Alternating Current- 2 Practice Questions - DPP for NEET

 Page 1


1. (a) For resonant frequency to remain same
LC should be const.  LC = const
Þ LC = L' × 2C  '
2
L
L Þ=
2. (b) At resonance, 
LCR
circuit behaves as purely resistive
circuit, for purely resistive circuit power factor = 1
3. (a) If the current is wattless than power is zero. Hence
phase difference f = 90°
4. (c) 46
L
V = volts, 40
C
V = volts, 8
R
V = volts
E.M.F . of source 
22
8 (46 40) 10 V= + -=
volts
5. (c) Resonant frequency 
1
2 LC
=
p
does not depend on
resistance.
6. (a) At resonance 
LCR
series circuit behaves as pure
resistive circuit. For resistive circuit 0 f=
o
7. (b)
2 22
cos
RR
Z
RL
f==
+w
2 2 22
12
cos 0.30
(12) 4 (60) (0.1)
= Þ f=
+´p´´
8. (a)
11
2
ff
LCC
= Þµ
p
9. (b) In non resonant circuits
impedance 
2
2
1
,
11
Z
C
L
R
=
æö
+ w-
ç÷
w
èø
with rise in
frequency 
Z
decreases i.e. current increases so circuit
behaves as capacitive circuit.
10. (c)
101
cos 60
202
R
Z
f= = = Þf=
o
11. (d)
12. (a)
1
2
2
tan tan 45
CL
fL
XX fC
RR
-p
- p
f= Þ=
o
1
2(2)
C
f fLR
Þ=
p p+
13. (b) Resonance frequency
36
11
2500
8 10 20 10
LC
--
w===
´ ´´
rad/sec
Resonance current 
220
5
44
V
A
R
= ==
14. (c)
3
... ...
100 10
cos cos
3 22
r ms r ms
PVi
-
p
= ´ ´ f= ´´
2 31
10 10 1 10
0.025
2 24
--
´
= ´== watt
15. (b) R = X
L
 = 2X
C
22
LC
Z R (X X) = +-
= 
22
C CC
(2X) (2X X) +-
= 
22
CC
4XX +
f
Z
R
X –X
C L
C
5R
5X
2
==
LC CC
C
X X 2XX
tan
R 2X
--
f==
1
tan
2
f=  ;  
1
1
tan
2
-
æö
f=
ç÷
èø
16. (d) Phase angle 90 f=
o
, so power cos0
rms rms
P VI = f=
17. (b)
2 2
(30)
90
10
rms
V
PW
R
= ==
18. (c) At :
CL
AXX >
At :
CL
BXX =
At :
CL
CXX <
19. (d) The instantaneous values of emf and current in
inductive circuit are given by 
0
sin E E at = and
0
sin
2
ii
p æö
= w-
ç÷
èø
respectively . .
So, 00
sin sin
2
inst
P EiE tit
p æö
= = w ´ w-
ç÷
èø
00
sin cos Ei tt = ww
00
1
sin2
2
Eit =w
(sin 2 2sin cos ) t tt w= ww
Hence, angular frequency of instantaneous power is
2w
.
Page 2


1. (a) For resonant frequency to remain same
LC should be const.  LC = const
Þ LC = L' × 2C  '
2
L
L Þ=
2. (b) At resonance, 
LCR
circuit behaves as purely resistive
circuit, for purely resistive circuit power factor = 1
3. (a) If the current is wattless than power is zero. Hence
phase difference f = 90°
4. (c) 46
L
V = volts, 40
C
V = volts, 8
R
V = volts
E.M.F . of source 
22
8 (46 40) 10 V= + -=
volts
5. (c) Resonant frequency 
1
2 LC
=
p
does not depend on
resistance.
6. (a) At resonance 
LCR
series circuit behaves as pure
resistive circuit. For resistive circuit 0 f=
o
7. (b)
2 22
cos
RR
Z
RL
f==
+w
2 2 22
12
cos 0.30
(12) 4 (60) (0.1)
= Þ f=
+´p´´
8. (a)
11
2
ff
LCC
= Þµ
p
9. (b) In non resonant circuits
impedance 
2
2
1
,
11
Z
C
L
R
=
æö
+ w-
ç÷
w
èø
with rise in
frequency 
Z
decreases i.e. current increases so circuit
behaves as capacitive circuit.
10. (c)
101
cos 60
202
R
Z
f= = = Þf=
o
11. (d)
12. (a)
1
2
2
tan tan 45
CL
fL
XX fC
RR
-p
- p
f= Þ=
o
1
2(2)
C
f fLR
Þ=
p p+
13. (b) Resonance frequency
36
11
2500
8 10 20 10
LC
--
w===
´ ´´
rad/sec
Resonance current 
220
5
44
V
A
R
= ==
14. (c)
3
... ...
100 10
cos cos
3 22
r ms r ms
PVi
-
p
= ´ ´ f= ´´
2 31
10 10 1 10
0.025
2 24
--
´
= ´== watt
15. (b) R = X
L
 = 2X
C
22
LC
Z R (X X) = +-
= 
22
C CC
(2X) (2X X) +-
= 
22
CC
4XX +
f
Z
R
X –X
C L
C
5R
5X
2
==
LC CC
C
X X 2XX
tan
R 2X
--
f==
1
tan
2
f=  ;  
1
1
tan
2
-
æö
f=
ç÷
èø
16. (d) Phase angle 90 f=
o
, so power cos0
rms rms
P VI = f=
17. (b)
2 2
(30)
90
10
rms
V
PW
R
= ==
18. (c) At :
CL
AXX >
At :
CL
BXX =
At :
CL
CXX <
19. (d) The instantaneous values of emf and current in
inductive circuit are given by 
0
sin E E at = and
0
sin
2
ii
p æö
= w-
ç÷
èø
respectively . .
So, 00
sin sin
2
inst
P EiE tit
p æö
= = w ´ w-
ç÷
èø
00
sin cos Ei tt = ww
00
1
sin2
2
Eit =w
(sin 2 2sin cos ) t tt w= ww
Hence, angular frequency of instantaneous power is
2w
.
DPP/ P 47
130
20. (d) The voltage 
L
V and 
C
V are equal and opposite so
voltmeter reading will be zero.
Also 30 , 25
LC
R XX =W = =W
So 
22
240
8
30
()
LC
VV
iA
R
R XX
= = ==
+-
21. (d) Since quality factor,  
1L
Q
RC
=
22. (d)
23. (a)
24. (c) Reactance Z = 
( )
2
2
LC
X XR -+
= 
( )
2 2
80 50 40 -+
= 50 W
Power factor = cos f = 
R
Z
 = 
40
50
= 0.8
I
rms
 = 
rms
V
Z
 = 
200V
50W
 = 4 AA
Power current = I
rms
.cos f = 4 × 0.8
= 3.2 A
 Wattless current = I
rms
.sin f = 4 × 0.6
= 2.4 A
Sol. 25-27
25. (d)    
DV
L
DV
C
DV
R
L C max
( V V) D +D =
CL
V V 7.4 2.6 4.8 volt D -D =-=
26. (c)
22
m R max C L max
E (V ) (V V ) = D + D -D
       = 
22
(8.8) (4.8) + = 10 volt
27. (a) If f ­ then (DV
L
)
max
 ­
28. (a) Capacitive reactance XC = 
1
C w
. When capacitance
(C) increase, the capacitive reactance decreases. Due
to decrease in its values, the current in the circuit will
increase 
22
C
E
I=
RX
æö
ç÷
ç÷
+
èø
 and hence brightness of
source (or electric lamp) will also increase.
29. ( b) The phase angle for the LCR circuit is given by
1
tan
LC
L
XX
C
RR
w-
-
w
f==
where X
L
, X
C
 are inductive reactance and capacitive
reactance respectively when X
L
 > X
C
 then tan f is
positive i.e. f is positive (between 0 and p/2). Hence
emf leads the current.
30. (a) If resistor is used in controlling ac supply, electrical
energy will be wasted in the form of heat energy across
the resistance wire. However, ac supply can be
controlled with choke without any wastage of energy .
This is because, power factor (cos f) for resistance is
unity and is zero for an inductance. [P = EI cos f].