MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 15

The equation of an alternating voltage is

$V = 220\sin \left [\omega t + \dfrac {\pi}{6}\right ]$ and the equation for current is

$I = 10\sin \left (\omega t + \dfrac {\pi}{6}\right )$ . The impedance (in ohm) of the circuit is

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$11$
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$44$
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$20$
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$22$

Find the resonant frequency and

$Q-factor$ of a series

$LCR$ circuit with

$L = 3.0\ H, C = 27\mu F$ and

$R = 7.4\Omega$ .

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$111\ rad/s, 45$
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$111\ rad/s, 90$
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$55.5\ rad/s, 45$
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$55.5\ rad/s, 90$

In an inductor, the current l varies with time t as i=5A+16 (A/s) t. If induced emf in the inductor is 5 mV, the self inductance of the inductor is

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$3.75\times 10^{-3}H$
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$3.75\times 10^{-4}H$
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$3.125\times 10^{-3}H$
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$3.125\times 10^{-4}H$

A magnet is moving towards the coil along the axis and the emf induced tin the coil is 'e' if the coil also starts moving towards the magent with same speed. the induced emf will be

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$\frac{e}{2}$
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e
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2e
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4e

A condenser of capacity

$C$ is charged to a potential difference of

$V_1$ . The plates of the condenser are then connected to an ideal inductor of inductance

$L$ . The current through the inductors when then potential difference across the condenser reduces to

$V_2$

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$\frac{\mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)}{\mathrm{L}}$
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$\frac{C\left(V_{1}^{2}+V_{2}^{2}\right)}{L}$
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$\left(\frac{C\left(V_{1}^{2}-V_{2}^{2}\right)}{L}\right)^{1 / 2}$
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$\left(\frac{a v_{1}-v_{2} p^{2}}{L}\right)^{1 / 2}$

For an alternating current of frequency

$\dfrac { 5 }{ \pi } Hz$ in

$L-C-R$ series circuit with

$L-1H,C-1{ \mu }{ F }_{ R }-100 { \Omega }$ impedance is :-

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${ 100 }{ \Omega }$
0%
${ 100 }{\sqrt \pi \Omega }$
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${ 100 }{\sqrt 2\pi \Omega }$
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${ 100 }{ z\Omega }$

yf the frequency of the source e.m.f. in an AC circuit is

$n$ , the power varies wath a frequency

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$n$
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2 $n$
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$n / 2$
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zero

For a series LCR circuit at resonance, the statement which is not true:-

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Peak energy stored by a capacitor $=$ peak
energy stored by an inductor
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Average power $=$ apparent power
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Wattless current is zero
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Power factor is one

An alternating source of frequency

$\nu$ is connected to a series LCR-circuit. The graph that best represents the variation of the current

$I$ with the frequency of the applied ac is

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be

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maximum in situation (a)
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maximum in situation (b)
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maximum in situation (c)
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the same in all situation

An LCR circuit as shown in the figure is connected to a voltage soruce

$V_{ac}$ whose frequency one be varied.
The frequency, at which the voltage across the resistor is maximum, is :

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

As the frequency of an alternating current increases, the impedance of the circuit

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increase continuously
0%
decreases continuously
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remains constant
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None of these

Two inductors

${L}_{1}$ and

${L}_{2}$ are connected in parallel and a time varying current

$i$ flows as shown. The ratio of currents

${i}_{1}/{i}_{2}$ at any time

$t$ is

0%
${L}_{1}/{L}_{2}$
0%
${L}_{2}/{L}_{1}$
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$\cfrac { { L }_{ 1 }^{ 2 } }{ { \left( { L }_{ 1 }+{ L }_{ 2 } \right) }^{ 2 } }$
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$\cfrac { { L }_{ 2 }^{ 2 } }{ { \left( { L }_{ 1 }+{ L }_{ 2 } \right) }^{ 2 } }$

In a

$R - L - C$ series circuit shown, the readings of voltmeters

$V _ { 1 }$ and

$V _ { 2 }$ are

$100 V$ and

$120 V .$ Choose the correct statement

$( s ).$

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Voltage across resistor, inductor and capacitor are $50{ V },86.6{ V }$ and $206.6{ V }$ respectively.
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Voltage across resistor, inductor and capacitor are $10 V , 90 V$ and $30 V$ respectively.
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Power factor of the circuit is $\dfrac { 5 } { 13 }$
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None of these

The resistance of a coil for dc is

$10$ ohm. When ac is sent through it, its resistance would be

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$A= 10 ohm$
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$A< 10 ohm$
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$A > 10 ohm$
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$A= 5 ohm$

On decreasing the angular frequency of A.C. source used in

$L-C-R$ series circuit, the capacitive reactance____________ and inductive reactance _________ respectively.

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Increases, Decreases
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Increases, Increases
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Decreases, Increases
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Decreases, Decreases

Two long straight wires equal cross-sectional radii a are located parallel to each other in air .The distance between their axes equal b. Find the mutual capacitance of the wire per unit length under the condition b>>a

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$C=\dfrac { { 2\pi \varepsilon }_{ 0 } }{ ln\left( b/a \right) }$
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$C=\dfrac { { \pi \varepsilon }_{ 0 } }{ ln\left( a/b \right) }$
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$C=\pi /{ \varepsilon }_{ 0 }In\left( b/a \right)$
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None of these

An alternating emf is applied across a parallel combination of a resistance

$R,$ capacitance

$C$ and an inductance

$L.$ If

$I _ { R } , I _ { L } , I _ { C }$ are the currents through

$R ,$

$L$ and

$C ,$ respectively, then the diagram which correctly represents, the phase relationship among

$I _ { R } , I _ { L } , I _ { C }$ and source emf

$E,$ is given by

The equation P of an ac is represented by I=0.6 sin 100

$\pi t$ . The frequency of ac is

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$50 \pi$
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$50$
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$100 \pi$
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$100$

In a series LCR circuit

$R=200\Omega$ and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On talking out the capacities from the circuit the current lags behind the voltage by

${ 30 }^{ \circ }$ On taking out the induct or from the circuit the current leads the voltage by

${ 30 }^{ \circ }$ The power dissipated in the LCR circuit is

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305 W
0%
210 W
0%
0 W
0%
242 W

In an A.C circuit, a resistance of R ohm is connected in series with an indutance L.If phase angle between voltage and current be

${ 45 }^{ 0 }$ , the value of inductive reacatance will be

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R/4
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R/2
0%
R
0%
R/5

The average power of A.C. lost per cycle is given by?

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$\dfrac{1}{2}E_0i_0\sin\phi$
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$\dfrac{1}{2}E_0i_0\cos\phi$
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$\dfrac{1}{2}E_0i_0\tan\phi$
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$\dfrac{1}{2}E_0i_0\phi$

In the R-C series A.C. circuit, current is by $\delta$ than voltage.

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$ahead,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{1}{{\omega CR}}} \right)$
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$behind,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{1}{{\omega CR}}} \right)$
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$ahead,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{{\omega C}}{R}} \right)$
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$behind,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{{\omega C}}{R}} \right)$

A choke coil is preferred over a resistor in ac circuit because

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it is cheaper than a resistor
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it lasts longer than a resistor
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the resistor may over heat and melt
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it brings about a voltage drop with negligible power consumption

The resonance frequency of a certain RLC series circuit is

$\omega_o$ . A source of angular frequency

$2 \omega_o$ is inserted into the circuit.After transients die out, the angular frequency of current oscillation is

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$\dfrac {\omega_o}{2}$
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$\omega_o$
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$2 \omega_o$
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$1.5 \omega_o$

An A.C. circuit consists of a resistance and a choke in series. The resistance is of

$220$

$\Omega$ and choke is of

$0.7$ henry. The power absorbed from

$220$ volts and

$50$ Hz, source connected with the circuit, is?

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$120.08$ watt
0%
$109.97$ watt
0%
$100.08$ watt
0%
$98.08$ watt

In the circuit diagram shown,

$X_c= 100\Omega$ ,

$X_L= 200\Omega$ and

$R= 100\Omega$ . The effective current through the source is

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$2A$
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$\sqrt2$ A
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$0.5$ A
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$2\sqrt2$ A

An ac source of frequency

$\omega$ when fed into an RC series circuit, current is recorded to be I. If now frequency is charged to

$\frac {\omega}{3}$ (keeping voltage same) the circuit is found to be I/2.The ration of reactance to resistance at original frequency

$\omega$ is

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2
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$\frac {1}{2}$
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$\frac { 1 }{ \sqrt { 2 } }$
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$\sqrt {3\5}$

The growth of current in two series LR circuits

$A$ and

$B$ is shown in

$C$ . Let

$L_1, R_1$ and

$R_2$ be the corresponding inductor, resistor values in the two circuits. Then
1567994_4ce77f6ddd804970be8ebfd2c9f6d879.1567994-Q

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$R_1 > R_2$
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$R_1 = R_2$
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$L_1 > L_2$
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$L_1 < L_2$

In a series

$LCR$ circuit which is connected to an

$AC$ voltage source, choose the incorrect statement

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Algebraic sum of instantaneous voltage across $L, C, R$ is a variable
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$V_{Linst} + V_{Cinst} + V_{Rinst} = V_{source inst}$
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Voltage across inductor, capacitor, resistance behave as a vector
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Current is same in inductor, capacitor and resistance

In fig.

$10.43$ , which voltmeter reads zero, when

$\omega$ is equal to the resonant frequency of series

$LCR$ circuit :

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$V_1$
0%
$V_2$
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$V_3$
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None of these

In the series LCR circuit (Fig. 7.33) , the voltmeter and ammeter readings are :

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

Explanation

Here ,

$V_L = V_C$ . They are in opposite phase . Hence , they will cancel each other . Now the resultant potential difference is equal to the applied difference

$= 100\,V$

$Z = R$

$( \therefore \, X_L = X_C )$

$\therefore$

$I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{V_{rms}}{R} = \dfrac{100}{50} = 2\,A$

In an AC series circuit, the instantaneous current is zero when the instantaneous voltage is maximum. Connected to the source may be a

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pure inductor
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pure capacitor
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pure resistor
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combination of an inductor and a capacitor

In the series LCR circuit, the voltmeter and ammeter readings are respectively.

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$V=250V, I=4A$
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$V=150V, I=2A$
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$V=1000V, I=5A$
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$V=100V, I=2A$

Explanation

Ref. image 1

As, current leads the voltage across capacitor by

$\pi/2$ , while current lags from voltage in inductor by,

$\pi/2$ . Also current and voltage are in same phase across resistance

Ref. image 2

as sum of net voltage = Source voltage

Sum of

$V_L , V_C$ &

$V_R = \sqrt{V_R^2 + (V_L - V_C)^2}$

$V_L = 200 V, \, V_C = 200 V, V_R = V$

$\Rightarrow \sqrt{V^2 + (200 - 200)^2} = 100 volt$

$\Rightarrow V = 100 volt$

Also

$V_L = iX_L, V_C = i X_C \, \left\{\begin{matrix} X_L = inductive \, reactance\\ X_C = capacitive \, reactance\end{matrix}\right.$

Also

$V_L = V_C = 200$

$\Rightarrow X_L = X_C$

Now, net impedance of the circuit

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

$X_L = X_C$

$\Rightarrow Z = R$ and

$R = 50 \Omega$ given

$\Rightarrow Z = 50 \Omega$

Npw, current in the circuit

$i = \dfrac{source \, voltage}{impedance} = \dfrac{100}{Z}$

$\Rightarrow i = \dfrac{100}{50} = 2A$

Hence, value of

$V = 100\, volt \,\, A = 2A$

Option D is correct.
2045359_1705013_ans_496754f83a134c0880dc9284b59f0e04.png

The alternating current is given by the equation

$i = 10\sin \left ( 100 \pi t + \dfrac{\pi}{6} \right ).$ The current attains its first maximum at t is :

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$\dfrac{1}{600}s$
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$\dfrac{1}{50}s$
0%
$\dfrac{1}{100}s$
0%
$\dfrac{1}{300}s$

Explanation

Given,

$i = 10\sin \left(100\pi t + \dfrac{\pi}{6}\right)$

$t = 0$

$i = 10\sin \dfrac{\pi}{6} = 5$

$\dfrac{di}{dt} = 10 \cos \left(100\pi t + \dfrac{\pi}{6} \right) (100 \pi)$

$\left[ \dfrac{di}{dt}\right]_{t=0} = 10\cos \left(\dfrac{\pi}{6}\right) (100) \pi >0$

$\dfrac{di}{dt}$ at

$t = 0$ is +ve, so sin function is increasing at

$t = 0$ . therefore next maxima will be

$i = 10$

$10 = 10\sin \left(100\pi t + \dfrac{\pi}{6}\right)$

$1 = \sin \left(100\pi t + \dfrac{\pi}{6}\right)$

$\dfrac{\pi}{2} = 100\pi t + \dfrac{\pi}{6}$

$100\pi t = \dfrac{2\pi}{ 6} = \dfrac{\pi}{3}$

$t = \dfrac{1}{300}s$ .

In a radio receiver, the short wave and medium wave station are tuned by using the same capacitor but coils of difference inductance

$L_s$ and

$L_m$ respectively then

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$L_s > L_m$
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$L_s < L_m$
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$L_s = L_m$
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$None\ of\ these$

Explanation

$v=\dfrac {c}{\lambda}\Rightarrow v_m=\dfrac {c}{\lambda_m}$ and

$v_s =\dfrac {c}{\lambda_s}$

$\because \lambda_m > \lambda_s \Rightarrow v_m < v_s$
Also

$v_m=\dfrac {1}{2\pi \sqrt {L_m C}}$ and

$v_s =\dfrac {1}{2\pi \sqrt {L_s C}}$

$\Rightarrow \dfrac {v_m}{v_s}=\sqrt {\dfrac {L_s}{L_m}}\Rightarrow L_s < L_m$

How can you decrease current in an alternating current circuit without any loss of power?

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By using pure inductor
0%
By using pure resistance
0%
By using resistance and inductor
0%
By using resistance and capacitor

In an A.C. circuit, a resistance of

$3 \Omega$ , an inductance coil of

$4 \Omega$ and a condenser of

$8 \Omega$ are connected in series with an A.C. source of

$50$ volts (R.M.S). The average power loss in the circuit will be

0%
$600$ watt
0%
$500$ watt
0%
$400$ watt
0%
$300$ watt

In an LCR series circuit , at resonance

0%
the current and voltage are in phase
0%
the impedance is maximum
0%
the current is minimum
0%
the quality factor is independent of $R$
0%
the current leads the voltage by $\dfrac{\pi}{2}$

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 15 - 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 15 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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