» 1105 Hz
Phase angle at resonance is always 0°.
2. Re sis tance of arc lamp,
R
V
I
=
DC
= =
40
10
4W
Impedance of series combination,
Z
V
I
= = =
AC
200
10
20W
Power factor = f = = cos
R
Z
4
20
=
1
5

Alternating Current
25
AIEEE Corner
Sub jec ti ve Ques tions (Level-1)
1. (a) X L fL
L
= = w p 2
= ´ ´ ´ 2 50 2 3.14
=628W
(b) X L
L
=w Þ L
X
L
=
w
= =
´ ´
X
f
L
2
2
2 50 p 3.14
= 6.37 mH
(c) X
C fC
C
= =
1 1
2 w p
=
´ ´ ´ ´
-
1
2 50 2 10
6
3.14

= = 1592 1.59 k W W
(d) X
C
C
=
1
w
Þ C
X
C
=
1
w
=
´ ´ ´
=
1
2 50 2 3.14
1.59mF
2. (a) Z R X X
L C
= + -
2 2
( )
= + -
æ
è
ç
ç
ö
ø
÷
÷
R L
C
2
2
1
w
w
= + ´ -
´ ´
æ
è
ç
ç
ö
ø
÷
÷ -
( ) 300 400
1
400 8 10
2
6
2
0.25
=367.6W
I
V
Z
0
0
120
= = =
367.6
0.326 A
(b) f =
-
-
tan
1
X X
R
L C
= » - °
-
tan
1
300
212.5
35.3
As X X
C L
> voltage will lag behind
current by 35.3°.
(c) V I R
R
= = ´
0
300 0.326 = 97.8 V,
V I X
L L
= =
0
32.6V
V I X
C C
= = ´
0
0.326 312.5
= » 101.875 V 120 V
3. (a) Power fac tor at res o nance is al ways 1,
as Z R = , Power factor = f = = cos
R
Z
1.
(b) P
I E E
R
=
f
=
0 0 0
2
2 2
cos
=
´
=
( ) 150
3 150
75
2
W
(c) Because resonance is still maintained,
average power consumed will remain
same, i.e., 75 W.
4. (a) As voltage is lag behind current,
inductor should be added to the circuit
to raise the power factor.
(b) Power factor = f = cos
R
Z
Þ Z
R
=
f
=
cos
60
0.720
=
250
3
W
X Z R
C
= -
2 2
=
æ
è
ç
ö
ø
÷
-
250
3
60
2
2
( )
=58W
C
X
C
=
1
w
=
1
2pf X
C
=
´ ´ ´
1
2 50 58 3.14
=54mF
For resonance,
w
r
LC
=
1

Þ    L
C
r
=
1
2
w
=
1
2
2
( ) pf C
Þ      L =
´ ´ ´ ´
-
1
2 50 54 10
2 6
( ) 3.14
=0.185 H
131
5. V t t ( ) sin ( / ) = + 170 6280 3 p volt
i t t ( ) sin ( / ) = + 8.5 6280 2 p amp.
(b) f = =
´
=
w
p 2
6280
2
1000
3.14
Hz
=1 kHz
(c) f = - =
p p p
2 3 6
Þ cos cos f = =
p
6
3
2
As phase of i is greater than V, current is
(d)Clearly the circuit is capacitive in
nature, we have
cos f =
R
Z
Þ
3
2
=
R
Z
Þ Z R =
2
3
Also,  Z
V
i
= = =
0
0
170
20
8.5
W
R Z = =
3
2
10 3 W
Again, Z R X
C
= +
2 2
Þ X Z R
C
= -
2 2
= - = 400 300 10W
X
C X
C
C
= Þ =
´
1 1 1
6280 10 w w
=1592 . mF
6. I
V
X
V
L
L
= =
w
\ I =
´
=
60
100 5
0.12 A
\ I =
´
= ´
-
60
1000 5
10
2
1.2 A
\ I =
´
= ´
-
60
10000 5
10
3
1.2 A
7. V t
R
= ( )cos [( ) ] 2.5 V rad/s 950
(a) I
V
R
R
=
=
( ) cos [( 2.5 V rad/s) ] 950
300
t
=( cos[( ) ] 8. mA) rad/s 33 950 t
(b) X L
L
= = ´ w 950 0.800
=760W
(c) V I X t
L L
= +
0
2 cos( / ) w p
Þ  V I X t
L L
= -
0
sin w
= - 6.33 rad / s sin [( ) ] 950 t V
8. Given, L = 0.120 H, R =240W, C =7.30 F m ,
I
rms
.450 A =0 , f = 400 Hz
X L fL
L
= = w p 2
= ´ ´ ´ 2 400 3.14 0.120
=301.44W
X
C f C
C
= =
1 1
2 w p
=
´ ´ ´ ´
-
1
2 400 10
6
3.14 7.3
=54.43W
(a) cos
( )
f= =
+ -
R
Z
R
R X X
L C
2 2

=
+ -
240
240
2 2
( ) ( ) 301.44 54.43
=0.697
f= » °
-
cos ( . ) .
1
0697 458
(b) Z R X X
L C
= + -
2 2
( )
= + - ( ) ( ) 240
2 2
301.44 54.43 =344 W
(c) V I Z
rms rms
0.450 = = ´ 344
=154.8 V »155 V
(d) P V I
av
= f
rms rms
cos
= ´ ´ = 155 0.450 0.697 48.6W
(e) P I R
R
= = ´
rms
0.450)
2 2
240 ( = 48.6 W
(f) and (g) Average power associated with
inductor and capacitor is always zero.
132
170V
V
–170V
O
0.25 0.50 0.75 1.00 1.25 t (ms)
i
O
1
—
12
1
—
3
7
—
12
5
—
6
13
—
12
4
—
3
t (ms)
Page 4

Introductory Exercise 25.1
1.    R
V
I
= = =
DC
100
10
10W
Z
V
I
= = =
AC
150
10
15W
X Z R
L
= -
2 2
= - ( ) ( ) 15 10
2 2
=5 5W
L
X X
f
L L
= =
w p 2
=
´ ´
5 5
2 50 3.14
»0.036 H
\ V IX
L L
= = 50 5 V
=111.8 V
2. For phase an gle to be zero,
X X
L C
=
Þ       w
w
L
C
=
1

Þ L
C f C
= =
1 1
2
2 2
w p ( )
=
´
-
1
360 10
2 6
( )
»7.7 H
As   X X
L C
=
\      Z R =
I
V
Z
= = =
120
20
6 A
In tro duc tory Ex er cise 25.2
1. Res o nat ing fre quency,
w
r
L C
= =
´ ´
-
1 1
2 1 0
6
0 . 0 3
=
1 0
6
4

f
r
r
= =
´ ´
w
p 2
10
2 6
4
3.14
» 1105 Hz
Phase angle at resonance is always 0°.
2. Re sis tance of arc lamp,
R
V
I
=
DC
= =
40
10
4W
Impedance of series combination,
Z
V
I
= = =
AC
200
10
20W
Power factor = f = = cos
R
Z
4
20
=
1
5

Alternating Current
25
AIEEE Corner
Sub jec ti ve Ques tions (Level-1)
1. (a) X L fL
L
= = w p 2
= ´ ´ ´ 2 50 2 3.14
=628W
(b) X L
L
=w Þ L
X
L
=
w
= =
´ ´
X
f
L
2
2
2 50 p 3.14
= 6.37 mH
(c) X
C fC
C
= =
1 1
2 w p
=
´ ´ ´ ´
-
1
2 50 2 10
6
3.14

= = 1592 1.59 k W W
(d) X
C
C
=
1
w
Þ C
X
C
=
1
w
=
´ ´ ´
=
1
2 50 2 3.14
1.59mF
2. (a) Z R X X
L C
= + -
2 2
( )
= + -
æ
è
ç
ç
ö
ø
÷
÷
R L
C
2
2
1
w
w
= + ´ -
´ ´
æ
è
ç
ç
ö
ø
÷
÷ -
( ) 300 400
1
400 8 10
2
6
2
0.25
=367.6W
I
V
Z
0
0
120
= = =
367.6
0.326 A
(b) f =
-
-
tan
1
X X
R
L C
= » - °
-
tan
1
300
212.5
35.3
As X X
C L
> voltage will lag behind
current by 35.3°.
(c) V I R
R
= = ´
0
300 0.326 = 97.8 V,
V I X
L L
= =
0
32.6V
V I X
C C
= = ´
0
0.326 312.5
= » 101.875 V 120 V
3. (a) Power fac tor at res o nance is al ways 1,
as Z R = , Power factor = f = = cos
R
Z
1.
(b) P
I E E
R
=
f
=
0 0 0
2
2 2
cos
=
´
=
( ) 150
3 150
75
2
W
(c) Because resonance is still maintained,
average power consumed will remain
same, i.e., 75 W.
4. (a) As voltage is lag behind current,
inductor should be added to the circuit
to raise the power factor.
(b) Power factor = f = cos
R
Z
Þ Z
R
=
f
=
cos
60
0.720
=
250
3
W
X Z R
C
= -
2 2
=
æ
è
ç
ö
ø
÷
-
250
3
60
2
2
( )
=58W
C
X
C
=
1
w
=
1
2pf X
C
=
´ ´ ´
1
2 50 58 3.14
=54mF
For resonance,
w
r
LC
=
1

Þ    L
C
r
=
1
2
w
=
1
2
2
( ) pf C
Þ      L =
´ ´ ´ ´
-
1
2 50 54 10
2 6
( ) 3.14
=0.185 H
131
5. V t t ( ) sin ( / ) = + 170 6280 3 p volt
i t t ( ) sin ( / ) = + 8.5 6280 2 p amp.
(b) f = =
´
=
w
p 2
6280
2
1000
3.14
Hz
=1 kHz
(c) f = - =
p p p
2 3 6
Þ cos cos f = =
p
6
3
2
As phase of i is greater than V, current is
(d)Clearly the circuit is capacitive in
nature, we have
cos f =
R
Z
Þ
3
2
=
R
Z
Þ Z R =
2
3
Also,  Z
V
i
= = =
0
0
170
20
8.5
W
R Z = =
3
2
10 3 W
Again, Z R X
C
= +
2 2
Þ X Z R
C
= -
2 2
= - = 400 300 10W
X
C X
C
C
= Þ =
´
1 1 1
6280 10 w w
=1592 . mF
6. I
V
X
V
L
L
= =
w
\ I =
´
=
60
100 5
0.12 A
\ I =
´
= ´
-
60
1000 5
10
2
1.2 A
\ I =
´
= ´
-
60
10000 5
10
3
1.2 A
7. V t
R
= ( )cos [( ) ] 2.5 V rad/s 950
(a) I
V
R
R
=
=
( ) cos [( 2.5 V rad/s) ] 950
300
t
=( cos[( ) ] 8. mA) rad/s 33 950 t
(b) X L
L
= = ´ w 950 0.800
=760W
(c) V I X t
L L
= +
0
2 cos( / ) w p
Þ  V I X t
L L
= -
0
sin w
= - 6.33 rad / s sin [( ) ] 950 t V
8. Given, L = 0.120 H, R =240W, C =7.30 F m ,
I
rms
.450 A =0 , f = 400 Hz
X L fL
L
= = w p 2
= ´ ´ ´ 2 400 3.14 0.120
=301.44W
X
C f C
C
= =
1 1
2 w p
=
´ ´ ´ ´
-
1
2 400 10
6
3.14 7.3
=54.43W
(a) cos
( )
f= =
+ -
R
Z
R
R X X
L C
2 2

=
+ -
240
240
2 2
( ) ( ) 301.44 54.43
=0.697
f= » °
-
cos ( . ) .
1
0697 458
(b) Z R X X
L C
= + -
2 2
( )
= + - ( ) ( ) 240
2 2
301.44 54.43 =344 W
(c) V I Z
rms rms
0.450 = = ´ 344
=154.8 V »155 V
(d) P V I
av
= f
rms rms
cos
= ´ ´ = 155 0.450 0.697 48.6W
(e) P I R
R
= = ´
rms
0.450)
2 2
240 ( = 48.6 W
(f) and (g) Average power associated with
inductor and capacitor is always zero.
132
170V
V
–170V
O
0.25 0.50 0.75 1.00 1.25 t (ms)
i
O
1
—
12
1
—
3
7
—
12
5
—
6
13
—
12
4
—
3
t (ms)
Objective Questions (Level-1)
1. In an AC cir cuit, cosf is called power
fac tor.
2. DC am me ter mea sures charge flow ing in
the cir cuit per unit time, hence it
mea sures av er age value of cur rent, but
av er age value of AC over a long time is
zero.
3.     Z R X X
L C
= + -
2 2
( )
= + -
æ
è
ç
ç
ö
ø
÷
÷
R L
C
2
2
1
w
w
Hence, for X X
L C
< , Z decreases with
increase in frequency and for X X
L C
> , Z
increases with increase in frequency.
4. As volt age leads cur rent and f <
p
2
, hence
ei ther cir cuit con tain s in du c tance and
re sis tance or con tain s in du c tance,
ca pac i tan ce and re sis tance with X X
L C
> .
5. RMS value of sine wave AC is 0.707I
0
,
but can be dif fer ent for dif fer ent types of
AC’s.
6. P I E
v v
= f= cos 0
7. Z R X X
L C
= + -
2 2
( )
8. P
V I
=
0 0
2
[V
0
and I
0
are peak volt age and
cur rent through re sis tor only]
9. V
V
rms
= =
0
2
170 V
f = =
´
w
p 2
120
2 3.14
»19 Hz
10. Cur r ent is max i mum at
w w = =
r
L C
1
=
´ ´
-
1
8 10
6
0.5
11. P
I E
=
f
0 0
2
cos
=
´
´ =
-
100 100
2 3
10
3
cos
p
2.5W
12. X
C
C
= =¥
1
w
if w = 0, i.e., for DC
13. V t = 10 100 cos p
at t =
1
600
s,
V =10 100
1
600
cos p
= = ´ = 10
6
10
3
2
5 3 cos
p
V
14. For purely re sis tive cir cuit f = 0.
15. X
C
C
=
1
w
Þ X
C
µ
1
w
or X
f
C
µ
1

16.    sinf= =
X
Z
1
3

Þ      f=
é
ë
ê
ù
û
ú
-
sin
1
1
3
17. f= = f=
3
2 2
0
0 0
p
, cos P
I E
18. R
V
I
= =
DC
DC
100W
Z
V
I
= = =
AC
AC
0.5
100
200W
X Z R
L
= - =
2 2
100 3 W
L
X X
f
L L
= = =
´ w p p 2
100 3
2 50
=
æ
è
ç
ç
ö
ø
÷
÷
3
p
H
19. I
V
X
CV
C
rms
rms
rms
= = w
= ´ ´ ´
-
100 1 10
200 2
2
6
I
rms
mA =20
20. V V V
R L
= + = +
2 2 2 2
20 15 ( ) ( )
=25 V, V
0
25 2 = V
21. P
I V
=
f
=
0 0
2
0
cos
Þ cos f =0
Þ    f= ° 90
22. R is in de pend ent of fre quency.
23. L is very high so that cir cuit con sumes
less power.
133
Page 5

Introductory Exercise 25.1
1.    R
V
I
= = =
DC
100
10
10W
Z
V
I
= = =
AC
150
10
15W
X Z R
L
= -
2 2
= - ( ) ( ) 15 10
2 2
=5 5W
L
X X
f
L L
= =
w p 2
=
´ ´
5 5
2 50 3.14
»0.036 H
\ V IX
L L
= = 50 5 V
=111.8 V
2. For phase an gle to be zero,
X X
L C
=
Þ       w
w
L
C
=
1

Þ L
C f C
= =
1 1
2
2 2
w p ( )
=
´
-
1
360 10
2 6
( )
»7.7 H
As   X X
L C
=
\      Z R =
I
V
Z
= = =
120
20
6 A
In tro duc tory Ex er cise 25.2
1. Res o nat ing fre quency,
w
r
L C
= =
´ ´
-
1 1
2 1 0
6
0 . 0 3
=
1 0
6
4

f
r
r
= =
´ ´
w
p 2
10
2 6
4
3.14
» 1105 Hz
Phase angle at resonance is always 0°.
2. Re sis tance of arc lamp,
R
V
I
=
DC
= =
40
10
4W
Impedance of series combination,
Z
V
I
= = =
AC
200
10
20W
Power factor = f = = cos
R
Z
4
20
=
1
5

Alternating Current
25
AIEEE Corner
Sub jec ti ve Ques tions (Level-1)
1. (a) X L fL
L
= = w p 2
= ´ ´ ´ 2 50 2 3.14
=628W
(b) X L
L
=w Þ L
X
L
=
w
= =
´ ´
X
f
L
2
2
2 50 p 3.14
= 6.37 mH
(c) X
C fC
C
= =
1 1
2 w p
=
´ ´ ´ ´
-
1
2 50 2 10
6
3.14

= = 1592 1.59 k W W
(d) X
C
C
=
1
w
Þ C
X
C
=
1
w
=
´ ´ ´
=
1
2 50 2 3.14
1.59mF
2. (a) Z R X X
L C
= + -
2 2
( )
= + -
æ
è
ç
ç
ö
ø
÷
÷
R L
C
2
2
1
w
w
= + ´ -
´ ´
æ
è
ç
ç
ö
ø
÷
÷ -
( ) 300 400
1
400 8 10
2
6
2
0.25
=367.6W
I
V
Z
0
0
120
= = =
367.6
0.326 A
(b) f =
-
-
tan
1
X X
R
L C
= » - °
-
tan
1
300
212.5
35.3
As X X
C L
> voltage will lag behind
current by 35.3°.
(c) V I R
R
= = ´
0
300 0.326 = 97.8 V,
V I X
L L
= =
0
32.6V
V I X
C C
= = ´
0
0.326 312.5
= » 101.875 V 120 V
3. (a) Power fac tor at res o nance is al ways 1,
as Z R = , Power factor = f = = cos
R
Z
1.
(b) P
I E E
R
=
f
=
0 0 0
2
2 2
cos
=
´
=
( ) 150
3 150
75
2
W
(c) Because resonance is still maintained,
average power consumed will remain
same, i.e., 75 W.
4. (a) As voltage is lag behind current,
inductor should be added to the circuit
to raise the power factor.
(b) Power factor = f = cos
R
Z
Þ Z
R
=
f
=
cos
60
0.720
=
250
3
W
X Z R
C
= -
2 2
=
æ
è
ç
ö
ø
÷
-
250
3
60
2
2
( )
=58W
C
X
C
=
1
w
=
1
2pf X
C
=
´ ´ ´
1
2 50 58 3.14
=54mF
For resonance,
w
r
LC
=
1

Þ    L
C
r
=
1
2
w
=
1
2
2
( ) pf C
Þ      L =
´ ´ ´ ´
-
1
2 50 54 10
2 6
( ) 3.14
=0.185 H
131
5. V t t ( ) sin ( / ) = + 170 6280 3 p volt
i t t ( ) sin ( / ) = + 8.5 6280 2 p amp.
(b) f = =
´
=
w
p 2
6280
2
1000
3.14
Hz
=1 kHz
(c) f = - =
p p p
2 3 6
Þ cos cos f = =
p
6
3
2
As phase of i is greater than V, current is
(d)Clearly the circuit is capacitive in
nature, we have
cos f =
R
Z
Þ
3
2
=
R
Z
Þ Z R =
2
3
Also,  Z
V
i
= = =
0
0
170
20
8.5
W
R Z = =
3
2
10 3 W
Again, Z R X
C
= +
2 2
Þ X Z R
C
= -
2 2
= - = 400 300 10W
X
C X
C
C
= Þ =
´
1 1 1
6280 10 w w
=1592 . mF
6. I
V
X
V
L
L
= =
w
\ I =
´
=
60
100 5
0.12 A
\ I =
´
= ´
-
60
1000 5
10
2
1.2 A
\ I =
´
= ´
-
60
10000 5
10
3
1.2 A
7. V t
R
= ( )cos [( ) ] 2.5 V rad/s 950
(a) I
V
R
R
=
=
( ) cos [( 2.5 V rad/s) ] 950
300
t
=( cos[( ) ] 8. mA) rad/s 33 950 t
(b) X L
L
= = ´ w 950 0.800
=760W
(c) V I X t
L L
= +
0
2 cos( / ) w p
Þ  V I X t
L L
= -
0
sin w
= - 6.33 rad / s sin [( ) ] 950 t V
8. Given, L = 0.120 H, R =240W, C =7.30 F m ,
I
rms
.450 A =0 , f = 400 Hz
X L fL
L
= = w p 2
= ´ ´ ´ 2 400 3.14 0.120
=301.44W
X
C f C
C
= =
1 1
2 w p
=
´ ´ ´ ´
-
1
2 400 10
6
3.14 7.3
=54.43W
(a) cos
( )
f= =
+ -
R
Z
R
R X X
L C
2 2

=
+ -
240
240
2 2
( ) ( ) 301.44 54.43
=0.697
f= » °
-
cos ( . ) .
1
0697 458
(b) Z R X X
L C
= + -
2 2
( )
= + - ( ) ( ) 240
2 2
301.44 54.43 =344 W
(c) V I Z
rms rms
0.450 = = ´ 344
=154.8 V »155 V
(d) P V I
av
= f
rms rms
cos
= ´ ´ = 155 0.450 0.697 48.6W
(e) P I R
R
= = ´
rms
0.450)
2 2
240 ( = 48.6 W
(f) and (g) Average power associated with
inductor and capacitor is always zero.
132
170V
V
–170V
O
0.25 0.50 0.75 1.00 1.25 t (ms)
i
O
1
—
12
1
—
3
7
—
12
5
—
6
13
—
12
4
—
3
t (ms)
Objective Questions (Level-1)
1. In an AC cir cuit, cosf is called power
fac tor.
2. DC am me ter mea sures charge flow ing in
the cir cuit per unit time, hence it
mea sures av er age value of cur rent, but
av er age value of AC over a long time is
zero.
3.     Z R X X
L C
= + -
2 2
( )
= + -
æ
è
ç
ç
ö
ø
÷
÷
R L
C
2
2
1
w
w
Hence, for X X
L C
< , Z decreases with
increase in frequency and for X X
L C
> , Z
increases with increase in frequency.
4. As volt age leads cur rent and f <
p
2
, hence
ei ther cir cuit con tain s in du c tance and
re sis tance or con tain s in du c tance,
ca pac i tan ce and re sis tance with X X
L C
> .
5. RMS value of sine wave AC is 0.707I
0
,
but can be dif fer ent for dif fer ent types of
AC’s.
6. P I E
v v
= f= cos 0
7. Z R X X
L C
= + -
2 2
( )
8. P
V I
=
0 0
2
[V
0
and I
0
are peak volt age and
cur rent through re sis tor only]
9. V
V
rms
= =
0
2
170 V
f = =
´
w
p 2
120
2 3.14
»19 Hz
10. Cur r ent is max i mum at
w w = =
r
L C
1
=
´ ´
-
1
8 10
6
0.5
11. P
I E
=
f
0 0
2
cos
=
´
´ =
-
100 100
2 3
10
3
cos
p
2.5W
12. X
C
C
= =¥
1
w
if w = 0, i.e., for DC
13. V t = 10 100 cos p
at t =
1
600
s,
V =10 100
1
600
cos p
= = ´ = 10
6
10
3
2
5 3 cos
p
V
14. For purely re sis tive cir cuit f = 0.
15. X
C
C
=
1
w
Þ X
C
µ
1
w
or X
f
C
µ
1

16.    sinf= =
X
Z
1
3

Þ      f=
é
ë
ê
ù
û
ú
-
sin
1
1
3
17. f= = f=
3
2 2
0
0 0
p
, cos P
I E
18. R
V
I
= =
DC
DC
100W
Z
V
I
= = =
AC
AC
0.5
100
200W
X Z R
L
= - =
2 2
100 3 W
L
X X
f
L L
= = =
´ w p p 2
100 3
2 50
=
æ
è
ç
ç
ö
ø
÷
÷
3
p
H
19. I
V
X
CV
C
rms
rms
rms
= = w
= ´ ´ ´
-
100 1 10
200 2
2
6
I
rms
mA =20
20. V V V
R L
= + = +
2 2 2 2
20 15 ( ) ( )
=25 V, V
0
25 2 = V
21. P
I V
=
f
=
0 0
2
0
cos
Þ cos f =0
Þ    f= ° 90
22. R is in de pend ent of fre quency.
23. L is very high so that cir cuit con sumes
less power.
133
24. tan f =
X
R
L
Þ tan 45
100
° =
X
L
Þ       X
L
=100W
wL =100W
L =
´ ´
»
100
2 10
16
3
3.14
mH
25. The min i mum time taken by it in reach ing
from zero to peak value =
T
4
= =
´
= =
1
4
1
4 50
1
200
5
f
ms
26. f = ° 60
P
I V
=
f
=
´ ´
0 0
2
4 220
1
2
2
cos
=220 W
JEE Corner
Assertion and Reasons
1. X
C
and X
L
can be greater than Z be cause
Z R X X
L C
= + -
2 2
( )
Hence, V IX
C C
= and V IX
L L
= can be
greater than V IZ = .
2. At res o nan ce X X
L C
= , with fur ther
in crease in fre quency, X
L
in creases but
X
C
de creases hence volt age will lead
cur rent.
3. f
L C
r
=
1
2p
, if di elec tric slab is in serted
be tween the plates of the ca pac i tor, its
ca pac i tan ce will in crease, hence, f
r
will
decrease.
4. q= Area under graph
= ´ ´ + + ´ ´ +
1
2
4 2 3
1
2
4 2 4 ( ) ( )
=22C
Average current = = =
q
t
22
6
3.6 A
5. On in sert in g fer ro mag n etic rod in sid e the
in duc t or, X
L
and hence V
L
in crease. Due
to this cur rent will in crease if it is lag ging
and vice-versa.
6. V V V
R L C
= = Þ R X X
L C
= =
Hence, f =0 and I is maximum.
as Z R X X
L C
= + -
2 2
( ) is minimum.
7. I I I
L C
= - = 0
8. P I R = = ´ =
rms
W
2 2
2 10 20 ( )
9. In duc tor coil re sists vary in g cur rent.
10. I
E
R L
0
0
2 2 2
=
+ w
, f =
-
tan
1
w L
R

11. At res o nance, cur rent and volt age are in
same phase and I
V
R
0
0
= . Hence, I
0
de pends on R .
Objective Questions (Level-2)
Sin gle Cor rect Op tions
1. For par al l el cir cuit
f =
é
ë
ê
ù
û
ú
-
t a n
/
/
1
1
1
X
R
L
=
-
t a n
1
4
3
= ° 53
2. Cur rent will re main same in se ries cir cuit
given by
I I t = -f
0
sin( ) w
= -
æ
è
ç
ö
ø
÷
-
I t
X
R
L
0
1
sin tan w
3. R R R
L
= + =
1
10 W
X L
L
= = w 10 W,
X
C
C
= =
1
10
w
W
134
```

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## DC Pandey Solutions for NEET Physics

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