MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 9

Figure shows a series LCR circuit connected to a variable frequency 230 V source.
The source frequency which drives the circuit in resonance is

0%
4 Hz
0%
5 Hz
0%
6 Hz
0%
8 Hz

Explanation

Here,

$L = 5H, C = 80\mu F = 80\times 10^{-6}F, R = 40\Omega$

$V_{rms}= 230 V$

The resonant angular frequency is

$\omega_r = \dfrac{1}{\sqrt{LC}} = \sqrt { \dfrac{1}{5\times 80\times 10^{-6}}} = 50 \, rad \, s^{-1}$

$\therefore \, \nu= \dfrac{\omega_r}{2\pi} = \dfrac{50}{2\pi} = 8 Hz$

A series LCR circuit with R = 22

$\Omega$ , L = 1.5 H and C = 40

$\mu$ F is connected to a variable frequency 220 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

0%
2000 W
0%
2200 W
0%
2400 W
0%
2500 W

Explanation

When the frequency of the supply equals to the natural frequency of circuit, resonance occurs.

$\therefore \, Z \, = \, R \, = 22 \, \Omega \, and \, I_{rms} \, = \, \dfrac{V_{rms}}{Z} \, = \dfrac{220}{22} \, = \, 10 \, A$
Average power transferred per cycle,

$P \, = \, V_{rms} \, I_{rms} \, cos \, 0^{\circ} \, = \, 220 \, \times \, 10 \, \times \, 1 \, = \, 2200 \, W$

To reduce the resonant frequency in an

$LCR$ series circuit with a generator.

0%
The generator frequency should be reduced
0%
Another capacitor should be added in parallel to the first
0%
The iron core of the inductor should be removed
0%
Dielectric in the capacitor should be removed

Explanation

Resonant frequency in series LCR circuit is

$v_r \, = \, \dfrac{1}{2\pi \sqrt {LC}}$
If capacitance C increases the resonant frequency will reduce, which can be achieved by adding another capacitor in parallel to the first.

An LC circuit contains a 20 mH inductor and a 50 .

$\mu$ F capacitor with an initial charge of 10 mC. The resistance of the circuit is.negligible. Let the instant at which the circuit which is closed be t =At what time the energy stored is completely magnetic?

0%
t = 0
0%
t = 1.54 ms
0%
t = 3.14 ms
0%
t = 6.28 ms

Explanation

At time

$t \, = \, \dfrac{T}{4},$ energy stored is completely mangnetic.

$\therefore \, Time \, t \, = \, \dfrac{T}{4} \, = \, \dfrac{2\pi\sqrt{LC}}{4} \, \Rightarrow \, t \, = \, \dfrac{\pi\sqrt{20 \, \times \, 10^{-3} \, \times \, 50 \, \times \, 10^{-6}}}{2}$

$t \, = \, \dfrac{\pi\sqrt{1000 \, \times \, 10^{-9}}}{2} \, = \, \dfrac{\pi\sqrt{1 \, \times \, 10^{-6}}}{2} \, = \, 1.57 ms$

In the given circuit, initially

${ K }_{ 1 }$ is closed and

${ K }_{ 2 }$ is open. Then

${ K }_{ 1 }$ is opened and

${ K }_{ 2 }$ is closed. If

${ q }_{ 1 }^{ \prime }$ and

${ q }_{ 2 }^{ \prime }$ are charges on

${ C }_{ 1 }$ and

${ C }_{ 2 }$ and

${ V }_{ 1 }$ and

${ V }_{ 1 }$ are the voltage respectively, then

0%
charge on ${ C}_{ 1 }$ gets redistributed such that ${ V }_{ 1 }$ = ${ V }_{ 2 }$
0%
charge on ${ C}_{ 1 }$ gets redistributed such that ${ q }_{ 1 }^{ \prime }$ = ${ q }_{ 2 }^{ \prime }$
0%
charge on ${ C}_{ 1 }$ gets redistributed such that ${ C }_{ 1 }{ V }_{ 1 }$ = ${ C }_{ 2 }{ V }_{ 2 }$ = ${ C }_{ 1 }V$
0%
charge on ${ C}_{ 1 }$ gets redistributed such that ${ q }_{ 1 }^{ \prime }$ + ${ q }_{ 2 }^{ \prime }$ = 2q

Explanation

From the figure, when

${K}_{1}$ is closed and

${K}_{2}$ is open,

${C}_{1}$ is charged to potential

$V$ acquiring a total charge

$q={C}_{1} V$
When

${K}_{1}$ is open and

${K}_{2}$ is closed, battery is cut off

${C}_{1}$ and

${C}_{2}$ are in parallel. The charge on

${C}_{1}$ is shared between

${C}_{1}$ and

${C}_{2}$ such that

${V}_{1}={V}_{2}$
As there is no loss of charge

${q}_{1}'+{q}_{2}'=q$
1096026_945107_ans_90894cba1eb14b3e9b4d567fd5b6a197.png

Two inductors of inductance L each are connected in series with opposite magnetic fluxes. The resultant inductance is (Ignore mutual inductance)

0%
zero
0%
L
0%
2L
0%
3L

Explanation

When the two inductors are connected in series to one another, they are simply been added up and the direction of the magnetic flux does not have any affect on the net inductance of the two inductors.

$L'=L_1+L_2$

$L'=L+L$

$L'=2L$

So, option

$C$ is correct.

A voltage of peak value 283 V and varying frequency is applied to series LCR combination in which R = 3

$\Omega$ , L = 25 mH and C = 400

$\mu$ F. Then the frequency (in Hz) of the source at which maximum power is dissipated in the above is

0%
51.5
0%
50.7
0%
51.1
0%
50.3

Explanation

Here,

$V_0 \, = \, 283 \, V, \, R \, = \, 3\Omega, \, L \, = \, 25 \, \times \, 10^{-3} \, H$

$C \, = \, 400 \, \mu F \, = \, 4 \, \times \, 10^{-4} F$
Maximum power is dissipated at resonance, for which

$\nu \, = \, \dfrac{1}{2\pi \sqrt{LC}} \, = \, \dfrac{1 \, \times \, 7}{2 \, \times \, 22 \, \sqrt{25 \, \times \, 10^{-3} \, \times \, 4 \, \times \, 10^{-4}}}$

$= \, \dfrac{7 \, \times \, 10^3}{44\sqrt{10}} \, = \, 50.3 \, Hz$

The natural frequency

$(\omega_0)$ of oscillations in LC Circuit is given by:

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

Explanation

Let natural frequency

$={\omega }_{ 0 }$

Condition -

${ X }_{ C }={ X }_{ L }$

$\dfrac { 1 }{ {\omega }_{ 0 }C } ={ \omega }_{ 0 }L$

$\Rightarrow { \omega }_{ 0 }^2=\dfrac { 1 }{ LC }$

$\Rightarrow { \omega }_{ 0 }=\sqrt { \dfrac { 1 }{ LC } }$

A transmitter transmits at a wave length of 300 meters. A capacitor of a capacitance 9.6

$\mu$ F is being used. The value of the inductance for the resonant circuit is approximately

0%
2.5 mH
0%
2.5 $\mu$ H
0%
2.5 nH
0%
2.5 pH

Explanation

V = f

$\lambda$ where V is velocity and f is frequency

f = 1/

$2 \pi \sqrt{LC}$

$L=\dfrac{({3 \times 10^2})^2}{4 \pi^2 \times 9.6 \times 10^{-6} \times ({3 \times 10^8)}^2}$

$L=2.5 \times 10^{-9}$

$L=2.5 nH$

$50W, 100V$ lamp is to be connected to an AC mains of

$200V, 50Hz$ . What capacitance is essential to be put in series with the lamp?

0%
9.2
0%
8
0%
10
0%
12

Explanation

Power

$=50w$

$V=100v\\I=\cfrac{P}{V}=.5A$

The phase: potential across the capacitor must be lag by

$90°$

$(200)^2=100^2+V_2^2\\V_c=173.2V\\V_c=I_c\times x_e$

$x_e=346.4\Omega\\ X_c=\cfrac{1}{2\pi f_c}\\C=9.29 {\mu} f$

A source of constant voltage

$V$ is connected to a resistance

$R$ and two ideal inductors

$L_{1}$ and

$L_{2}$ through a switch

$S$ as shown. There is no mutual inductance between the two inductors. The switch

$S$ is initially open. At

$t=0$ , the switch is closed and current beings to flow. Which of the following options is/are correct?

0%
After a long time, the current through $L_{1}$ will be $\dfrac {V\ \ L_{2}}{R\ L_{1}+L_{2}}$
0%
After a long time, the current through $L_{2}$ will be $\dfrac {V\ \ L_{2}}{R\ L_{1}+L_{2}}$
0%
The ratio of the currents through $L_{1}$ and $L_{2}$ is fixed at all times $(t\ >\ 0)$
0%
At $t=0$ , the current through the resistance $R$ is $\dfrac {V}{R}$

Explanation

After a long time current through

${ L }_{ 1 }$ will be,

$\cfrac { V }{ R } \times \cfrac { { L }_{ 2 } }{ { L }_{ 1 }+{ L }_{ 2 } }$

After a long time current through

${ L }_{ 2 }$ will be ,

$\cfrac { V }{ R } \times \cfrac { { L }_{ 1 } }{ { R }_{ 2 } }$

Time can be inferred from that

${ I }_{ L1 }:{ I }_{ L2 }={ L }_{ 2 }:{ L }_{ 1 }$

The ratio of current is fixed all the time in

${ L }_{ 1 }$ and

${ L }_{ 2 }$ .

If R is resistance, L is inductance, C is capacitance, H is latent heat, and s is specific heat, then match the quantity given in Column I with the dimensions given in Column II.

Column I	Column II
i. LC	a. $MA^{-2}L^2T^{-2}$
ii. LR	b. $ML^2T^{-3}A^{-2}$
iii. H	c. $T^2$
iv. R	d. $M^2L^4T^{-5}A^{-4}$

0%
i. c, ii.b, iii.a, iv.d
0%
i. a, ii.d, iii.a, iv.b
0%
i. c, ii.d, iii.a, iv.b
0%
i. c, ii.d, iii.a, iv.d

Explanation

$L = \dfrac{\phi}{L} = \dfrac{BA}{L} = \dfrac{FA}{q v L}= \dfrac{(MLT^{-2})(L^2)}{(AT)(LT^{-1}) (A)}$

$[L] = ML^2 T^{-2} A^{-2}$

$[R] = ML^2 T^{-3} A^{-2}$

i) LC we know

$w = \dfrac{1}{\sqrt{LC}}$

$[LC] = \dfrac{1}{[w^2]} = \dfrac{1}{(T^{-1})^2} = T^2$

ii)

$[LR] = ML^2 T^{-2} A^{-2} \times ML^2 A^{-2} T^{-3}$

$= M^2 L^4T^{-5} A^{-4}$

iii)

$H = [L] = ML^2 T^{-2} A^{-2}$

iv)

$R = ML^2 T^{-3} A^{-2}$

$100mH$ inductor, a

$25\mu F$ capacitor and a

$15\Omega$ resistor are connected in series to a

$120V$ ,

$50Hz$ ac source. Calculate
(a) impedance of the circuit at resonance
(b) current at resonance
(c) resonant frequency

0%
(a) $16\Omega$ (b) $9A$ (c) $100Hz$
0%
(a) $15\Omega$ (b) $8A$ (c) $100Hz$
0%
(a) $15\Omega$ (b) $8A$ (c) $120Hz$
0%
(a) $15\Omega$ (b) $7A$ (c) $100Hz$

Explanation

$(a)$ Reason occurs when

$X_C=X-L$

$\therefore z=15\Omega\\(b)I=\cfrac{V}{R}=\cfrac{120}{15}=8A\\(c)X_C=X_L\\\cfrac{1}{\omega_C}=\omega_L\\\omega^2=\cfrac{1}{LC}\\ \omega=\cfrac{1}{\sqrt{LC}}\\f=\cfrac{2\pi}{\sqrt{LC}}\times\cfrac{1}{2\pi}\\=100Hz$

Find the maximum value of current when inductance of two henry is connected to

$150V,50$ cycle supply

0%
$\cfrac { 3 }{ 2\sqrt { 2 } A }$
0%
$\cfrac { 5 }{ 2\sqrt { 2 } A }$
0%
$\cfrac { 6 }{ 2\sqrt { 2 } A }$
0%
$\cfrac { 7 }{ 2\sqrt { 2 } A }$

Explanation

$V_L=I\omega L\\I=\cfrac{150}{\sqrt2}\times\cfrac{1}{50\times2}\\ \cfrac{150}{2\sqrt2\times50}=\cfrac{3}{2\sqrt2}A$

$10\Omega$ resistance,

$5mH$ coil and

$10\mu F$ capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency:

0%
is halved
0%
is doubled
0%
remains unchanged
0%
is quadrupled

Explanation

At resonance

$f=\cfrac{1}{2\pi\sqrt{LC}}$

Hence no effect of R. The resonance frequency remains same when R is changed.

A resistance

$(R)=12\Omega$ ; inductance

$(L)=2$ henry and capacitive reactance

$C=5mF$ are connected in series to an ac generator, then:

0%
at resonance, the circuit impedance is zero
0%
at resonance, the circuit impedance is $12\Omega$
0%
the resonance frequency of the circuit is $1/2\pi$
0%
at resonance, the inductive reactance is less than the capacitive reactance

Explanation

At resonance

$X_L=X_C\\ \therefore z=R$

So, circuit impedence is

$12\Omega$

Find the impedance of AC circuit at resonance shown in the adjacent figure.

0%
$100\sqrt{2}\Omega$
0%
$200\Omega$
0%
$100\Omega$
0%
$200\sqrt{2}\Omega$

An ac source of angular frequency

$\omega$ 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

$\omega/3$ (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance at the original frequency

$\omega$ will be:

0%
$\sqrt { \cfrac { 3 }{ 5 } }$
0%
$\sqrt { \cfrac { 5 }{ 3 } }$
0%
$3/5$
0%
$5/3$

Explanation

According to given problem

$I=\cfrac { V }{ Z } =V/{ \left[ { R }^{ 2 }+{ (1/C{ \omega })^{ 2 } }^{ } \right] }^{ 1/2 }...(1)\quad$
and

$\cfrac { I }{ 2 } =\cfrac { V }{ { \left[ { R }^{ 2 }+{ (3/C{ \omega }^{ 2 } }^{ } \right] }^{ 2 } } .....(2)$
substituting the value of

$I$ from equation (1) in (2)

$4\left( { R }^{ 2 }+\cfrac { 1 }{ { C }^{ 2 }{ \omega }^{ 2 } } \right) ={ R }^{ 2 }+\cfrac { 9 }{ { C }^{ 2 }{ \omega }^{ 2 } } \quad$ ie.,

$\cfrac { 1 }{ { C }^{ 2 }{ \omega }^{ 2 } } =\cfrac { 3 }{ 5 } { R }^{ 2 }\quad$
so that

$\cfrac { X }{ R } =\cfrac { (1/C\omega ) }{ R } =\cfrac { { \left[ (3/5){ R }^{ 2 } \right] }^{ 1/2 } }{ R } =\sqrt { \cfrac { 3 }{ 5 } }$

Determine the characteristic impedance of a transmission line which has a capacitance of 35pF/ft and an inductance of 0.25

$\mu H/ft$

0%
84.5
0%
95
0%
65
0%
66

Explanation

Given: C = 35pF/ft; L = 0.25

$\mu H/ft$
Using the equation relation

${ Z }_{ 0 }$ , L and C,

${ Z }_{ 0 }=\sqrt { L/C }$
Substituting numerical values,

${ Z }_{ 0 }=\sqrt { \dfrac { 0.25\times { 10 }^{ -6 } }{ 35\times { 10 }^{ -12 } } } =84.5\Omega$

In an LCR circuit having

$L=8.0H,C-0.5\mu F$ and

$R=100\Omega$ in series, the resonance frequency is:

0%
$600$ rad/sec
0%
$600Hz$
0%
$500$ rad/sec
0%
$500Hz$

Explanation

$X_L=X_C\\2\pi fL=\cfrac{1}{2\pi fC}\\ f^2=\cfrac{1}{4\pi^2C\times L}\\=79Hz\\ \omega=2\pi\times79\\=500 rad/sec$

The reciprocal of impedance is called

0%
reactance
0%
admittance
0%
inductance
0%
susceptance

In LCR circuit the capacitance is changed from C to 4C. For the same resonant frequency, the inductance should be changed from

$L$ to

0%
$2L$
0%
$L/2$
0%
$L/4$
0%
$4L$

In a series

$LCR$ circuit

$K=200\ \Omega$ and the voltage and frequency of the main supply are

$220\ V$ and

$50\ Hz$ respectively. On taking out the capacitor from the circuit, the current leads the voltage by

${30}^{o}$ . On taking out the indicator from the circuit the current leads the voltage by

${30}^{o}$ . The power dissipated in the

$LCR$ circuit is :

0%
$342\ W$
0%
$305\ W$
0%
$209\ W$
0%
$242\ W$

Explanation

$P=\cfrac { { V }_{ rms }^{ 2 } }{ R } cos\phi =\cfrac { { 220 }^{ 2 } }{ 200 } cos30°=209W$

In the series

$LCR$ circuit as shown in figure, the voltmeter and ammeter readings are:

0%
$V=100\ volt, I=2\ amp$
0%
$V=100\ volt, I=5\ amp$
0%
$V=1000\ volt, I=2\ amp$
0%
$V=300\ volt, I=1\ amp$

Explanation

As, Voltage across capacitor

$=$ Voltage across inductor

Circuit is in Resonance,

$I=\cfrac{V}{R}$

$\Rightarrow I=\cfrac{100}{50}=2$

$A$

$V=V_{in}=100$

$V$

For series L-C-R AC circuit shown in figure, the readings of

${V}_{2}$ and

${V}_{3}$ are same and each equal to

$100\ V$ . Then

0%
The reading ${V}_{1}$ is $200\ V$
0%
The reading of ${V}_{2}$ is $0$
0%
The circuit is in resonant mode
0%
The inductive and capacitive reactance are equal

Explanation

${ V }_{ 2 }={ V }_{ 3 }\\ { V }_{ 2 }-{ V }_{ 3 }=0\\ According\quad to\quad Phasor\quad diagram\quad \\ V=\sqrt { { V }_{ 1 }^{ 2 }+{ \left( { V }_{ 2 }-{ V }_{ 3 } \right) }^{ 2 } } \\ V={ V }_{ 1 }\\ { V }_{ 1 }=200V\\ In\quad this\quad condition\quad PF\quad is\quad unity\quad an\quad circuit\quad is\quad in\quad resonant\quad mode\quad and\quad clearly\quad \\ Inductive\quad voltage\quad { V }_{ 2 }={ V }_{ 3 }\quad capacitive\quad voltage\\ i{ X }_{ L }=i{ X }_{ C }\\ { X }_{ L }={ X }_{ C }\\ Inductive\quad reactance\quad =\quad Capactive\quad Reactance$

973503_1022792_ans_4c0c219621b54700ae975a9013d6c334.png

The characteristic impedance of a co-axial cable is of order of:

0%
$50\Omega$
0%
$200\Omega$
0%
$270\Omega$
0%
none of these

Explanation

Characteristic impedence for coax cable has main standard of

$50r$ , because it gives the minimum low for a given weight.

If instantaneous current in a circuit is given by

$l = (2 + 3 sin$

$\omega t)A$ , then the effective value of resulting current in the circuit is:

0%
$\sqrt\frac{17}{2}A$
0%
$\sqrt\frac{2}{17}A$
0%
$\sqrt\frac{3}{\sqrt2}A$
0%
$3\sqrt2A$

Explanation

$I_{effective}=\sqrt { { 2 }^{ 2 }+({ \cfrac { 3 }{ \sqrt { 2 } } ) }^{ 2 } }$

$=\sqrt { 4+\cfrac { 9 }{ 2 } } =\sqrt { \cfrac { 17 }{ 2 } } A$

Which of the following option is correct for an ideal capacitor connected to a sinusoidal voltage source over a complete cycle?

0%
Neither the average power nor the average current is zero
0%
Average voltage is zero but the average power is non zero
0%
Both the average power and average current is zero
0%
Average power is zero, but the average current is non zero

A resistance of $10\Omega$ , a
capacitance of $0.1\mu F$ and an inductance of $2mH$ are connected in
series across a source of alternating
emf of variab frequency. At what frequency does maximum current flow?

0%
$11.25kHz$
0%
$23.76kHz$
0%
$35.46kHz$
0%
$46.72kHz$

The frequancy of oscillation of current in the inductor is-

0%
$\dfrac { 1 }{ 3\sqrt { LC } }$
0%
$\dfrac {1}{6\pi \sqrt {LC}}$
0%
$\dfrac { 1 }{\sqrt { LC } }$
0%
$\dfrac {1}{2\pi \sqrt {LC}}$

Explanation

Frequency of oscilation

$=\cfrac { 1 }{ \sqrt { 3C(3L) } } =\cfrac { 1 }{ 3\sqrt { LC } }$
1171676_1024264_ans_4233520337954ef89169d4fe11238971.png

The equation of an alternating voltage is

$V=100\sqrt {2}\sin {100\pi t}$ volt. The RMS value of voltage and frequency will be respectively

0%
$100V,50Hz$
0%
$50V,100Hz$
0%
$150V, 50Hz$
0%
$200V, 50Hz$

Explanation

Angular frequency is nothing but

$coefficient$ of

$t$ in the equation,

so we have

$\omega =100\pi$ or

$2\pi f=100\pi$ , we get

$f=50Hz$

rms voltage is

$\dfrac{\text{peak voltage}}{\sqrt {2}}=100 \sqrt{2}/ \sqrt{2}=100 V$ .

Option A is correct.

An alternating power supply of

$220 V$ is applied across a series circuit of resistance

$10\sqrt{3}\Omega$ , capacitive reactance

$40 \Omega$ and inductive reactance

$30 \Omega$ . The respective current in the circuit for zero and infinite frequencies are

0%
$2A,\dfrac{1}{2}A$
0%
$0A,10A$
0%
$10 A,0A$
0%
$0A,0A$

Explanation

Case I: When the frequency is zero.

$f=0$ implies

$\omega = 0$

Therefore, the capacitive reactance becomes infinite

$X_C= \dfrac{1}{\omega C} \to \infty$

As a result the impedance(Z) of the circuit becomes infinite and the current becomes zero.

Case I: When the frequency is infinite.

$f \to \infty$ implies

$\omega \to \infty$

Therefore, the inductive reactance becomes infinite

$X_L= \omega L \to \infty$

As a result the impedance(Z) of the circuit becomes infinite and the current becomes zero.

A coil has resistance

$30 \Omega$ and inductive reactance

$20 \Omega$ at

$50 Hz$ frequency.If an ac source of

$200 V,100Hz$ is connected across the coil ,the current in the coil in the coil will be

0%
$\dfrac{20}{\sqrt{13}}A$
0%
$2.0 A$
0%
$4.0A$
0%
$8.0A$

Explanation

Given,

$X_L=20\Omega$

frequency

$f=50Hz$

So,

$2\pi fL=20$

$\Rightarrow 2\pi\times 50L=20\Rightarrow L=\dfrac{1}{5\pi}$

When the frequency is

$100Hz$

$X_L=2\pi\times\dfrac{1}{5\pi}\times100=40\Omega$

Impedance

$Z= \sqrt{R^2+X_L^2}=\sqrt{900+1600}=50\Omega$

$V=200volt$

Thus the current

$I=\dfrac{200}{50}=4A$

Figure shows a long straight conductor carrying current i. A square loop of side a is kept at a distance r from it. Which of the following is correct?

0%
Mutual Inductance $M = \dfrac{\mu_0 a}{2 \pi} ln(1 + \dfrac{a}{r})$
0%
If loop is pulled with constant speed v along x-axis, emf induced at the instant is $\dfrac{\mu_0 i a^2 v}{2 \pi r (r + a)}$
0%
If loop is pulled with constant speed v, no emf is induced
0%
If i increases with time, the loop is replied away from the straight conductor.

A L - C resonant circuit contains a 200 pF capacitor and a

$100 \mu$ H inductor it is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is

0%
377 mm
0%
266 m
0%
377 cm
0%
3.77 cm

Explanation

Given that,

$C=200pF$

$C=200\times {{10}^{-12}}F$

$L=100\mu H$

$L=100\times {{10}^{-6}}H$

We know that,

Frequency of wave = $\nu =\dfrac{1}{2\pi \sqrt{LC}}$

$\nu =\dfrac{1}{2\times 3.14\sqrt{200\times {{10}^{-12}}\times 100\times {{10}^{-6}}}}$

$\nu =\dfrac{1}{2\times 3.14\times 1.4142\times {{10}^{-7}}}$

Now, the wavelength is

$\lambda =\dfrac{3\times {{10}^{8}}}{\nu }$

$\lambda =\dfrac{3\times {{10}^{8}}}{\dfrac{1}{2\times 3.14\times 1.4142\times {{10}^{-7}}}}$

$\lambda =3\times {{10}^{8}}\times 2\times 3.14\times 1.4142\times {{10}^{-7}}$

$\lambda =26.643\times 10$

$\lambda =266.43\,m$

Hence, the wavelength of radiated electromagnetic wave is $266$ m.

In a series L-C circuit , if

$L= 10^{-3}$ H and

$C=3 \times 10^{-7}$ F is connected to a

$100V-50Hz$ a.c. source, the impedance of the circuit is

0%
$\cfrac{10^5}{3\pi} - 10\pi$
0%
$0.1\pi - 3\times 10^{-5}\pi$
0%
$\cfrac{10^5}{3\pi} - \cfrac{\pi}{10}$
0%
None of these

Explanation

Impedance

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

Where

$X_L = \omega L$

$X_C = \cfrac{1}{\omega C}$

and

$\omega = 2 \pi f$

$=2 \pi \times 50 = 100\pi$

$\therefore X_L = 100 \pi \times 10^{-3} = \cfrac{\pi}{10}$

$X_C = \cfrac{1}{100 \pi \times 3 \times 10^{-7} } = \cfrac{10^5}{3\pi}$

$R=0$

$\therefore Z=\sqrt{(\cfrac{\pi}{10} - \cfrac{10^5}{3\pi})^2}$

$= \cfrac {10^5}{3\pi} - \cfrac{\pi}{10}$ As, (

$\cfrac {10^5}{3\pi} > \cfrac{\pi}{10}$ )

The sinusoidal potential difference

$V_1$ shown in figure applied across a resistor

$R$ produces heat at a rate

$W$ . what is the rate of heat dissipation when the square waves potential different

$V_2$ as shown in figure is applied across the resistor?

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

In L-C-R series circuit consists of a resistance of 10

$\Omega$ a capacitor of reactance 6.0

$\Omega$ and an inductor coil. The circuit is found to resonate when put across a 300 V, 100 Hz supply. The inductance of coil is (Take

$\pi$ = 3)

0%
0.1 H
0%
0.01 H
0%
0.2 H
0%
0.02 H

In a series LCR circuit,the inductive reactance is twice the resistance and the capacitance reactance is

${\frac{1}{3}^{rd}}$ the inductive reactance. The power factor of the circuit is:

0%
$0.5$
0%
$0.6$
0%
$0.8$
0%
$1$

Explanation

$\begin{array}{l}\omega L = 2R\\\frac{1}{{\omega C}} = \frac{1}{3}\left( {\omega L} \right)\\\omega L - \frac{1}{{\omega C}} = 2R - \frac{{2R}}{3} = \frac{{4R}}{3}\\\tan \phi = \frac{{4R}}{3} \times \frac{1}{R} = \frac{4}{3}\\\cos \phi = \frac{1}{{\sqrt {1 + {{\tan }^2}\phi } }} = \frac{1}{{\sqrt {1 + \frac{{{4^2}}}{{{3^2}}}} }} = \frac{3}{5} = 0.6\end{array}$

For the series LCR circuit shown in the figure, what is the resonance frequency and the amplitude of the current at the resonating frequency.

0%
$2500 rad s^{-1}$ and $5\sqrt 2 A$
0%
$2500 rad s^{-1}$ and $5A$
0%
$2500 rad s^{-1}$ and $\frac{5}{\sqrt 2} A$
0%
$25 rad s^{-1}$ and $5\sqrt 2 A$

Explanation

$\textbf{Hint: At resonance current and voltage are in same phase}$

$\textbf{Step1: Find resonating angular frequency}$

$\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{8\times 10^{-3}\times 20 \times 10^{-6}}} = 2500 rad/sec$

$\textbf{Step2: Find amplitude of current at resonance}$

Resonance current

$= \dfrac{V}{R} = \dfrac{220}{44} = 5A$

$\textbf{Hence option B correct}$

In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected. When switch

$'S'$ is closed then there is a phase difference of

$\dfrac {\pi}{4}$ between current and applied voltage and voltage across resister is

$\dfrac {100}{\sqrt {2}}V$ . When switch is open current and applied voltage are in same phase. Neglecting resistance of connecting wire answer the following questions:
Resonance frequency of circuit is

0%
$50\ Hz$
0%
$25\ Hz$
0%
$75\ Hz$
0%
Data insufficient for calculation

In the L -C circuit shown in figure,the current is in direction shown in the figure and charges on the capacitor plates have sign shown in the figure A this time:-

0%
i as well as Q increasing
0%
i as well as Q decreasing
0%
i is increasing but Q is decreasing
0%
i decreasing but Q is increasing

The voltage across a pure inductor is represented in figure. Which one of the following curves in the figure will represent the current?

$2$

$F$ capacitor is initially charged to

$20$

$V$ and then shorted across a

$5$

$mH$ inductor. The angular frequency of oscillation is.

0%
$1000$ ${rad/s}$
0%
$100$ ${rad/s}$
0%
$10000$ ${rad/s}$
0%
$10$ ${rad/s}$

Explanation

$\omega_o=\cfrac{1}{\sqrt{LC}}$

$\omega_o=\cfrac{1}{\sqrt{2\times 5\times 10^{-3}}}$

$\omega_o=10$

${rad/sec}$

A inductance capacitance circuit is in the state of resonance. if the

$C = 0.1 \mu F$ and

$L = 0.25$ Henry. Neglecting ohmic resistance of circuit what is the frequency of oscillations

0%
$1007 \ Hz$
0%
$100 \ Hz$
0%
$109 \ Hz$
0%
$500 \ Hz$

Explanation

From the formula, the frequency of oscillation is

$f=\dfrac{1}{2\pi\sqrt{LC}}$

$=\dfrac{1}{2\pi\sqrt{0.25\times 0.1\times 10^{-6}}}$

$=1007\,Hz$

In a series LCR circuit

$V_l = 3V_R$ and

$V_C = 2V_R$
Volatge across capacitance

0%
110 V
0%
220 V
0%
330 V
0%
440 V

In the given circuit the maximum current can be

0%
$\dfrac{E_0}{R}$
0%
$\dfrac{E_0}{\omega L}$
0%
$E_0\omega C$
0%
$\frac{E_0}{\Bigg[ \Big\lgroup \omega L - \dfrac{1}{\omega C} \Big\rgroup^2 + R^2 \Bigg ]^{1/2}}$

Impedance of the following circuit will be:

0%
$150\ \Omega$
0%
$200\ \Omega$
0%
$250\ \Omega$
0%
$340\ \Omega$

In case of

$AC$ circuits, impedance is given by the ratio of
I. Peak values of voltage and current
II. rms values of voltage and current
III. Instantaneous values of voltage and current.

0%
I only
0%
II only
0%
I and II
0%
I, II and III

A loop is formed by two parallel conduction connected by a solenoid with inductance

$L = 10^3 H$ and a conducting rod of mass

$m=1 gm$ which can freely slide (without friction) over a pair of conductors. The conductors are located in a horizontal plane in a uniform magnetic field

$B = 4 T$ in the direction shown. The distance between conductors is equal to

$t = 1 m.$ At the moment

$t = 0$ , the rod is imparted an initial velocity

$V_0$ to the right. Determine angular frequency (in rad/s) of oscillation of rod.

0%
$4$
0%
$6$
0%
$8$
0%
$10$

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.