Alternating Current- 1 Practice Questions - DPP for NEET

 Page 1


1. ( b) The coil has inductance L besides the resistance 
R
.
Hence for ac it’s impedance resistance 
22
L
RX + will
be larger than it’s impedance
R
for dc.
2. (b)
0
. ..
4
22
22
r ms
i
i === ampere e
3. (d) The current takes sec
4
T
to reach the peak value.
In the given question 
21
200 sec
100
T
T
p
= pÞ=
\
 Time to reach the peak value 
1
sec
400
=
4. (b) 50 c/s or Hz
5. (b)
00
2
cos cos
p
= w=
t
EE tE
T
2 501
10cos 10cos 5 3
600 6
p´´p
= == volt.
6. (c)
22
2 2 1/2 12
12
1
()
2 2
rms
ii
i ii
+
= =+
7. (c) Hot wire ammeter reads rms  value of current. Hence
its peak value 2
rms
i =´ = 14.14 amp
8. (b)
0
120
84.8
1.414 2
rms
V
VV = ==
9. (d) Peak value to r. m. s. value means, current becomes
1
2
 times.
If peak is at t = 0, current is of the form
0 00
1
cos100 cos100
2
iit iit = pÞ´=p
3
1
cos cos100 sec 2.5 10 sec.
4 400
tt
-
p
Þ = pÞ= =´
10. (b)
22
,
LL
Z R XXL = + =w and 2 f w=p
2 2 22
4 Z R fL \ = +p
11. (b) The applied voltage is given by 
22
RL
V VV =+
22
(200) (150) 250 V= += volt
12. (b)
2 22 2 22
120
0.016
100 4 60 20
V
iA
RL
= ==
+w +p´´
13. (a) The voltage across a L-R combination is given by
2 22
RL
V VV =+
22
400 144 256 16
LR
V VV = - = - == volt.
14. (b) Reading of ammeter,
rmso
rms
c
V VC
i
X 2
w
==
= 
6
200 2 100 (1 10 )
2
-
´ ´´
= 2 × 10
–2
 A = 20 m A
15. (a)
~
R
C
V
R
V
R
Let the applied voltage be V , volt.
Here,V
R
 = 12 V , V
C
 = 5V
V = 
22
RC
VV + = 
( ) ()
22
125 + = 13 V . V.
16. (c) 2 60 0.7
L
ZX = = p´´
120 120
0.455
2 60 0.7
\===
p´´
i
Z
ampere e
17. (d) Current will be max at first time when
100 pt + p/3 = p/2 Þ 100 pt = p/6 Þ t = 1/600 s.
18. (d)
22 22
Z R X R (2 fL) p =+ =+
2
2
0.4
(30) 2 50 900 1600 50
æö
= + p´ ´ = + =W
ç÷
èøp
V 200
i 4 ampere
Z 50
= ==
1 9. (d) In purely inductive circuit voltage leads the current by 
90
o
.
20. (a) Current through the bulb i = 
P 60
6A
V 10
==
60 W, 10 W
i
10 V
V
L
100 V , 50 Hz 
i
L
Page 2


1. ( b) The coil has inductance L besides the resistance 
R
.
Hence for ac it’s impedance resistance 
22
L
RX + will
be larger than it’s impedance
R
for dc.
2. (b)
0
. ..
4
22
22
r ms
i
i === ampere e
3. (d) The current takes sec
4
T
to reach the peak value.
In the given question 
21
200 sec
100
T
T
p
= pÞ=
\
 Time to reach the peak value 
1
sec
400
=
4. (b) 50 c/s or Hz
5. (b)
00
2
cos cos
p
= w=
t
EE tE
T
2 501
10cos 10cos 5 3
600 6
p´´p
= == volt.
6. (c)
22
2 2 1/2 12
12
1
()
2 2
rms
ii
i ii
+
= =+
7. (c) Hot wire ammeter reads rms  value of current. Hence
its peak value 2
rms
i =´ = 14.14 amp
8. (b)
0
120
84.8
1.414 2
rms
V
VV = ==
9. (d) Peak value to r. m. s. value means, current becomes
1
2
 times.
If peak is at t = 0, current is of the form
0 00
1
cos100 cos100
2
iit iit = pÞ´=p
3
1
cos cos100 sec 2.5 10 sec.
4 400
tt
-
p
Þ = pÞ= =´
10. (b)
22
,
LL
Z R XXL = + =w and 2 f w=p
2 2 22
4 Z R fL \ = +p
11. (b) The applied voltage is given by 
22
RL
V VV =+
22
(200) (150) 250 V= += volt
12. (b)
2 22 2 22
120
0.016
100 4 60 20
V
iA
RL
= ==
+w +p´´
13. (a) The voltage across a L-R combination is given by
2 22
RL
V VV =+
22
400 144 256 16
LR
V VV = - = - == volt.
14. (b) Reading of ammeter,
rmso
rms
c
V VC
i
X 2
w
==
= 
6
200 2 100 (1 10 )
2
-
´ ´´
= 2 × 10
–2
 A = 20 m A
15. (a)
~
R
C
V
R
V
R
Let the applied voltage be V , volt.
Here,V
R
 = 12 V , V
C
 = 5V
V = 
22
RC
VV + = 
( ) ()
22
125 + = 13 V . V.
16. (c) 2 60 0.7
L
ZX = = p´´
120 120
0.455
2 60 0.7
\===
p´´
i
Z
ampere e
17. (d) Current will be max at first time when
100 pt + p/3 = p/2 Þ 100 pt = p/6 Þ t = 1/600 s.
18. (d)
22 22
Z R X R (2 fL) p =+ =+
2
2
0.4
(30) 2 50 900 1600 50
æö
= + p´ ´ = + =W
ç÷
èøp
V 200
i 4 ampere
Z 50
= ==
1 9. (d) In purely inductive circuit voltage leads the current by 
90
o
.
20. (a) Current through the bulb i = 
P 60
6A
V 10
==
60 W, 10 W
i
10 V
V
L
100 V , 50 Hz 
i
L
DPP/ P 46
128
V = 
22
RL
VV +
(100)
2
 = (10)
2
 + 
2
L
V Þ VV
L
 = 99.5 V olt
AlsoV
L
 = iX
L
 = i ´ (2pvL)
Þ 99.5 = 6 ´ 2 ´ 3.14 ´ 50 ´ L = Þ L = 0.052 H.
21. (a) Phase angle 
L 2 200 14
tan
R 300 3
w p´
f= = ´=
p
1
4
tan
3
-
\ f=
22. (a) The root mean square voltage is effective voltage.
23. (c) E = 141 sin (628t),
0
141
100 and 2 628
1.41 2
rms
E
E Vf = = = p=
100 f Hz Þ=
24. (d) Time taken by the current to reach the maximum value
3
11
5 10 sec
4 4 4 50
T
t
v
-
= = = =´
´
and 
0
2 102
rms
ii== = 14.14 amp
25. (a) As in case of ac,
( )
0
sin = w -f VVt
The peak value 
0
220 2 311V V==
and as in case of ac,
0
; 220V
2
rms rms
V
VV ==
26. (b) In case of ac,
0
2 2 622
311
av
VVV = = ´=
p pp
27. (a) As 
314
2 ,2 314i.e., 50Hz
2
fff w=p p= ==
´p
28. (d) When ac flows through an inductor current lags behind
the emf., by phase of p/2, inductive reactance,
.2 .,
L
X L fL = w =p so when frequency increases
correspondingly inductive reactance also increases.
29. (c) Like direct current, an alternating current also produces
magnetic field. But the magnitude and direction of the
field goes on changing continuously with time.
30. (b) We can use a capacitor of suitable capacitance as a
choke coil, because average power consumed per cycle
in an ideal capacitor is zero. Therefore, like a choke
coil, a condenser can reduce ac without power
dissipation.