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CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 12 - MCQExams.com
CBSE
Class 12 Medical Physics
Alternating Current
Quiz 12
A sinusoidally varying potential difference has amplitude
170
V
.
What is its rms value?
Report Question
0%
170
V
0%
170
V
0%
0
0%
−
120
V
0%
−
170
V
Explanation
Δ
V
r
m
s
=
Δ
V
m
a
x
/
√
2
=
170
/
√
2
=
120
V
The reactance of a capacitor of capacitance C is X. If both the frequency and capacitance be doubled, then new reactance will be
Report Question
0%
X
0%
2
X
0%
4
X
0%
X
4
Explanation
X
c
=
X
When both capacitance frequency are doubled the reactance will become
X
′
C
=
1
2
π
×
(
2
f
)
×
(
2
C
)
=
1
4
×
1
2
π
f
C
=
X
4
In an A.C circuit, the resistance
R
=
0.2
Ω
. At a certain instant,
V
A
−
V
B
=
0.5
V
,
I
=
0.5
A
, and the current is increasing at the rate of
Δ
I
Δ
t
=
8
A
/
s
. The inductance of the coil is:
Report Question
0%
0.05
H
0%
0.1
H
0%
0.2
H
0%
none of these
Explanation
Given
I
=
0.5
A
From Ohm's Law,
V
R
=
(
0.5
)
×
(
.2
)
=
0.1
V
From Kirchoff's Voltage Law,
V
L
=
V
a
−
V
b
−
V
R
=
0.5
−
0.1
=
0.4
V
For an inductor,
V
L
=
L
d
i
d
t
0.4
=
L
×
8
L
=
0.05
H
An AC source of angular frequency
ω
is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to
ω
/
3
(but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency
ω
is
Report Question
0%
√
3
5
0%
√
5
3
0%
3
5
0%
5
3
Explanation
I
=
V
√
R
2
+
(
1
w
c
)
2
, for w
for
w
3
,
I
2
=
V
√
R
2
+
(
3
w
c
)
2
2
√
R
2
+
9
w
2
C
2
=
1
√
R
2
+
1
w
2
C
2
4
R
2
+
4
w
2
c
2
=
R
2
+
9
w
2
c
2
3
R
2
=
5
w
2
c
2
3
R
2
=
5
(
x
c
)
2
[at w]
√
3
5
=
x
c
R
When a series combination of
L
and
R
are connected with a
10
V
,
50
H
Z
A.C. source, a current 1A flows in the circuit. The voltage leads the current by a phase angle of
π
/
3
radian. Then the resistance is
Report Question
0%
5
Ω
0%
10
Ω
0%
15
Ω
0%
20
Ω
Explanation
V
r
m
s
=
10
V
,
I
r
m
s
=
1
A
i.e.
t
a
n
π
3
=
x
L
x
R
√
3
=
x
L
R
also
10
z
=
1
10
2
=
R
2
×
X
2
L
10
2
=
R
2
+
3
2
100
=
4
R
2
R
=
5
Ω
A 100 volt A.C. source of frequency 500 hertz is connected to a L-C-R circuit with L
=
8.1 millihenry, C
=
12.5 microfarad and R
=
10 ohm, all connected in series. The potential difference across the resistance will be:
Report Question
0%
10V
0%
100V
0%
50V
0%
500V
Explanation
Impedance of series RLC circuit is given by:
Z
=
√
R
2
+
(
2
π
f
L
−
1
2
π
f
C
)
2
Z
=
√
10
2
+
(
2
π
×
500
×
8.1
×
10
−
3
−
1
2
π
×
500
×
12.5
×
10
−
6
)
2
Z
≈
10
Ω
Current flowing through the circuit is:
I
=
V
Z
=
10
A
Potential difference across the resistor, by ohm's Law is:
V
R
=
I
R
=
10
×
10
=
100
V
An AC source of variable frequency is applied across a series L-C-R circuit. At a frequency double the resonance frequency, the impedance is
√
10
times the minimum impedance. The inductive reactance is
Report Question
0%
R
0%
2
R
0%
3
R
0%
4
R
Explanation
Let
ω
0
be the resonance frequency
minimum impedance is at
ω
0
Z
m
i
n
=
X
R
=
R
At
ω
=
2
ω
0
,
Z
=
√
10
Z
m
i
n
√
R
2
+
(
ω
L
−
1
ω
C
)
2
=
√
10
×
R
R
2
+
(
2
ω
0
L
−
1
2
ω
0
×
C
)
2
=
10
×
R
2
because at
ω
0
,
ω
0
×
L
=
1
ω
0
C
,
R
2
+
(
2
ω
0
L
−
ω
0
L
2
)
2
=
10
×
R
2
(
3
2
ω
0
L
)
2
=
9
×
R
2
3
2
×
ω
0
L
=
3
×
R
X
L
=
ω
L
=
2
ω
0
L
=
4
R
(at
ω
=
2
ω
0
)
At a frequency
ω
0
the reactance of a certain capacitor equals that of a certain inductor. If frequency is changed to 2
ω
0
. The ratio of reactance of the inductor to that of the capacitor is
Report Question
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4 : 1
0%
√
2
:
1
0%
1
:
2
√
2
0%
1 : 2
Explanation
at
w
o
,
w
o
L
1
w
o
C
=
1
i.e.
w
2
o
L
C
=
1
now at
2
w
o
,
2
w
o
L
1
2
w
o
C
=
X
1
L
X
1
C
4
w
2
o
L
C
=
X
1
L
X
1
C
4
1
=
X
1
L
X
1
C
An ideal inductor takes a current of
10
A
when connected to a
125
V
,
50
H
z
AC supply. A pure resistor across the same source takes
12.5
A
. If the two are connected in series across a
100
√
2
V
,
40
H
z
supply, the current through the circuit will be
Report Question
0%
10
A
0%
12.5
A
0%
20
A
0%
25
A
Explanation
Inductor
X
L
=
125
10
=
12.5
Ω
Resistor
X
R
=
125
12.5
=
10
Ω
X
′
L
=
12.5
(
ω
2
ω
1
)
Ω
=
12.5
(
40
50
)
Ω
=
10
Ω
both are connected in series
z
=
√
10
2
+
(
10
)
2
=
10
√
2
Ω
V
r
m
s
=
100
√
2
V
I
r
m
s
=
100
√
2
10
√
2
=
10
A
A circuit containing resistance
R
1
, Inductance
L
1
and capacitance
C
1
connected in series resonates at the same frequency
n
as a second combination of
R
2
,
L
2
and
C
2
. If the two are connected in series, then the circuit will resonate at:
Report Question
0%
n
0%
2
n
0%
√
L
2
C
2
L
1
C
1
0%
√
L
1
C
1
L
2
C
2
Explanation
Condition of resonance
ω
=
1
√
L
C
So
ω
1
=
1
√
L
1
C
1
ω
2
=
1
√
L
2
C
2
Since both frequency are same
L
1
C
1
=
L
2
C
2
Equivalent capacitance of
C
1
a
n
d
C
2
in series =
C
1
C
2
C
1
+
c
2
Equivalent inductance of
L
1
and
L
2
=
L
1
+
L
2
Resonance for combination =
√
(
C
1
+
C
2
)
C
1
C
2
(
L
1
+
L
2
)
Using
L
1
C
1
=
L
2
C
2
, resonance of combination =
1
√
L
1
C
1
So the frequency of resonance remains same =
n
In the given circuit,
R
is a pure resistor,
L
is a pure inductor, S is a 100V, 50Hz AC source and A is an AC ammeter. With either
K
1
or
K
2
alone closed, the ammeter reading is
I
. If the source is changed to 100 V, 100 Hz, the ammeter reading with
K
1
alone closed and with
K
2
alone closed will be respectively
Report Question
0%
I
,
I
2
0%
I
,
2
I
0%
2
I
,
I
0%
2
I
,
I
2
Explanation
I
=
100
R
=
100
2
π
(
50
)
(
L
)
now at
100
H
z
new I when
K
1
is close only
I
1
=
100
R
=
I
and new I when only
K
2
is closed
I
2
=
100
2
π
×
(
100
)
×
L
=
I
2
The self-inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of : (Take
π
2
= 10)
Report Question
0%
4
μ
F
0%
8
μ
F
0%
1
μ
F
0%
2
μ
F
Explanation
L
=
10
H
,
f
=
50
H
z
,
w
=
50
×
2
π
w
=
1
√
L
C
C
=
1
L
w
2
C
=
1
10
×
π
2
×
10
4
=
10
−
6
f
=
1
μ
F
A 120V, 60Hz a.c. power is connected 800
Ω
non-inductive resistance and unknown capacitance in series. The voltage drop across the resistance is found to be 102V, then voltage drop across capacitor is
Report Question
0%
8V
0%
102V
0%
63V
0%
55V
Explanation
For RC circuits,
V
2
=
V
2
R
+
V
2
C
120
2
=
102
2
+
V
2
C
V
2
C
=
63.214
≈
63
V
An alternating e.m.f. of
E
=
100
sin
(
100
π
t ) is connected to a choke of negligible resistance and current oscillations of amplitude 1 A are produced. The inductance of the choke should be
Report Question
0%
100
H
0%
1
π
H
0%
1
H
0%
π
H
Explanation
Angular frequency
ω
=
100
π
Reactance
X
L
=
100
π
×
L
I
o
=
1
A
E
o
=
100
V
∴
L=\dfrac{1}{\pi}H
A coil has an inductance of 0.7H and is joined in series with a resistance of 220
\Omega
. When an alternating e.m.f. of 220V at 50 c.p.s. is applied to it, then the wattless component of the current in the circuit is
Report Question
0%
5 ampere
0%
0.5 ampere
0%
0.7 ampere
0%
7 ampere
Explanation
Impedance of the circuit is given by:
Z = \sqrt {R^2 + (2 \pi f L)^2}
Z = \sqrt {220^2 + (2 \pi \times 50 \times 0.7)^2}
Z \approx 220\sqrt 2\ \Omega
Power factor is given by:
\cos \phi = \dfrac{R}{Z} = \dfrac{\pi}{4}
Total current in the circuit flowing is:
I = \dfrac{V}{Z} = \dfrac{220}{220\sqrt 2} = \dfrac{1}{\sqrt 2} A
Wattless component of current in the circuit is:
I_a = I \sin (\phi)
I_a = \dfrac{1}{\sqrt 2} \times \dfrac{1}{\sqrt 2} = 0.5 A
In the given circuit the readings of the voltmeter
V_1
and the ammeter
A
are:
Report Question
0%
220 V, 2.2 A
0%
110 V; 1.1A
0%
220 V, 1.1 A
0%
110 V; 2.2 A
Explanation
w=100\pi\Omega
we observe that
w^2=\dfrac{1}{LC}
i.e. this frequency is resonance frequency
all the voltage drop across resistor.
\therefore v_1=220V
and
i=\dfrac{V_1}{R}
=\dfrac{220}{100}
=2.2A
Assertion: A resistance is connected to an ac source. Now a capacitor is included in the series circuit. The average power absorbed by the resistance will remain same.
Reason: By including a capacitor or an inductor
in the circuit average power across resistor does
not change.
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0%
A and R both are true and R is correct explanation of A
0%
A and R both are true but R is not the correct explanation of A
0%
A is true R is false
0%
A is false and R is true
Explanation
After connecting the capacitor Irms will charge because impedance is changed.
\therefore
A is false
A certain choke coil of negligible resistance draws a current of 8A, when connected to a supply of 100 volts, at 50Hz. A certain non-inductive resistance, under the same conditions carries a current of 10 A. If the two are transferred to a supply system working at 150 V, at 40 Hz, the total current they will take if joined in series is
Report Question
0%
1.06 A
0%
10.6 A
0%
0.106 A
0%
100.6 A
Explanation
In choke coil,
I=8A, Erms=100V
\omega=100\pi \ rad/s
\dfrac{100}{100\pi\times L}=8=L=\dfrac{1}{8\pi}H
In resistance,
I=10 \ amp
\therefore R=\dfrac{100}{10}=10 \ \Omega
New source
V=150 V, f= 40 Hz,
\omega'=40\times 2\pi
Connected in series,
z=\sqrt{10^2+(\dfrac{80\pi}{8\pi})}
=10\sqrt{2}\Omega
Irms=\dfrac{150}{10\sqrt{2}}
=10.61 \ A
A capacitor has a resistance of
1200 M\Omega
and capacitance of
22 \mu F
. When connected to an AC supply of frequency 80Hz, the alternating voltage supply required to drive a current of 10 virtual amperes is
Report Question
0%
904\sqrt{2} V
0%
904 V
0%
904/\sqrt{2} V
0%
452 V
Explanation
X_c=1200\times 10^6\Omega
C=22\mu F
w=160\pi\ rad/s
\dfrac{V_o}{X_c}=10A
V_o=\dfrac{10}{160\pi(22\times 10^{-6})}=904.29V
The potential difference across a
2 H
inductor as a function of time is shown in figure. At time
t=0
, current is zero. Current at
t=2
second is:
Report Question
0%
1 A
0%
3 A
0%
4 A
0%
5 A
Explanation
t=0, i=0
t=2, i=?
L=2H
V=L\dfrac{dI}{dt}
V=2\dfrac{dI}{dt}
\int_{0}^{2}Vdt=2\int_{0}^{I}dI
area of shaded portion
\dfrac{1}{2}\times 2\times 10=2\times I
I=5A
A sinusoidal voltage
e = 150\ sin (100
\pi t) V
is applied to a series connection of resistance
R
and inductance
L
.
Select TRUE statement/s from the following :
Report Question
0%
Steady state current leads the voltage by
30^{\circ}
0%
The instantaneous voltages across the resistance and inductance become equal at
t =
\dfrac{1}{300}s
0%
Voltage leads the current by
30^{\circ}
0%
The maximum current in the circuit is
0.75\ A
An inductor of inductance
L= 200\ \text{mH}
and the resistors
R_{1}
and
R_{2}
, each of which is equal to
4\ \Omega
, are connected to a battery of emf
12\ \text{V}
through a switch
S
as shown in the Figure. The switch
S
is closed and after the steady state is reached, the switch
S
is opened. The current in
R_{1}
then is
(
in
A)
Report Question
0%
6e^{-20t}\ \text{from B to A}
0%
3e^{-40t}\ \text{from A to B}
0%
3e^{-40t}\ \text{from B to A}
0%
6e^{-20t}\ \text{from A to B}
Explanation
When S is closed at t = 0 s, current in
LR_{2}
branch
i_{2}=3\left ( 1-e^{-\dfrac{R_{2}}{L}t} \right ),\dfrac{R_{2}}{L}=\dfrac{4}{2\times 10^{-1}}=20\ s^{-1}
=3(1-e^{-20t})A
Current i, through
R_{1}=3A
When S is opened, current in R
_{1}
reduces to zero just after opening.
But in
L-R_{2}
, it decreases exponentially
\therefore
Current through
R_{1}=3e^{-40t}
A, from B to A
The frequency at which the impedance of the circuit becomes maximum is
Report Question
0%
\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}+\dfrac{R^{2}}{L^{2}}}
0%
\dfrac{1}{2\pi }\dfrac{1}{\sqrt{LC}}
0%
\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}
0%
\dfrac{1}{2\pi }\dfrac{R}{L}
Explanation
Magnitude of admittance
Y=\dfrac{1}{Z}
is given by
|\displaystyle \mathrm{Y}|=\dfrac{[\mathrm{R}^{2}\mathrm{X}_{\mathrm{c}}^{2}+(\mathrm{R}^{2}+\mathrm{X}_{L}^{2}-X_{L}X_{\mathrm{c}})^{2}]}{\mathrm{X}_{\mathrm{c}}(\mathrm{R}^{2}+\mathrm{X}_{L}^{2})}\dfrac{1}{2}
Now,
|Z|
is maximum or
|Y|
is minimum at
\omega =\omega _{0}
such that
R^{2}+X_{L}^{2}-X_{L}X_{C}=0
where
X_L = w_oL
and
X_C = \dfrac{1}{w_o C}
Or
R^{2}+\omega _{0}^{2}L^{2}=\omega _{0}L\times \dfrac{1}{\omega _{0}C}=\dfrac{L}{C}
or
\omega _{0}^{2}=\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}
or
\omega _{0}=\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}
Hence resonance frequency
f_{0}=\dfrac{w_o}{2\pi}=\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}
A series RLC circuit is made as shown in the figure with an AC source of 60 V, 20 Hz. Then
Report Question
0%
the rms current through the resistor R is 4.2 A
0%
the effective potential difference between P and Q should be 42 V.
0%
the instantaneous current leads the source voltage by
45^{\circ}
0%
the instantaneous current lags behind the applied voltage by
45^{\circ}
Explanation
Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{100+100}=10\sqrt{2}\Omega
\therefore I=\dfrac{60}{10\sqrt{2}}=6\times 0.707=4.2A
\therefore V_{R}=4.2\times 10=42V
The phase difference
\phi =tan^{-1}\left ( \dfrac{X_{L}-X_{C}}{R} \right )=tan^{-1}1=45^{\circ}
As
X_L > X_C
, Source voltage leads the current by
45^{\circ}
The reading shown in AC Voltmeter V when S is moved to B is:
Report Question
0%
120 V
0%
100 V
0%
140 V
0%
160 V
Explanation
Phase difference
=\phi =45
cos \phi = \dfrac{R}{\sqrt {X_L^2+R^2}} = \dfrac{1}{\sqrt 2}
Solving,
X_L=R
\implies X_L=100\Omega
Impedance is
Z = \sqrt {R^2+X_L^2} = 100\sqrt 2
Current in the circuit:
I_o=\dfrac{e}{Z} = \dfrac{200}{(\sqrt{2})100}=\sqrt{2}A
I_{rms}=1A
AC voltmeter shows RMS reading. Hence,
V_{rms}=I_{rms} R =(100\Omega)(1A)
=100V
In the circuit diagram shown,
X_{C}=100\Omega , X_{L}=200 \Omega
&
R=100 \Omega .
The effective current through the source is:
Report Question
0%
2 A
0%
2 \sqrt{2} A
0%
0.5 A
0%
\sqrt{0.4} A
Explanation
I_{R}=\dfrac{V}{R}=\dfrac{200}{100}=2A
{I}'=\dfrac{V}{X_{L}-X_{C}}=\dfrac{200}{100}=2A
I=\sqrt{I{_{R}}^{2}+{I}'^{2}}=2\sqrt{2} Amp.
In a series L, C circuit , which of the following represents variation of magnitude of reactance (X) with frequency (f) ?
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0%
0%
0%
0%
Explanation
X=\left ( \omega L- \dfrac{1}{\omega C} \right ).
This is of the form
y=x-\dfrac{1}{x}
which gives us the graph as shown.
For the values of frequency between 0 and 1 we take its absolute value and take the mirror image of the part which is below the x axis.
A current source sends a current i
= i_0 cos (\omega t).
When connected across an unknown load gives a voltage output of,
v = v_0 sin (\omega t + \pi/4)
across that load. Then voltage across the current source may be brought in phase with the current through it by
Report Question
0%
connecting an inductor in series with the load
0%
connecting a capacitor in series with the load
0%
connecting an inductor in parallel with the load
0%
connecting a capacitor in parallel with the load.
Explanation
Here, current leads the voltage by a phase of
\pi/4
.
Thus unknown load is a capacitor.
The current can be brought in phase with current through it by connecting it i series with an inductor of same impedance as that of the capacitor.
Switch S is closed at t =After sufficiently long time an iron rod is inserted into the inductor L. Then, the bulb :
Report Question
0%
glows more brightly
0%
gets dimmer
0%
glows with the same brightness
0%
gets momentarily dimmer and then glows more brightly
Explanation
When the iron rod is inserted into the inductor, the inductance of the coil increases.
As a result potential difference across the inductor increases, potential difference across the resistor( bulb ) decreases, so the bulb becomes dimmer.
The instantaneous potential difference between points A and B is (Phase angle is
37^0
)
Report Question
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8 \sin (50\pi t+37\dfrac{\pi }{180})
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8 \sin (50\pi t-37\dfrac{\pi }{180})
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10 \sin (50\pi t )
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10 \cos (50\pi t )
Explanation
Impedance
z=\sqrt{(8-2)^{2}+(8)^{2}}= 10\Omega
current lags voltage by
37^{\circ}
,then
i=\dfrac{10}{10} \sin (50\pi t-37^{\circ} )
V_{AB}=i\times R=8\sin (50\pi t-37^{\circ} )
The current flowing through the resistor in a series LCR a.c. circuit, is
I =\varepsilon / R.
Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning)
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0%
equal to I
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more than I
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less than I
0%
zero
Explanation
In figure (1)
Z = R
X_{L} = X_{C}
If
I_{1} = \dfrac{V_{0}}{X_{C}} sin(\omega t+\pi/2)
then
I_{2} = \dfrac{V_{0}}{X_{L}} sin(\omega t-\pi/2)
and Total current in circuit two
I = I_{1} + I_{2}=0
An ac voltage source
V = V_0\sin(\omega t)
is connected across resistance
R
and capacitance
C
as shown in figure. It is given that
R = \dfrac{1}{\omega C}
. The peak current is
I_0
. If the angular frequency of the voltage source is changed to
\dfrac{\omega}{\sqrt{3}}
then the new peak current in the circuit is:
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\dfrac{I_0}{2}
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\dfrac{I_0}{\sqrt{2}}
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\dfrac{I_0}{\sqrt{3}}
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\dfrac{I_0}{3}
Explanation
The peak value of the current is:
I_0 = \dfrac{V_0}{\sqrt{R^2+\dfrac{1}{{\omega}^2C^2}}} = \dfrac{V_0}{\sqrt{2}R}
When the angular frequency is changed to
\dfrac{\omega}{\sqrt{3}}
, the new peak value is:
I'_0 = \dfrac{V_0}{\sqrt{R^2+\dfrac{3}{{\omega}^2C^2}}}=\dfrac{V_0}{\sqrt{4R^2}}=\dfrac{V_0}{2R}
\therefore I'_0 = \dfrac{I_0}{\sqrt{2}}
A coil has resistance
30\ \text{ohm}
and inductive reactance
20\ \text{Ohm at}\ 50\ \text{Hz}
frequency. If an ac source, of
200\ \text{volt},\ 100\ \text{Hz},
is connected across the coil, the current in the coil will be :
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4.0\ \text{A}
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8.0\ \text{A}
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\dfrac {20}{\sqrt {13}}\ \text{A}
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2.0\ \text{A}
Explanation
For a given coil
R:Resistance
=30\Omega
{ X }_{ L }:
Inductive Reactance
=\omega L\Omega =\left( 2\pi F \right) L\Omega
At
50㎐,{ X }_{ L }=20\Omega
\therefore 2\pi \times 50\times L=20
\therefore L=\cfrac { 20 }{ 100\pi }
At
100㎐
{ X }_{ L }=2\pi \times 100\times L=2\pi \times 100\times \cfrac { 20 }{ 100\pi } =40\Omega
At
100㎐
z=impedance=\sqrt { { R }^{ 2 }+{ { X }_{ L } }^{ 2 } }
\therefore Z=\sqrt { { 40 }^{ 2 }+{ 30 }^{ 2 } } =50\Omega
Current at
200V,100㎐
is given as
I=\cfrac { V }{ z } =\cfrac { 200 }{ 50 } =4A
Therefore option A is correct
Regarding the given circuit, the correct statement is:
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(
V_a - V_b
) is increasing with time
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(
V_a - V_b
) is decreasing with time
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(
V_a - V_b) =
10V
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(
V_a - V_b) =
zero
Explanation
Q_{c} = 5(1-e^{-\dfrac{t}{2}})
I_{L} = 5(1-e^{-\dfrac{t}{2}})
V_{a} - V_{b} = 2I_{L} - \dfrac{Q _{c}}{0.5}
= 10(1-e^{-\dfrac{t}{2}}) - \dfrac{5}{0.5}(1-e^{-\dfrac{t}{2}})
= 0
Calculate
\omega
and
T
.
Report Question
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6.28\times{10}^{-3}rad/s
;
{10}^{-3}s
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6.28\times{10}^{3}rad/s
;
{10}^{-3}s
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7.28\times{10}^{3}rad/s
;
{10}^{-3}s
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7.28\times{10}^{-3}rad/s
;
{10}^{-3}s
Explanation
Oscillation frequency
f = 10^3 \ Hz
Angular frequency
w = 2\pi f = 2\pi\times 10^3 = 6.28\times 10^3 \ rad/s
Time period
T = \dfrac{1}{f} = \dfrac{1}{10^3} = 10^{-3} \ s^{-1}
When
100
volt
DC
source is applied across a coil, a current of
1A
flows through it. When
100V
AC
source of
50Hz
is applied to the same coil, only
0.5A
current flows. Calculate the inductance of the coil.
Report Question
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\displaystyle (\pi/ \sqrt 3)H
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\displaystyle (\sqrt 3/ \pi)H
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\displaystyle (2/\pi)H
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None of these
Explanation
\displaystyle I_{dc}=\frac{V_{dc}}{R}
\displaystyle I=\frac{100}{R}\Rightarrow R=100\Omega
\displaystyle I_{ac}=\frac{V_{ac}}{\sqrt{R^2+X^2_L}}
\displaystyle 0.5=\frac{100}{\sqrt{(100)^2+X^2_L}}
or
\displaystyle X_L=100\sqrt 3 \Omega =(2\pi fL)
\displaystyle \therefore L=\frac{100\sqrt 3}{2\pi f}
\displaystyle =\frac{100\sqrt 3}{2\pi (50)}
\displaystyle = \left(\frac{\sqrt 3}{\pi}\right)H
A coil, a capacitor and an
AC
source of rms voltage
24V
are connected in series. By varying the frequency of the source, a maximum rms current of
6A
is observed. If coil is connected to a dc battery of emf
12
volt and internal resistance
4\Omega
, then current through it in steady state is :
Report Question
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2.4A
0%
1.8A
0%
1.5A
0%
1.2A
Explanation
\displaystyle I_{rms}=\frac{V}{R}
(at resonance)
\displaystyle 6=\frac{24}{R}
\displaystyle \therefore R=4\Omega
\displaystyle I_{dc}=\frac{V}{R+r}=\frac{12}{4+4}=1.5A
Identify the graph which correctly represents the variation of capacitive reactance
X_C
with frequency.
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0%
0%
0%
Explanation
X_C=\displaystyle \frac{1}{\omega C}=\frac{1}{2\pi fC}
or
X_C\propto \displaystyle\frac{1}{f}
i.e.,
X_C
versus
f
graph is a rectangular hyperbola.
A
50Hz
AC
source of
20V
is connected across
R
and
C
as shown in figure. The voltage across
R
is
12V
. The voltage across
C
is
Report Question
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8V
0%
16V
0%
10V
0%
Not possible to determine unless value of
R
and
C
are given.
Explanation
\displaystyle V_C=\sqrt{V^2-V^2_R}=\sqrt{(20)^2-(12)^2}
=16V
The frequency of oscillation of current in the inductance is
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\dfrac {1}{3\sqrt {LC}}
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\dfrac {1}{6\pi \sqrt {LC}}
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\dfrac {1}{\sqrt {LC}}
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\dfrac {1}{3\pi \sqrt {LC}}
Explanation
Step 1: Find equivalent inductance
Here, both inductor are connected in series
So, Equivalent inductance is given by
\mathrm{L}_{\mathrm{eq}}=\mathrm{L}+2\mathrm{~ L}=3L
\textbf{Step 2: Find equivalent Capacitance}
Here, both capacitors are connected in parallel
So, Equivalent capacitance is given by
\mathrm{C}_{\mathrm{eq}}=\mathrm{C}+2\mathrm{C}=3C
\textbf{Step 3: Find the frequency of oscillation.}
Hence, Frequency of oscillation is
\Rightarrow\mathrm{f}=\frac{1}{2\pi\sqrt{\mathrm{L}_{\mathrm{eq}}\mathrm{C}_{\mathrm{eq}}}}
\Rightarrow\frac{1}{6\pi\sqrt{\mathrm{LC}}}
Therefore, The correct option is 'B' is
\frac{1}{6 \pi \sqrt{L C}}
.
Consider an
L-C-R
circuit as shown in figure, with an
AC
source of peak value
V_0
and angular frequency
\omega
. Then the peak value of current through the
AC
source is :
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\displaystyle \frac{V_0}{\sqrt{\displaystyle \frac{1}{R^2}+\left(\omega L-\displaystyle\frac{1}{\omega C}\right)^2}}
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\displaystyle V_0 \sqrt {1/ {R^2}+\left(\omega C-\displaystyle \frac{1}{\omega L}\right)^2 }
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\displaystyle \frac{V_0}{\sqrt {R^2+\left(\omega L-\displaystyle \frac{1}{\omega C}\right)}^2}
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None of these.
Explanation
As R, L , C are in parallel, equivalent impedance is given by:
\dfrac{1}{Z} = \dfrac{1}{Z_R } + \dfrac{1}{Z_L} + \dfrac{1}{Z_C} = \dfrac{1}{R} + \dfrac{1}{j L \omega} + j C \omega
\left | \dfrac{1}{Z} \right| = \left | \dfrac{1}{R} + \dfrac{1}{j L \omega} + j C \omega \right | = \sqrt{ \dfrac{1}{R^2} + \left( \dfrac{1}{L \omega} - C \omega \right)^2 }
Peak value of current is given by:
I_{peak} = \left | \dfrac{V_0}{Z} \right |= { V_0}{ \sqrt{ \dfrac{1}{R^2} + \left( \dfrac{1}{L \omega} - C \omega \right)^2 } }
When an alternating voltage of
220V
is applied across a device
P
, a current of
0.25A
flows through the circuit and it leads the applied voltage by an angle
\pi/2
radian. When the same voltage source is connected across another device
Q
, the same current is observed in the circuit but in phase with the applied voltage. What is the current when the same source is connected across a series combination of
P
and
Q
?
Report Question
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\displaystyle \frac{1}{4\sqrt 2}A
lagging in phase by
\pi/4
with voltage.
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\displaystyle \frac{1}{4\sqrt 2}A
leading in phase by
\pi/4
with voltage.
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\displaystyle \frac{1}{\sqrt 2}A
leading in phase by
\pi/4
with voltage.
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\displaystyle \frac{1}{4\sqrt 2}A
leading in phase by
\pi/2
with voltage.
Explanation
In first case,
\displaystyle X_C=\frac{V}{I}=\frac{220}{0.25}=880\Omega
In the second case,
\displaystyle R=\frac{V}{I}=\frac{220}{0.25}=880\Omega
In the combination of
P
and
Q
\displaystyle \tan\phi=\frac{X_C}{R}=1
\therefore \phi=45^o
Since the circuit is capacitive, current leads the voltage. Further,
\displaystyle Z=\sqrt{R^2+X^2_C}=880\sqrt 2\Omega
\displaystyle I=\frac{V}{Z}=\frac{220}{880\sqrt 2}=\frac{1}{4\sqrt 2}A
A signal generator supplies a sine wave of
200V
,
5kHz
to the circuit as shown in the figure. Then choose the wrong statement:
Report Question
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The current in the resistive branch is
0.2A
.
0%
The current in the capacitive branch is
0.126A
.
0%
Total line current is
=0.283A
.
0%
Current in both the branches is same.
Explanation
\displaystyle I_R=\frac{V_{rms}}{R}=\frac{200}{100}=0.2A
\displaystyle X_C=\frac{1}{2\pi fC}=\frac{1}{(2\pi)(5\times 10^3)\left(\frac{1}{\pi}\times 10^{-6}\right)}
=100\Omega
\therefore \displaystyle I_C=\frac{V_{rms}}{X_C}=\frac{200}{100}=0.2A
I_C
is
90^o
ahead of the applied voltage and
I_R
is in phase with the applied voltage. Hence, there is a phase difference of
90^o
between
I_R
and
I_C
too.
\therefore \displaystyle I=\sqrt{I^2_R+I^2_C}
=\displaystyle \sqrt{(0.2)^2+(0.2)^2}
=0.283A
An
LCR
circuit contains resistance of
100\Omega
and a supply of
200\space V
at
300\space rad/s
angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by
60^{\small\circ}
. If on the other hand, only inductor is taken out, the current leads by
60^{\small\circ{}}
with the applied voltage. The current flowing in the circuit is
Report Question
0%
1\space A
0%
1.5\space A
0%
2\space A
0%
2.5\space A
Explanation
if only capacitance is remove,
Z=R+jL\omega \quad \Rightarrow \quad \left| Z \right| cos(60)=R=100
\Rightarrow \quad \left| Z \right| =200\quad \Omega
\Rightarrow \quad L\omega =\left| Z \right| sin(60)=173.205
if only inductor is taken out,
Z=R-j\frac { 1 }{ C\omega } \quad \Rightarrow \quad \left| Z \right| cos(-60)=R=100
\Rightarrow \quad \left| Z \right| =200\quad \Omega
\Rightarrow \quad \frac { 1 }{ C\omega } =\left| Z \right| sin(-60)=-173.205
therefore actual impedance=
R+jL\omega - j \frac{1}{C\omega} = 100 \Omega
therefore current in circuit=
\dfrac{200 V }{100 \Omega} = 2A
Resonance occurs in a series
LCR
circuit when the frequency of the applied emf is
1000\space Hz
.
Report Question
0%
When frequency
= 900\space Hz
, then the current through the voltage source will be ahead of emf of the source
0%
The impedance of the circuit is minimum at
f = 1000\space Hz
0%
At only resonance the voltages across
L
and
C
differ in phase by
180^{\small\circ}
0%
If the value of
C
is doubled, resonance occurs at
f = 2000\space Hz
Explanation
f_0 = \frac{1}{2 \pi \sqrt{LC}} = 1000 Hz
when
f=900Hz< f_0 \Rightarrow \omega < \frac{1}{\sqrt{ LC}} \Rightarrow L \omega < \frac{ 1}{C \omega } \Rightarrow X_L < X_C \Rightarrow
circuit is capacitive. therefore the current leads the voltage of the source.
total impedance of circuit =
\sqrt { R^2 + (X_L - X_C )^2} \Rightarrow
impedance will be minimum when
X_L-X_C=0
i.e., impedance will be minimum at resonance frequency of
1000 Hz
voltage across L and C is always differ in phase by 180 degree .
resonant frequency when C is doubled =
\frac{1} { 2 \pi \sqrt{ L 2 C }} = \frac{1} {\sqrt{2}} \frac{1}{2 \pi \sqrt{LC}} = \frac{1000}{\sqrt{2} } Hz
A capacitor of
10\space \mu F
and an inductor of
1\space H
are joined in series. An ac of
50\space Hz
is applied to this combination. What is the impedance of the combination?
Report Question
0%
\displaystyle\frac{5(\pi^2 - 5)}{\pi}\space \Omega
0%
\displaystyle\frac{10(10 - \pi^2)}{\pi}\space \Omega
0%
\displaystyle\frac{10(\pi^2 - 5)}{\pi}\space \Omega
0%
\displaystyle\frac{5(10 - \pi^2)}{\pi}\space \Omega
Explanation
X_L = L \omega = L 2 \pi f = 100 \pi \Omega
X_C = \frac{ 1 } { C \omega } = \frac{ 1 } { C 2 \pi f } = 1000/ \pi \Omega
impedance =
\left | X_L - X_C \right | = 100 \dfrac{ 10- {\pi}^2 }{\pi} \Omega
A transmitter transmits at a wavelength of
300\space m
. A condenser of capacitance
2.4\space \mu F
is being used. The value of the inductance for the resonant circuit is approximately
Report Question
0%
10^{-4}\space H
0%
10^{-6}\space H
0%
10^{-8}\space H
0%
10^{-10}\space H
Explanation
f = \dfrac{c}{ \lambda} = 3 \times 10^8 / 300
\omega = 2\pi f = \dfrac{1}{\sqrt{L C }}
L = \dfrac{ 1}{C 4 {\pi}^{2} f^2 }= 10^{-8} H
Impedance of box
P
at the above frequency is
Report Question
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70\Omega
0%
77\Omega
0%
90\Omega
0%
100\Omega
Explanation
\displaystyle X_C=\frac{1}{\omega C}=\frac{1}{\left(\displaystyle \frac{10^5}{7}\right)(10^{-6})}=70\Omega
\displaystyle Z_P=\sqrt{R^2_P+X^2_C}
=\displaystyle \sqrt{(32)^2+(70)^2}=77\Omega
The angular frequency for which he detects maximum current in the circuit is
Report Question
0%
10^5/7 rad/s
0%
10^4 rad/s
0%
10^5 rad/s
0%
10^4/7 rad/s
Explanation
\displaystyle \omega =\frac{1}{\sqrt {LC}}
=\displaystyle \frac{1}{\sqrt{4.9\times 10^{-3}\times 10^{-6}}}
\displaystyle =\frac{10^5}{7} rad/s
A choke coil of resistance
5\Omega
and inductance
0.6\space H
is in series with a capacitance of
10\space \mu F
. If a voltage of
200\space V
is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitance are
I_0, V_0
and
V_1
respectively. We have
Report Question
0%
I_0 = 40\space A
0%
V_0 = 9.8\space kV
0%
V_1 = 9.8\space kV
0%
V_1 = 19.6\space kV
Explanation
At resonance current,
I_0= \dfrac{ V}{R} = \dfrac{ 200} {5 } = 40 A
At resonance frequency voltage across L
= V_0
, and voltage across C
= V_1
are always equal and out of phase by 180 degree.
Therefore, if
f=65
Hz ,
V_0 = j I_0 L \omega =I_0 L 2 \pi f = j 9800 V
V_1 = -j \dfrac{I_0}{ C \omega } = -j \dfrac{ I_0}{ C 2 \pi f} = -j \ 9800 V
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Practice Class 12 Medical Physics Quiz Questions and Answers
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