Loading [MathJax]/jax/output/CommonHTML/jax.js
MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 8 - MCQExams.com
CBSE
Class 12 Medical Physics
Alternating Current
Quiz 8
A resistor R, inductor L and a capacitor C are connected in series to an oscillator of frequency v. If the resonant frequency is
v
r
, then the current lags behind the voltage, when:
Report Question
0%
v
=
0
0%
v
<
v
r
0%
v
>
v
r
0%
v
=
v
r
Explanation
The current will lag behind the voltage when reactance of inductor is more than the reactance of capacitor i.e.,
ω
L
>
1
ω
c
or
ω
>
1
√
L
C
o
r
2
π
v
>
1
√
L
C
v
>
1
2
π
√
L
C
o
r
v
>
v
r
where
v
r
is the resonant frequency.
In L-C-R circuit,
f
=
50
π
H
z
,
V
=
50
V
,
R
=
300
Ω
. If
L
=
1
H
and
C
=
20
μ
C
, then the voltage across capacitor is :
Report Question
0%
50
V
0%
20
V
0%
Zero
0%
30
V
Explanation
For an L-C-R circuit, the impedance
(
Z
)
is given by
∵
Z
=
√
R
2
+
(
X
L
−
X
C
)
2
where,
X
L
=
ω
L
=
2
π
f
1
and
X
C
=
1
ω
C
=
1
2
π
f
C
Given,
f
=
50
π
H
z
,
R
=
300
Ω
and
L
=
1
H
and
C
=
20
μ
C
=
20
×
10
−
6
C
Z
=
√
(
300
)
2
+
(
2
π
×
50
π
×
1
−
1
2
π
×
50
π
×
20
×
10
−
6
)
Z
=
√
90000
+
(
100
−
500
)
2
⇒
Z
=
√
90000
+
160000
=
√
250000
⇒
Z
=
500
Ω
Hence, the current in the circuit is given by
i
=
V
Z
=
50
500
=
0.1
A
Voltage across capacitor is
V
C
=
i
X
C
=
i
2
π
f
C
=
0.1
2
π
×
50
π
×
20
×
10
−
4
=
0.1
×
10
6
100
×
20
⇒
V
C
=
50
V
A coil of inductive reactance
1
/
√
3
Ω
and resistance
1
Ω
is connected to
200
V
,
50
H
z
A.C. supply. The time lag between maximum voltage and current is
Report Question
0%
1
600
s
0%
1
200
s
0%
1
300
s
0%
1
500
s
Explanation
ω
=
2
π
f
=
314
r
a
d
/
s
t
a
n
ϕ
=
1
1
/
√
3
→
ϕ
=
π
6
ω
t
=
π
6
t
=
1
600
s
An inductive coil has a resistance of 100 When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by
45
o
. The inductance of the coil is
Report Question
0%
1
10
π
0%
1
20
π
0%
1
40
π
0%
1
60
π
Explanation
Given :
f
=
1000
Hz
Resistance of inductive coil
X
L
=
100
Ω
Thus inductance of coil
L
=
X
L
2
π
f
∴
L
=
100
2
π
×
1000
=
1
20
π
Henry
An AC source is connected in parallel with an L-C-R circuit as shown. Let
I
S
,
I
L
,
I
C
and
I
R
denote the currents through and
V
S
,
V
L
,
V
C
and
V
R
voltages across the corresponding components. Then :
Report Question
0%
I
S
=
I
L
+
I
C
+
I
R
0%
V
S
=
V
L
+
V
C
+
V
R
0%
(
I
L
,
I
C
,
I
R
)
<
I
S
0%
I
L
,
I
C
may be greater than
I
S
Explanation
For parallel RLC circuit,
I
S
=
√
I
2
R
+
(
I
L
−
I
C
)
2
In case of resonance,
I
L
=
I
C
So,
I
S
=
I
R
Here
I
L
and
I
C
can be much greater than
I
R
Option D is correct.
At inductance
1
H
is connected in series with an
A
C
source of
220
V
and
50
H
z
. The inductive resistance (in ohm) is :
Report Question
0%
2
π
0%
50
π
0%
100
π
0%
1000
π
Explanation
Inductive reactants
X
L
=
w
L
=
2
π
v
L
=
2
π
×
50
×
1
=
100
π
.
In an LCR circuit having L
=
8
H, C
=
0.5
μ
F
and
R
=
100
Ω
in series, the resonance frequency rad/s is?
Report Question
0%
600
0%
200
0%
250
/
π
0%
500
Explanation
From the formula,
ω
=
1
√
L
C
Given,
L
=
8
H,
C
=
0.5
μ
F
=
0.5
×
10
−
6
F
∴
ω
=
1
√
8
×
0.5
×
10
−
6
⇒
ω
=
1
2
×
10
−
3
=
500
rad/s
In an
L
−
C
−
R
series circuit, the values of
R
,
X
L
and
X
C
are
120
Ω
,
180
Ω
and
130
Ω
, what is the impedance of the circuit?
Report Question
0%
120
Ω
0%
130
Ω
0%
180
Ω
0%
330
Ω
Explanation
Given,
R
=
120
Ω
X
L
=
180
Ω
and
X
C
=
130
Ω
The impedance of
L
−
C
−
R
circuit
Z
=
√
R
2
+
(
X
L
−
X
C
)
2
Z
=
√
(
120
)
2
+
(
180
−
130
)
2
Z
=
130
Ω
.
A series resonant circuit contains
L
=
5
π
m
H
,
C
=
200
π
μ
F
and
R
=
100
μ
. If a source of emf
e
=
200
s
i
n
1000
π
t
is applied, then the rms current is:
Report Question
0%
2
A
0%
200
√
2
A
0%
100
√
2
A
0%
1.41
A
Explanation
Source emf is given as
e
=
200
s
i
n
1000
π
t
Comparing it with
e
=
e
o
sin
w
t
We get
ω
=
1000
π
⇒
f
=
ω
2
π
=
500
H
z
Also,
e
0
=
200
V
At resonance
Z
=
R
So, current flowing
i
0
=
e
o
R
=
200
100
=
2
A
Rms value of current
i
r
m
s
=
i
0
√
2
=
2
√
2
=
√
2
=
1.41
A
The alternating voltage and current in an electric circuit are respectively given by
E
=
100
sin
100
π
t
,
I
=
5
sin
100
π
t
. The reactance of the circuit will be :
Report Question
0%
1
Ω
0%
0.05
Ω
0%
20
Ω
0%
Zero
Explanation
The general equations of alternating voltage and current in an AC circuit are as follows
E
=
E
0
sin
ω
t
....(i)
I
=
I
0
sin
ω
t
....(ii)
Given that,
E
=
100
sin
100
π
t
....(iii)
I
=
5
sin
100
π
t
....(iv)
Comparing equations (ii) and (iv), we get,
Peak value of current,
I
0
=
5
A
Comparing equations (i) and (iii), we get
Peak value of voltage,
E
0
=
100
V
Now, r.m.s. value of voltage is given by
E
r
m
s
=
E
0
√
2
=
100
√
2
V
Similarly, r.m.s. value of current is given by
I
r
m
s
=
I
0
√
2
=
5
√
2
A
Hence, reactance of the circuit is given by
Z
=
E
r
m
s
I
r
m
s
=
100
√
2
5
√
2
=
20
Ω
In non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency?
Report Question
0%
R
e
s
i
s
t
i
v
e
0%
C
a
p
a
c
i
t
i
v
e
0%
I
n
d
u
c
t
i
v
e
0%
N
o
n
e
o
f
t
h
e
s
e
Explanation
At resonant frequency
X
L
=
X
C
(
ω
L
=
1
ω
C
)
At frequencies higher than resonance frequencies
X
L
>
X
C
i.e., behaviour is inductive.
In a circuit,
L
,
C
and
R
are connected in series with an alternating voltage source of frequency
f
. The current leads the voltage by
45
o
. The value of
C
is :
Report Question
0%
1
π
f
(
2
π
f
L
+
R
)
0%
1
π
f
(
2
π
f
L
−
R
)
0%
1
2
π
f
(
2
π
f
L
+
R
)
0%
1
2
π
f
(
2
π
f
L
−
R
)
Explanation
The phase differenec
ϕ
between current and voltage is given by
tan
ϕ
=
X
L
−
X
C
R
tan
45
o
=
1
X
C
=
X
L
−
R
1
2
π
f
C
=
2
π
f
L
−
R
C
=
1
2
π
f
(
2
π
f
L
−
R
)
A
220
V main supply is connected to a resistance of
100
k
Ω
. The effective current is?
Report Question
0%
2.2
mA
0%
2.2
√
2
mA
0%
2.2
√
2
mA
0%
None of these
Explanation
Effective current is the rms value. Here,
220
V is the labelled value of AC which is also the rms value. Hence,
I
r
m
s
=
E
r
m
s
R
I
r
m
s
=
220
100
×
10
3
I
r
m
s
=
2.2
m
A
.
Which one of the following curves represents variation of current i with frequency f in series LCR circuit?
Report Question
0%
0%
0%
0%
Explanation
In series LCR circuit, the graph(c) represents the curve between current (i) and frequency(f).
For the given circuit, the natural frequency is given by :
Report Question
0%
1
2
π
√
L
C
0%
1
2
π
√
L
C
0%
√
L
C
0%
√
C
L
Explanation
Natural frequency
(
ω
)
is given by
f
=
1
2
π
⋅
1
√
L
C
For the given circuit
C
n
e
t
=
C
2
(in series)
L
n
e
t
=
2
L
(in series)
f
=
1
2
π
⋅
1
√
L
C
A 44mH inductor is connected to a 220V, 50Hz AC supply. Then, the inductive reactance of the inductor is :
Report Question
0%
3.82
Ω
0%
10.8
H
z
0%
13.82
Ω
0%
21.5
H
z
Explanation
Inductive reactance
X
L
is given by
X
L
=
ω
L
=
2
π
f
L
Given, f=50Hz, L=44mH
=
44
×
10
−
3
H
X
L
=
2
π
×
44
×
10
−
3
×
50
=
13.82
Ω
In the given circuit
R
in pure resistance and
X
is unknown circuit element. An AC voltage source is applied across
A
and
C
. If
V
A
B
=
V
A
C
, then
X
is
Report Question
0%
Pure resistance
0%
Pure inductance
0%
Combination of inductance and capacitance at resonance
0%
None of the above
Explanation
Since the voltage across the resistive element is same as the voltage applied, the voltage drop across
B
C
is zero. This is possible only when the ohmic value of the element connected across
B
C
is zero. So,
X
should be combination of inductance and capacitance at resonance.
The frequency of the output signal becomes ________ times by doubling the value of the capacitance in the LC oscillator circuit.
Report Question
0%
√
2
0%
1
√
2
0%
1
2
0%
2
Explanation
In the L-C circuits
For the first condition
f
1
=
1
2
π
√
1
L
C
.....(i)
For the second condition, when
C
=
doubling
f
2
=
1
2
π
√
1
2
L
C
....(ii)
From the equation (i), we get
f
2
=
1
2
π
√
1
L
C
×
1
√
2
⇒
f
2
=
f
1
×
1
√
2
So, the frequency of the output signal becomes
1
/
√
2
times by doubling the value of the capacitance in the L-C oscillator circuit.
In a series resonant circuit, the AC voltage across resistance R, inductor L and capacitor C are 5 V, 10 V and 10 V respectively. The AC voltage applied to the circuit will be
Report Question
0%
10 V
0%
25 V
0%
5 V
0%
20 V
Explanation
Given,
V
R
=
5
V
,
V
L
=
10
and
V
C
=
10
V
In the MR circuit, the AC voltage applied to the circuit will be
V
=
√
V
2
R
+
(
V
L
−
V
C
)
2
=
√
(
5
)
2
+
(
10
−
10
)
2
=
5
V
The power loss in an AC circuit can be minimized by.
Report Question
0%
Decreasing resistance and increasing inductance
0%
Decreasing inductance and increasimg resistance
0%
Increasing both inductance and resistance
0%
Decreasing both inductance and resistance
Explanation
In an AC circuit, power loss can be minimized by decreasing in resistance and by increasing in inductance.
The diagram given show the variation of voltage and current in an AC circuit. The circuit contains
Report Question
0%
Only a resistor
0%
Only a pure inductor
0%
Only a capcacitor
0%
A capacitor and and inductor
Explanation
The given circuit shows that the current lags the applied voltage. This is possible if the circuit has inductive element. So, the circuit contain a pure inductor.
What is the range of the characteristic impedance of a coaxial cable?
Report Question
0%
Between
150
Ω
to
600
Ω
0%
Between
50
Ω
to
70
Ω
0%
Between
0
Ω
to
50
Ω
0%
Between
100
Ω
to
150
Ω
Explanation
Characteristic impedance of a coaxial cable is between
50
Ω
to
70
Ω
.
In the given circuit the potential difference across resistance is
54
V and power consumed by it is
16
W. If AC frequency is
60
Hz find the value of L.
Report Question
0%
1
H
0%
2
H
0%
5
H
0%
4
H
Consider the
R
−
L
−
C
circuit given below. The circuit is driven by a
50
H
z
A
C
source with peak voltage
220
V
. If
R
=
400
Ω
,
C
=
200
μ
F
and
L
=
6
H
, the maximum current in the circuit is closest to
Report Question
0%
0.120
A
0%
0.55
A
0%
1.2
A
0%
5.5
A
Explanation
At maximum current
X
c
=
X
L
so Z = R
so
I
=
v
R
so I = 0.55 A
Consider two series resonant circuits with components
L
1
C
1
and
L
2
C
2
with same resonant frequency ,
ω
.When connected in series, the resonant frequency of the combination is
Report Question
0%
2
ω
0%
ω
2
0%
3
ω
0%
ω
Explanation
Initially,
ω
=
1
√
L
1
C
1
and
ω
=
1
√
L
2
C
2
Finally,
L
e
q
=
L
1
+
L
2
C
e
q
=
C
1
C
2
C
1
+
C
2
ω
′
=
1
√
L
e
q
C
e
q
from above equation, we will get
ω
′
=
ω
In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be
Report Question
0%
0 V, 8 A
0%
150 V, A 8
0%
150 V, 3 A
0%
0 V, 3 A
Explanation
Solution:-
Given:-
X
C
=
X
L
, V=240V, R=30
Ω
Since X_L=X_C, So,
Z=
√
R
2
+
(
X
L
−
X
C
)
2
=R
So, potential drop across C
&
L is zero and current is
→
i=
V
Z
=
V
R
=
240
30
=
8
A
i=8A
A
5
c
m
long solenoid having
10
ohm resistance and
5
m
H
inductance is joined to a
10
V
battery. At steady state, the current through the solenoid (in ampere) will be
Report Question
0%
5
0%
2
0%
1
0%
zero
Explanation
At steady state inductor behave like short circuit.
i
=
v
r
=
10
10
=
1
A
Consider
L
,
C
,
R
circuit as shown in figure, with a.c. source of peak value
V
and angular frequency
ω
. Then the peak value of current through the ac source.
Report Question
0%
V
√
1
R
2
+
(
ω
L
−
1
ω
C
)
2
0%
V
√
1
R
2
+
(
ω
C
−
1
ω
L
)
2
0%
V
√
R
2
+
(
ω
L
−
1
ω
C
)
2
0%
V
R
ω
C
√
ω
2
C
2
+
R
(
ω
2
C
2
−
1
)
2
Explanation
X
c
=
1
j
ω
c
X
L
=
j
ω
L
And
R
are in parallel,
Therefore, equivalent admittance of the circuit is
Y
=
1
X
c
+
1
X
L
+
1
R
.
Y
=
1
R
+
j
(
c
ω
−
1
L
ω
)
|
Y
|
=
√
1
R
2
+
(
ω
c
−
1
ω
L
)
2
Current
I
=
V
Y
=
V
√
1
R
2
+
(
ω
c
−
1
ω
L
)
2
Statement 1 : An inductance and a resistance are connected in series with an ac circuit. In this circuit the current and the potential difference across the resistance lags behind potential difference across the inductance by an angle of
π
/
2
.
Statement 2 : In a LR circuit voltage leads the current by phase angle which depends on the value of inductance and resistance both.
Report Question
0%
Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.
0%
Both statements 1 and 2 are true but statement 2 is not the correct explanation of statement 1.
0%
Statement 1 is true but statement 2 is false.
0%
Both statements 1 and 2 are false.
In A.C circuit having only capacitor, the current _____
Report Question
0%
lags behind the voltage by
π
2
in phase
0%
leads the voltage by
π
2
in phase
0%
leads the voltage by
π
in phase
0%
lags behind the voltage by
π
in phase
Explanation
Current in the capacitive A.C circuit leads the voltage by
π
2
in phase.
Correct answer is option B.
In the case of an inductor
Report Question
0%
voltage lags the current by
π
2
.
0%
voltage leads the current by
π
2
.
0%
voltage leads the current by
π
3
.
0%
voltage leads the current by
π
4
.
Explanation
In the purely inductive circuit,
Input voltage,
V
=
V
m
s
i
n
(
ω
t
)
. . . . . . .(1)
The induced emf across the inductor
e
=
−
L
d
I
d
t
The induced emf in the circuit is equal and opposite to the applied voltage
e
=
−
V
V
m
s
i
n
ω
t
=
L
d
I
d
t
d
I
=
V
m
L
s
i
n
ω
t
I
=
∫
d
I
=
V
m
ω
L
(
−
c
o
s
ω
t
)
I
=
V
m
X
L
s
i
n
(
ω
t
−
π
/
2
)
I
=
I
m
s
i
n
(
ω
t
−
π
/
2
)
. . . . . .(2)
where,
I
m
=
V
m
X
L
X
L
=
ω
L
(inductive reactants)
From equation (1) and (2)
Voltage leads the current by
90
0
.
The correct option is B.
Two coils have mutual inductance
0.005
H
. The current changes in the first coil according to equation
I
=
I
0
sin
ω
t
, where
I
0
=
10
A
and
ω
=
100
π
r
a
d
s
−
1
. The maximum value of emf in the second coil is
Report Question
0%
5
0%
5
π
0%
0.5
π
0%
π
Explanation
Current is given as
I
=
I
o
sin
w
t
=
10
sin
(
100
π
t
)
Rate of change of current
d
I
d
t
=
I
o
w
cos
w
t
=
1000
π
cos
(
100
π
t
)
Thus
d
I
d
t
|
m
a
x
=
1000
π
Given :
L
=
0.005
H
Maximum emf
E
m
a
x
=
L
d
I
d
t
|
m
a
x
=
0.005
×
1000
π
=
5
π
Which of the following graphs represents the correct variation of inductive reactane
X
L
with frequency
υ
?
Report Question
0%
0%
0%
0%
Explanation
Inductive reactance ,
X
L
=
ω
L
=
2
π
ν
L
X
L
=
(
2
π
L
)
×
ν
The above equation can be compared to the equation of a straight line.
y
=
m
×
x
Hence, inductive reactance increases linearly with frequency
An ideal inductor is in turn put across
220
V
,
50
H
z
and
220
V
,
100
H
z
supplies. The current flowing through it in the two cases will be then
Report Question
0%
equal
0%
different
0%
zero
0%
infinite
Explanation
The current in the inductor coil is given by
I
=
V
X
L
=
V
2
π
υ
L
Since frequency
υ
in the two cases is different, hence the current in two cases will be different.
The reciprocal of the impedance of an electric current is?
Report Question
0%
Admittance
0%
Susceptance
0%
Conductance
0%
Reactance
Explanation
As conductance is the complement of resistance, there is also a complementary expression of reactance or impedance, called susceptance
A capacitor 'C' is connected across a D.C. source, the reactance of capacitor will be _________.
Report Question
0%
ZERO
0%
HIGH
0%
LOW
0%
INFINITE
Explanation
Frequency of a source voltage
f
=
0
Reactance of capacitance
X
C
=
1
2
π
f
C
⟹
X
C
=
∞
Correct answer is option D.
A 5
μ
F capacitor is connected to a 200 V, 100 Hz ac source. The capacitive reactance is:
Report Question
0%
212
Ω
0%
312
Ω
0%
318
Ω
0%
412
Ω
Explanation
Given:
The capacitance of the capacitor is
5
μ
F
.
The voltage supplied is
200
V
.
The frequency of the voltage supplied is
100
H
z
.
To find:
The capacitive reactance of the capacitor.
The reactance of the capacitor is given by:
X
C
=
1
ω
C
=
1
2
π
f
C
⇒
1
2
π
×
100
×
5
=
318
Ω
Which of the following graphs represents the correct variation of capacitive reactance
X
C
with frequency
ν
?
Report Question
0%
0%
0%
0%
Explanation
Capacitive reactance ,
X
c
=
1
ω
C
=
1
2
π
ν
C
⇒
X
C
×
ν
=
c
o
n
s
t
a
n
t
With an increase in frequency,
X
C
decreases
Hence, the option (c) represents the correct graph.
An inductor of 30 mH is connected to a 220 V, 100 Hz ac source. The inductive reactance is then
Report Question
0%
10.58
Ω
0%
12.64
Ω
0%
18.85
Ω
0%
22.67
Ω
Explanation
Hence ,
L
=
30
m
H
=
30
×
10
−
3
H
V
r
m
s
=
220
,
υ
=
100
H
z
Inductive reactance,
X
L
=
2
π
υ
L
=
2
×
3.14
×
100
×
30
×
10
−
3
=
18.85
Ω
In a series LCR circuit, the plot of
I
m
vs
ω
is shown in the figure. The bandwidth of this plot will be then
Report Question
0%
Zero
0%
0.1
r
a
d
s
−
1
0%
0.2
r
a
d
s
−
1
0%
0.4
r
a
d
s
−
1
Explanation
Bandwidth is the frequency range at which the current amplitude is
1
√
2
times the magnitude of the current.
I
m
=
1
√
2
I
m
a
x
=
0.7
I
m
a
x
.
From figure,
Band width at this value is :
Δ
ω
=
1.2
−
0.8
=
0.4
r
a
d
s
−
1
In the series LCR circuit shown the impedance is:
Report Question
0%
200
Ω
0%
100
Ω
0%
300
Ω
0%
500
Ω
Explanation
Hence ,
L
=
1
H
,
C
=
20
μ
F
=
20
×
10
−
6
F
R
=
300
Ω
,
υ
=
50
π
H
z
The inductive reactance is
X
L
=
2
π
υ
L
=
2
×
π
×
50
π
×
1
=
100
Ω
The capacitive reactance is
X
C
=
1
2
π
υ
C
=
1
2
×
π
×
50
π
×
20
×
10
−
6
=
500
Ω
The impedance of the series LCR cicuit is
Z
=
√
R
2
+
(
X
C
−
X
L
)
2
=
√
(
300
)
2
+
(
500
−
100
)
2
Z
=
√
(
300
)
2
+
(
400
)
2
=
500
Ω
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
Report Question
0%
Only resistor
0%
Resistor and inductor
0%
Resistor and capacitor
0%
Only a capacitor
Explanation
We know,
Resistance (reactance) of the inductor and capacitor depends upon the frequency of the current.
X
c
=
1
2
π
f
C
X
l
=
2
π
f
L
I
R
M
S
=
V
R
M
S
√
R
2
+
(
X
L
−
X
C
)
2
The RMS current for a resistor will not be affected because its impedance is independent of frequency whereas in the case of a resistor and inductor also, the current will decrease with an increase in frequency.
whereas the increase in frequency in the case of the capacitor leads to an increase in the RMS current because in this the impedance decreases with an increase in frequency.
Hence Option
C
is correct answer.
Phase difference between voltage and current in a capacitor in an ac circuit is then
Report Question
0%
π
0%
π
/
2
0%
0
0%
π
/
3
Explanation
In a capacitive ac circuits , the voltage lags
behind the current in phase by
π
/
2
radian.
A circuit containing a 20
Ω
resistor and 0.1
μ
F
capacitor in series is connected to 230 V AC supply of angular frequency 100 rad
s
−
1
. The impedance of the circuit is
Report Question
0%
10
5
Ω
0%
10
4
Ω
0%
10
6
Ω
0%
10
10
Ω
Explanation
Hence,
R
=
20
Ω
,
C
=
0.1
μ
F
=
0.1
×
10
−
6
F
=
10
−
7
F
Impedance ,
Z
=
√
R
2
+
1
ω
2
C
2
√
20
2
+
1
(
100
)
2
×
(
10
−
7
)
2
=
√
400
+
10
10
=
10
5
Ω
A sinusoidal voltage of peak value 293 V and frequency 50 Hz is applied to a series LCR circuit in which R = 6
Ω
, L = 25 mH and C = 750
μ
F. The impedance of the circuit is:
Report Question
0%
7.0
Ω
0%
8.9
Ω
0%
9.9
Ω
0%
10.0
Ω
Explanation
Hence ,
R
=
6
Ω
,
L
=
25
m
H
=
25
×
10
−
3
H
C
=
750
μ
F
=
750
×
10
−
6
F
,
υ
=
50
H
z
X
L
=
2
π
υ
L
=
2
×
3.14
×
50
×
25
×
10
−
3
=
7.85
Ω
X
C
=
1
2
π
υ
C
=
1
2
×
3.14
×
50
×
750
×
10
−
6
=
4.25
Ω
∴
X
L
−
X
C
=
7.85
−
4.25
=
3.6
Ω
Impedence of the series LCR circuit is
Z
=
√
R
2
+
(
X
L
−
X
C
)
2
∴
Z
=
√
(
6
)
2
+
(
3.6
)
2
=
√
36
+
12.96
=
7.0
Ω
The resonant frequency of a series LCR circuit with L = 2.0H,C=32
μ
F and R= 10
Ω
is
Report Question
0%
20
H
z
0%
30
H
z
0%
40
H
z
0%
50
H
z
Explanation
Here,
L
=
2
H
C
=
32
μ
F
=
32
×
10
−
6
F
R
=
10
Ω
∴
ω
=
1
√
L
C
=
1
√
2
×
32
×
10
−
6
=
125
r
a
d
s
−
1
ν
r
=
ω
r
2
π
=
125
2
×
3.14
=
20
H
z
A 0.2
Ω
resistor and 15
μ
F capacitor are connected in series to a 220 V, 50 Hz ac source. The impedance of the circuit is
Report Question
0%
250
Ω
0%
268
Ω
0%
29.15
Ω
0%
291.5
Ω
Explanation
Here,
R
=
0.2
k
Ω
=
200
Ω
C
=
15
μ
F
=
15
×
10
−
6
F
ν
=
50
H
z
Capacitive reactance,
X
C
=
1
2
π
ν
C
=
1
2
×
3.14
×
50
×
15
×
10
−
6
=
212
Ω
The impedance of the RC circuit is
Z
=
√
R
2
+
X
2
C
=
√
(
200
)
2
+
(
212
)
2
=
291.5
Ω
A coil of inductance
L
=
300
m
H
and resistance
R
=
140
m
Ω
is connected to a constant voltage source, Current in the coil will reach to
50
of its steady state value after
t
is nearly equal to:
Report Question
0%
15
s
0%
0.75
s
0%
0.15
s
0%
1.5
s
Explanation
V
2
R
=
V
R
(
1
−
e
−
R
t
/
L
)
ℓ
−
R
t
L
=
1
2
R
L
t
=
ℓ
n
2
t
=
300
140
×
0.7
10
=
1.5
In a series resonant R-L-C circuit, if L is increased by
25
%
and C is decreased by
20
%
, then the resonant frequency will
Report Question
0%
Increases by
10
%
0%
Decreases by
10
%
0%
Remain unchanged
0%
Increases by
2.5
%
A series resonant LCR circuit has a quality factor (Q-factor) = 0.If R 2 k
Ω
, C = 0.1
μ
F, then the value of Inductance is
Report Question
0%
0.1 H
0%
0.064 H
0%
2 H
0%
5 H
Explanation
Quality factor
Q
=
1
R
√
L
C
o
r
L
C
=
(
Q
R
)
2
Here,
Q
=
0.4
,
R
=
2
K
Ω
=
2
×
10
3
ω
C
=
0.1
μ
F
=
0.1
×
10
−
6
F
∴
L
=
(
Q
R
)
2
C
∴
L
=
(
0.4
×
2
×
10
3
)
2
×
0.1
×
10
−
6
=
0.064
H
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Medical Physics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page