MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 10

A coil of

$10^{-2}\ H$ inductance carries a current

$I=2\sin (100t)A$ . When current is half of its peak value then at that instant the induced emf in coil :-

0%
$1V$
0%
$\sqrt{2}\ V$
0%
$\sqrt{3}\ V$
0%
$2V$

Explanation

$I=2\sin (100t)$

$\dfrac{I_{max}}{2}=2\sin (100t)$

$1=2\sin (100t)$

$\dfrac{dI}{dt}=2\times 100 \times \cos (100t)$

$=200 \times \cos \left ( \dfrac{\pi}{6} \right )$

$=200 \times \dfrac{\sqrt 3}{2}=100\sqrt 3$

$\epsilon =L \times \dfrac{di}{dt}$

$=10^{-2} \times 100\sqrt 3$

$=\sqrt 3V$

Current in resistance is

$A$ , then

0%
$V_{s}=15\ V$
0%
impedance of network is $15\ Omega$
0%
power factor of given circuit is $(0.46)$ lagging (current is lagging)
0%
NONE of the above

$\omega_t$ and

$\omega_2$ are half power frequencies of a series

$LCR$ circuit, then resonant frequency

$\omega$ , can be expressed as :

0%
$\omega_r = \dfrac{\omega_1+\omega_2}{2}$
0%
$\omega_r=\sqrt{\omega_1\omega_2}$
0%
$\omega_r= \sqrt{\dfrac{\omega_1^2+\omega_2^2}{2}}$
0%
$\omega_r = \dfrac{\omega_1\omega_2}{\omega_1+\omega_2}$

In the given circuit, the

$AC$ source has

$\omega=100\ rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

0%
The current through the circuit, $I$ is $0.3\ A$
0%
The current through the circuit, $I$ is $0.3\sqrt {2}\ A$
0%
The voltage across $100\Omega$ resistor $10\sqrt {2}V$
0%
The voltage across $50\Omega$ resistor $10\ V$

Explanation

Given that,

Resistance ${{R}_{C}}=100\Omega$

$\omega =100\,rad/s$

${{X}_{L}}=50\,H$

${{R}_{L}}=50\Omega$

${{X}_{C}}=100\mu F$

Now, the impedance is

${{z}_{1}}=\sqrt{{{R}^{2}}+X_{C}^{2}}$

${{Z}_{1}}=\sqrt{{{\left( 100 \right)}^{2}}+{{\left( 100 \right)}^{2}}}$

${{Z}_{1}}=100\sqrt{2}$

Now the current is

${{I}_{1}}=\dfrac{20}{100\sqrt{2}}\,A$

Now, the impedance is

${{Z}_{L}}=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}}$

${{Z}_{L}}=50\sqrt{2}$

Now, the current is

${{I}_{2}}=\dfrac{20}{50\sqrt{2}}\,A$

Now, the total current is

$I=\sqrt{I_{1}^{2}+I_{2}^{2}}$

$I=\sqrt{{{\left( \dfrac{20}{100\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{20}{50\sqrt{2}} \right)}^{2}}}$

$I=\dfrac{1}{\sqrt{10}}\,A$

Now, voltage across 100 Ω resistor is

${{V}_{100}}=\dfrac{20}{100\sqrt{2}}\times 100$

${{V}_{100}}=10\sqrt{2}\,V$

Now, voltage across 50 Ω resistor is

${{V}_{50}}=\dfrac{20}{50\sqrt{2}}\times 50$

${{V}_{50}}=10\sqrt{2}\,V$

Hence, the voltage across 50Ω resistor is $10\sqrt{2}\,V$

Rms value of current

$i = 3 + 4\sin \left( {\omega t + \frac{\pi }{3}} \right)$ is:

0%
$5A$
0%
$\sqrt {17} A$
0%
$\frac{5}{{\sqrt 2 }}A$
0%
$\frac{7}{{\sqrt 2 }}A$

Explanation

the RMS value of DC of current 3A be 3A

RMS value of

$4sin\left\{ \omega t+\frac { \pi }{ 3 } \right\} is\frac { 4 }{ \sqrt { 2 } }$

is add net RMS value

$I=\sqrt { 9+\frac { 16 }{ 2 } } =\sqrt { 17 } A$

If an alternating current

$i=i_{m} \sin \omega t$ is flowing through a capacitor then voltage drop

$\Delta V_{C}$ across capacitor

$C$ will be?

0%
$-\dfrac{i_{m}}{\omega C} \sin \omega t$
0%
$-\dfrac{i_{m}}{\omega C} \cos \omega t$
0%
$-\dfrac{i_{m}}{\omega C}\left(\sin \omega t +\dfrac{\pi}{4} \right)$
0%
$\dfrac{i_{m}}{\omega C}\left(\sin \omega t -\dfrac{\pi}{4} \right)$

Explanation

$\Delta { V }_{ C }=-{ X }_{ C }i=-\cfrac { 1 }{ C\omega } { I }_{ m }sin\omega t$

$=-\cfrac { { I }_{ m } }{ C\omega } sin\omega t$

In the given circuit the potential difference across resistance is

$54\ V$ and power consumed by it is

$16\ W$ . If

$AC$ frequency is

$60Hz$ find the value of

$L$ :-

0%
$1\ H$
0%
$2\ H$
0%
$5\ H$
0%
$4\ H$

The resistance

$R = 10\Omega$ , inductance L = 2 mH and capacitance

$c of 5 \mu F$ are connected in same to an AC source of frequency 50 Hz, then at resonance the impedance of circuit is

0%
zero
0%
$10\Omega$
0%
$1000\Omega$
0%
$10k\Omega$

Explanation

At resonance, the capacitive reactance and the inductive reactance of the series LCR circuit become equal to each other and cancels each other. Therefore, the total impedance of the circuit will be only due to resistance.

So __ $R = 10\Omega$

Which of the shown graphs may represent the reactance of a series L-C combination?

Explanation

The reactance of a series LC circuit is given by:-

$X=X_L-X_C=WL-\dfrac{1}{WC}$

$\Rightarrow X=2\pi fL-\dfrac{1}{2\pi fC}$

Where, W= angular frequency

L= Inductor , C= capacitor

The correct graph is given by,

So, reactance magnitude

$\rightarrow \infty$ as

frequency

$\rightarrow 0$ or

$\infty$

$X=0$ , when

$X_L=X_C$

Thus plot is : - Rectance

Refer image,

2081506_1117407_ans_227c8505ad1e4496b7907898cc5ac8c8.png

In a series circuit,

$R=300 \Omega, L=0.9 H,C=2.0 \mu F$ and

$\omega=1000 rad/s$ . The impedance of the circuit is

0%
$1300 \Omega$
0%
$900 \Omega$
0%
$500 \Omega$
0%
$400 \Omega$

Explanation

The impedance of the circuit is,

$z = \sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega L}\right)^2}$

$=\sqrt{(300)^2 + \left(1000 \times 0.9 - \dfrac{1}{1000 \times 2 \times 10^{-6}}\right)}$

$= \sqrt{300^2 + 400^2 } = 500 \Omega$

Two inductors

$L_{1}$ and

$L_{1}$ are connected in parallel and a time varying current

$i$ flows as shown. The ratio of current

$i_{1}/i_{2}$ at any time

$t$ is

0%
$\dfrac{L_{2}}{L_{1}}$
0%
$\dfrac{L_{1}}{L_{2}}$
0%
$\dfrac{L_{1}^{2}}{(L_{1}+L_{2})^{2}}$
0%
$\dfrac{L_{2}^{2}}{(L_{1}+L_{2})^{2}}$

In given LCR circuit, the voltage across the terminals of a resistance and current will be -

0%
$400V,2A$
0%
$800V,2A$
0%
$100V,2A$
0%
$100V,4A$

Explanation

Refer above image,

$\because V_L=V_C=400V$ .

Current passing trough inductor and capacitor is same then

$X_L=X_C$

$\therefore$ The formula for voltage across resister.

$Z=\sqrt{R^2+(X_L-X_C)^2}(\therefore Z_L=X_C)$

$Z=R$ (

$\therefore V=V_R$ or,

$V_R=100V$ )

$\therefore$ Source voltage is 100V.

$\therefore$ Voltage across resister is given by

$V=IZ(\because Z=R)(\because V_R=100)$

$100=I\times 50$ (

$\because Z =R=50\Omega$ )

$\therefore I=\dfrac{100}{5}=2A$

$\therefore V_R=100V$

$I=2A$

2081574_1118361_ans_d4c9f588aeb94777b0d715005300d8c5.png

An alternative current, L.R circuit comprises of an inductor, whose reactance

$X_L = 3R$ , where

$R$ is the resistance of the circuit. If a capacitor, whose reactance

$X_C = R$ is connected in series then what will be the ratio of the new and the old power factor?

0%
$\sqrt{2}$
0%
$\dfrac{1}{\sqrt{2}}$
0%
$2$
0%
$1$

The given graph shows volume with time in the source voltage and steady state current drawn by a series

$BLC$ circuit:-
Which of the following statements is/are correct?
(a) Current lags the voltage.
(b) Resistance in the circuit is

$250\sqrt{\Omega}$ .
(c) Reactance in the circuit is

$250\Omega$ .
(d) Average power dissipation in the circuit is

$20\sqrt{3\Omega}$

0%
Only a
0%
a & b
0%
a, b & c
0%
All

Explanation

D. All

$v=v_m(\sin wt+\phi )$

$i=i_m(\sin wt)$

$at,t=0,100=200\sin \phi \Rightarrow \phi =\dfrac{\pi}{6} rad$

current lags voltage by

$\dfrac{\pi}{6} rad$

$R=Z\cos \phi =\dfrac{V}{I}\cos \phi =\dfrac{200}{400m}\cos \dfrac{\pi}{6}$

$=250\sqrt{3}\Omega$

$X=Z\sin \phi =\dfrac{V}{I}\sin \phi =\dfrac{200}{400m}\sin \dfrac{\pi}{6}$

$=250\Omega$

$P_{av}=\dfrac{V_mI_m}{2}\cos \phi =20 \sqrt 3 W$

If the frequency of the AC source connected across a series LCR circuit is continuously in which the following is observed. Assume all other parameters are kept the same.

0%
rms current continously increases.
0%
rms current remains constant.
0%
rms current first increases, reaches a maximum and then decrease.
0%
rms current first decreases, reached a minimum and then increases.

Explanation

The frequency of the AC source connected across a series

$LCR$ circuit is continuously in the observed.

All other parameters are kept the same and rms current first increases, reaches a maximum and then decreases.

$\therefore$ Option

$C$ is correct answer.

In the adjoining figure the impedance of the circuit will be_

0%
$120\Omega$
0%
$50\Omega$
0%
$60\Omega$
0%
$90\Omega$

Explanation

According to the question,

$i_L = \dfrac{90}{30} = 3 \,A$ ,

$i_C = \dfrac{90}{20} = 4.5 \,A$

Net current through circuit

$i = i_C - i_L = 1.5 \,A$

$\therefore \, Z = \dfrac{V}{i} = \dfrac{90}{1.5} = 60\Omega$

An inductance of

$1mH$ , a condenser of

$10 \mu F$ and resistance of

$50\Omega$ are connected in series. The reactance of inductor and condensers are same. The reactance of either of them will be:-

0%
$100 \Omega$
0%
$30 \Omega$
0%
$3.2 \Omega$
0%
$10 \Omega$

Explanation

Given

$\omega L = \dfrac{1}{\omega C}$

$\Rightarrow \, \omega^2 = \dfrac{1}{LC}$
or

$\omega = \dfrac{1}{\sqrt{10^{-3} \times 10 \times 10^{-6}}} = \dfrac{1}{\sqrt{10^{-8}}} = 10^4$

$X_L = \omega L = 10^4 \times 10^{-3} = 10 \, \Omega$

The graph given below depict the dependence of two reactive impedances

$X_1$ and

$X_2$ on the frequency of the alternating e.m.f. applied individually to them. We can say that:

0%
$X_1$ is an inductor and $X_2$ is a capacitor
0%
$X_1$ is a resistor and $X_2$ is a capacitor
0%
$X_1$ is a capacitor and $X_2$ is an inductor
0%
$X_1$ is an inductor and $X_2$ is a resistor

Explanation

As we know,

$X_C = \dfrac{1}{C \times 2 \pi f}$

hence from the above figure

$X_!$ is a capacitor and

$X_L = L \times 2 \pi f$

hence

$X_2$ is an inductor.

An AC circuit draws

$5 A$ at

$160 V$ and the power consumption is

$600 W$ . Then the power factor us :-

0%
1
0%
0.75
0%
0.50
0%
Zero

Explanation

$\begin{array}{l} i=5A \\ v=160v \\ p=600w \\ p=V.I\cos \phi \\ \Rightarrow \cos \phi =\frac { 6 }{ 8 } =\frac { 3 }{ 4 } =0.75 \end{array}$

$\therefore$ Option

$B$ is correct.

An inductor of inductance 2.0 mH is connect across a charge capacitor of capacittance 5.0 uF , and the resulting LC circuit is set oscillating at natural frequency. Let Q denote the instantaneous charge on the capacitor, and I the current in the circuit. It is found the maximum value of Q is 200 uC .When Q=100uC the value of

$\dfrac{dI}{dt}$ is found to be

$10^{x}$ A/s. Find the value of x is then

0%
x=-4
0%
x=-5
0%
x=4
0%
x=5

Explanation

In an

$Le$ system

$he$ charge oscillation between

$I$ and

$Q$

Applying

$KVL$ gives,

$-L\dfrac {dI}{dt}+\dfrac {9}{c}=0$ where

$L=2\ mH$

$c=5\mu F$

also, maximum charge

$Q_0=200\ \mu c$

Thus, total energy of system,

$\mu =\dfrac {Q_0}{2c}=\dfrac {200Z^2}{2\times 5}=4000\ J$

when

$9=100\ u c$ , we get from

$(1)$ ,

$\dfrac {dI}{dt}dfrac {+9}{Lc}=\dfrac {+100}{2\times 10^{-3}\times 5}=+10^{-4}\ A/s$

$\dfrac {dI}{dt}=10^x$ in gives

$x=-4$

A coil has a resistance

$10 \Omega$ and an inductance of 0.4 henry. It is connected to an AC source of

$6.5 V , \frac {30} { \pi } Hz.$ The average power consumed in the circuit, is :

0%
$\cfrac {5} { 8} W$
0%
$\cfrac {4} {3} W$
0%
$\cfrac {3} {8} W$
0%
$\cfrac {6} {7} W$

The inductive reactance of an inductive coil with

$\dfrac{1}{\pi}$ henry and

$50$ Hz:-

0%
$\dfrac{50}{\pi}ohm$
0%
$\dfrac{\pi}{50}ohm$
0%
100 ohm
0%
50 ohm

Explanation

As we know,

$X_L = 2 \pi \nu L = 2 \times \pi \times 50 \times \dfrac{1}{\pi} = 100 \, \Omega$

In an electric circuit applied ac emf is e = 20 sin 300t volt and the current is i = 4 sin 300 t ampere. The reactance of the circuit is -

0%
5 ohm
0%
80 ohm
0%
$1.8 k\Omega$
0%
zero

Explanation

Given that,

Emf, $e=20\sin 300t$

Current, $i=4\sin 300t$

Reactance,

$R=\dfrac{emf}{I}$

$R = \dfrac{20}{4}=5\,ohm$

In the given circuit, the potential difference across the

$6 \mu F$ capacitor in steady state is

0%
$1\ V$
0%
$6\ V$
0%
$3\ V$
0%
$4\ V$

In general in an alternating current circuit for a complete cycle

0%
The average value of current is zero
0%
The average value of square of the current is zero
0%
Average power dissipation is zero
0%
All of these

The natural frequency of an LC circuit is 125 kHZz. when the capacitor is totally filled with a dielectric material, the natural frequency decreases by 25 kHz. Dielectric constant of the material is nearly.

0%
3.33
0%
2.12
0%
1.56
0%
1.91

The impedance of coaxial cable, when its inductance is 0.40

$\mu H$ and capacitance is 1

$\times { 10 }^{ -11 }$ F can be

0%
$2\times { 10 }^{ 2 }{ \Omega }$
0%
100 $\Omega$
0%
$3\times { 10 }^{ 3 }{ \Omega }$
0%
$3\times { 10 }^{ -2 }{ \Omega }$

Explanation

Usiing

$Z=\sqrt {\dfrac {L}{C}}$ , we get

$Z=\sqrt {\dfrac {0.40\times 10^{-6}}{10^{-11}}}=2\times 10^2 \Omega$

A capacitor is connected to an A.C generator.The ratio of reactance and impedance of capacitor is-

0%
1
0%
Less than 1
0%
greater than 1
0%
zero

When induced emf in inductor coil is

$50%$ of its maximum value then stored energy in inductor coil in the given circuit will be-

0%
$2.5 \,mJ$
0%
$5 \,mJ$
0%
$15 \,mJ$
0%
$20 \,J$

Explanation

For max. induced emf across inductor will be at

$t = 0$ , when it will behave as O.C.

$V = 2v = V_{max}$ (

$\coprod$ el circuit Voltage will be same)

$50%$ of its max. That means

$1$ volt.

When voltage of

$1$ volt appears across inductor
Then,

$2 = 1 \times i + 1$

$[I=1 Amp]$

Energy stored in inductor

$= \dfrac{1}{2} Li^2$

$= \dfrac{1}{2}\times 5 \times 1^2 \times 10^{-3}$

$E=2.5 mJ$

Even if a physical quantity depends upon three quantites,out of which two are dimensionally same, then the formula cannot be derived by the method of dimensions.This statement.

0%
may be true
0%
may be false
0%
must be true
0%
must be false

An inductive circuit contains resistance of 10 ohms and an inductance of 20 H. If an A.c voltage of 120 volt and frequency 60 Hz is applied to this circuit , the current would be nearly

0%
$0.16 amp$
0%
$0.3 amp$
0%
$0.48 amp$
0%
$0.80 amp$

Explanation

$Z=\sqrt { { R }^{ 2 }+{ (\omega L) }^{ 2 } } =\sqrt { { 10 }^{ 2 }+{ (120\pi \times 2) }^{ 2 } } =376\Omega \\ I=\cfrac { V }{ Z } =\cfrac { 120 }{ 376 } =0.3A$

The alternating current in a circular is described by the graph as shown in figure. The rms current obtained from the graph would be

0%
1.4 A
0%
2.2 A
0%
1.6 A
0%
2.6 A

Explanation

${i_{rms}} = \sqrt {\frac{{1 \times \frac{T}{2} + \frac{T}{2} \times 4}}{T}}$

$= \sqrt {\frac{{5T}}{{25}}} = \sqrt {2.5} = 1.58A$

Hence,

option

$(C)$ is correct answer.

A source of 220V is applied in an A.C. circuit. The value of resistance is

$220 \Omega$ . Frequency & inductance are 50 Hz and 0.7 H, then wattless current is

0%
$0.5 amp$
0%
$0.7 amp$
0%
$1.0 amp$
0%
none

Explanation

$\Rightarrow R = 2\Omega$

$\Rightarrow {V_{rms}} = 220V$

$\Rightarrow {V_ \circ } = 220\sqrt 2$

$\Rightarrow {X_L} = WL = 100\pi \times 0.7$

$\Rightarrow 100\dfrac{{22}}{7} \times \dfrac{7}{{10}} = 220\Omega$

$\therefore z = 220\sqrt 3$

also

$\tan \phi = \dfrac{{{V_L}}}{R} = 1$

$\phi = {45^ \circ }$

$\therefore$ wattless current

$= {i_ \circ }\sin \phi$

$= \dfrac{{220\sqrt 3 }}{{220\sqrt 2 }}\sin {45^ \circ }$

$= \dfrac{1}{{\sqrt 2 }} = 0.7A$

so we get

$0.7A$

Hence option

$B$ is correct.

An alternating emf of angular frequency

$\omega$ is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency :

0%
$\dfrac{\omega}{4 }$
0%
$\dfrac{\omega}{2}$
0%
$\omega$
0%
$2\omega$

Explanation

The instantaneous values of emf and current in inductive circuit are given by

$[E={{E}_{0}}\sin \omega t]$ and

$[i={{i}_{0}}\sin \left( \omega t-\frac{\pi }{2} \right)]$ respectively.

So,

$[{{P}_{inst}}=Ei={{E}_{0}}\sin \omega t\times {{i}_{0}}\sin \left( \omega t-\frac{\pi }{2} \right)]$

$[={{E}_{0}}{{i}_{0}}\sin \omega t\left( \sin \omega t\cos \frac{\pi }{2}-\cos \omega t\sin \frac{\pi }{2} \right)]$

$[={{E}_{0}}{{i}_{0}}\sin \omega t\ \cos \omega t]$

$[=\frac{1}{2}{{E}_{0}}{{i}_{0}}\sin 2\omega t]$

$[(\sin 2\omega t=2\sin \omega t\ \cos \omega t)]$

Hence, angular frequency of instantaneous power is

$[2\omega ].$

Let

$\ell, r, c$ and

$v$ represent inductance, resistance, capacitance and voltage, respectively. The dimension of

$\dfrac{\ell}{rcv}$ is

$SI$ units will be:

0%
$[LTA]$
0%
$[LA^{-2}]$
0%
$[A^{-1}]$
0%
$[LT^2]$

Explanation

$\left[\dfrac{\ell}{r}\right] = T$

$[CV] = AT$

So,

$\left[\dfrac{\ell}{rCV}\right] = \dfrac{T}{AT} = A^{-1}$

For LCR series resonant circuit_______

0%
impedence is maximum and current is minimum
0%
impedence is minimum and current is maximum
0%
impedence is minimum and current is minimum
0%
impedence is maximum and current is maximum

The three lowest consecutive resonant frequency of a system are 50 Hz, 150 Hz and 250 Hz. The system could be

0%
a tube of air closed at both ends
0%
a tube of air open at one end
0%
a tube of air open at both ends
0%
a vibrating with fixed ends

$LCR$ circuit is connected to a voltage source

$V_c$ whose frequency can be varied. The frequency at which the voltage across the resistor is maximum, is

0%
$902$ $Hz.$
0%
$143$ $Hz.$
0%
$23$ $Hz.$
0%
$345$ $Hz.$

When 100 volt DC source is applied across a coil, a current of 1 A flows through it. When 100 V AC source of 50 Hz is applied to the same coil, only 0.5 A current flows. Calculate the inductance of the coil.

0%
$(\pi/\sqrt{3})H$
0%
$(\sqrt{3}/\pi)H$
0%
$(2/\pi)H$
0%
None of these

In the given figure the impedance of the circuit will be

0%
$120ohm$
0%
$50ohm$
0%
$60ohm$
0%
$90ohm$

In an LCR circuit

$R=100 \Omega$ when capacitance C is removed, the current lags behind the voltage by

$\pi / 3$ . when inductance L is removed, the current leads the voltage by

$\pi / 3$ . the imoedance of the circuit is

0%
50 ohm
0%
100 ohm
0%
200 ohm
0%
400 ohm

Explanation

When

$C$ is removed circuit becomes

$RL$ circuit hence

$tan \dfrac{\pi}{3} = \dfrac{X_L}{R}$ ....(i)

When

$L$ is removed circuit becomes

$RC$ circuit hence

$tan \dfrac{\pi}{3} = \dfrac{X_C}{R}$ ......(ii)
From equation (i) and (ii) we obtain

$X_L = X_C$ . This is the condition of resonance and in resonance

$Z = R = 100 \, \Omega$

In the only capacitive circuit, A.C voltage is

0%
leading by $\pi$
0%
lagging behind by $\pi$
0%
lagging behind by $\pi/2$
0%
leading by $\pi/2$

A generator with an adjustable frequency of oscillation is connect to resistance .

$R=100\Omega$ inductance

${ L }_{ 1 }=1.7\quad mL\quad and\quad { L }_{ 2 }=2.3\quad mH\quad a$ and capacitance ,

${ C }_{ 1 }=\quad 4\mu F,\quad { C }_{ 2 }=2.5\mu F$ and

${ C }_{ 3 }=3.5\mu F$ . The resonant angular frequency of the circuit is

0%
0.5 rad/s
0%
$0.5\times { 10 }^{ 4 }\quad rad/s$
0%
2 rad/s
0%
$2\times { 10 }^{ -4 }\quad rad/s$

$L$ denotes the inductance of an inductor through which a current

$I$ is flowing, then the dimensional formula of

$LI^2$ is

0%
$[MLT^{-2}]$
0%
$[ML^2T^{-2}]$
0%
$[M^2L^2T^{-2}]$
0%
not expressible in terms of $M.L.T.$

In the circuit shown how soon will the coil current reach

$\eta$ fraction of the steady - state value

0%
$\frac { L } { R }$
0%
$\frac { L } { R } \ln \frac { \eta } { ( 1 - \eta ) }$
0%
$\frac { L } { R } \ln \frac { 1 } { ( 1 - \eta ) }$
0%
$\frac { L } { R } \ln ( 1 - \eta )$

In an A.C. circuit a capacitor of

$1 \mu F$ value is connected to a source of frequency 1000 rad/sec. The value of capacitive reactance will be

0%
$10 \Omega$
0%
$100 \Omega$
0%
$1000 \Omega$
0%
$10,000 \Omega$

Explanation

$\begin{array}{l} c=1\mu f \\ w=1000\, \, rad/\sec \\ { X_{ c } }=\frac { 1 }{ { cw } } =\frac { 1 }{ { 1\times { { 10 }^{ -6 } }\times 1000 } } =1000\Omega \end{array}$

$\therefore$ captive reactance

$=1000\Omega$

Hence,

Option

$C$ is correct answer.

The reactance of a coil when used in the domestic AC power supply (220 V 50 cycles) is 100 Ohm. The self inductance of the coil is nearly

0%
3.2 Henry
0%
0.32 Henry
0%
2.2 Henry
0%
0.22 Henry

Explanation

AS we know,

Reactance

$= 2 \pi \nu L \, \Rightarrow \, 100 \Omega = 2 \times \dfrac{22}{7} \times 50 \times L$

$\therefore \, L = 0.32$ henry

The rms speed of molecules of a gas is 200 m/s at

$27 ^ { \circ } C$ and 1 atmosphere pressure.The rms speed at

$127 ^ { \circ } C$ and double pressure is

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$\sqrt { \dfrac { 800 } { 3 } m }$
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$\dfrac { 100 \sqrt { 2 } } { 3 } m / s$
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$\dfrac { 400 } { \sqrt { 3 } } m / s$
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$\dfrac { 800 } { 3 } m / s$

Equivalent inductance of the circuit shown in the figure is

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$1.0 H$
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$1.75 H$
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$0.75 H$
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$0.25 H$

Above the resonant frequency,

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the frequency of the body is less than the frequency of the external periodic force.
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the frequency of the body is greater than the frequency of the external periodic force.
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the frequency of the body is exactly equal to the frequency of the external periodic force.
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both (A) and (B)

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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