Processing math: 31%

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

A sinusoidally varying potential difference has amplitude 170VWhat is its rms value?
  • 170V
  • 170V
  • 0
  • 120V
  • 170V
The reactance of a capacitor of capacitance C is X. If both the frequency and capacitance be doubled, then new reactance will be 
  • X
  • 2X
  • 4X
  • X4
In an A.C circuit, the resistance R=0.2Ω. At a certain instant, VAVB=0.5V, I=0.5A, and the current is increasing at the rate of  ΔIΔt=8A/s. The inductance of the coil is:

22742_1690eb8ece93443db207082050d1e6ba.png
  • 0.05H
  • 0.1H
  • 0.2H
  • none of these
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
  • 35
  • 53
  • 35
  • 53
When a series combination of L and R are connected with a 10V, 50HZ 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
  • 5Ω
  • 10Ω
  • 15Ω
  • 20Ω
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:
  • 10V
  • 100V
  • 50V
  • 500V
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
  • R
  • 2R
  • 3R
  • 4R
At a frequency ω0 the reactance of a certain capacitor equals that of a certain inductor. If frequency is changed to 2 ω0 . The ratio of reactance of the inductor to that of the capacitor is
  • 4 : 1
  • 2:1
  • 1:22
  • 1 : 2
An ideal inductor takes a current of 10A when connected to a 125 V, 50 Hz AC supply. A pure resistor across the same source takes 12.5 A. If the two are connected in series across a 1002V, 40 Hz supply, the current through the circuit will be
  • 10A
  • 12.5A
  • 20A
  • 25A
A circuit containing resistance R1, Inductance L1 and capacitance C1 connected in series resonates at the same frequency n as a second combination of  R2, L2 andC2. If the two are connected in series, then the circuit will resonate at:
  • n
  • 2n
  • L2C2L1C1
  • L1C1L2C2
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 K1 or K2 alone closed, the ammeter reading is I. If the source is changed to 100 V, 100 Hz, the ammeter reading with K1 alone closed and with K2 alone closed will be respectively
22757.png
  • I,I2
  • I,2I
  • 2I,I
  • 2I,I2
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 π2 = 10)
  • 4μF
  • 8μF
  • 1μF
  • 2μF
A 120V, 60Hz a.c. power is connected 800Ω 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
  • 8V
  • 102V
  • 63V
  • 55V
An alternating e.m.f. of E=100sin(100πt ) is connected to a choke of negligible resistance and current oscillations of amplitude 1 A are produced. The inductance of the choke should be
  • 100H
  • 1π H
  • 1H
  • πH
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
  • 5 ampere
  • 0.5 ampere
  • 0.7 ampere
  • 7 ampere
In the given circuit the readings of the voltmeter V_1 and the ammeter A are:
22776_8b23988a7c28401694d58d990970308c.png
  • 220 V, 2.2 A
  • 110 V; 1.1A
  • 220 V, 1.1 A
  • 110 V; 2.2 A
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.
  • A and R both are true and R is correct explanation of A
  • A and R both are true but R is not the correct explanation of A
  • A is true R is false
  • A is false and R is true
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
  • 1.06 A
  • 10.6 A
  • 0.106 A
  • 100.6 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
  • 904\sqrt{2} V
  • 904 V
  • 904/\sqrt{2} V
  • 452 V
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:

22794.png
  • 1  A
  • 3  A 
  • 4  A
  • 5  A
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 :
44745_d43dbfa042d142a7afcc354005514c3b.jpg
  • Steady state current leads the voltage by 30^{\circ}
  • The instantaneous voltages across the resistance and inductance become equal at t = \dfrac{1}{300}s
  • Voltage leads the current by 30^{\circ}
  • 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)
44770.jpg
  • 6e^{-20t}\ \text{from B to A}
  • 3e^{-40t}\ \text{from A to B}
  • 3e^{-40t}\ \text{from B to A}
  • 6e^{-20t}\ \text{from A to B}
The frequency at which the impedance of the circuit becomes maximum is
  • \dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}+\dfrac{R^{2}}{L^{2}}}
  • \dfrac{1}{2\pi }\dfrac{1}{\sqrt{LC}}
  • \dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}
  • \dfrac{1}{2\pi }\dfrac{R}{L}
A series RLC circuit is made as shown in the figure with an AC source of 60 V, 20 Hz. Then

44502.jpg
  • the rms current through the resistor R is 4.2 A
  • the effective potential difference between P and Q should be 42 V.
  • the instantaneous current leads the source voltage by 45^{\circ}
  • the instantaneous current lags behind the applied voltage by 45^{\circ}
The reading shown in AC Voltmeter V when S is moved to B is:
44749.jpg
  • 120 V
  • 100 V
  • 140 V
  • 160 V
In the circuit diagram shown, X_{C}=100\Omega , X_{L}=200 \Omega  & R=100 \Omega . The effective current through the source is:

75050.jpg
  • 2 A
  • 2 \sqrt{2} A
  • 0.5 A
  • \sqrt{0.4} A
In a series L, C circuit , which of the following represents variation of magnitude of reactance (X) with frequency (f) ? 
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

75637.jpg
  • connecting an inductor in series with the load
  • connecting a capacitor in series with the load
  • connecting an inductor in parallel with the load
  • connecting a capacitor in parallel with the load.
Switch S is closed at t =After sufficiently long time an iron rod is inserted into the inductor L. Then, the bulb :
74243.png
  • glows more brightly
  • gets dimmer
  • glows with the same brightness
  • gets momentarily dimmer and then glows more brightly
The instantaneous potential difference between points A and B is (Phase angle is 37^0)
74297.png
  • 8 \sin (50\pi t+37\dfrac{\pi }{180})
  • 8 \sin (50\pi t-37\dfrac{\pi }{180})
  • 10 \sin (50\pi t )
  • 10 \cos (50\pi t )
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)

75331.PNG
  • equal to I
  • more than I
  • less than I
  • zero
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:
75054.png
  • \dfrac{I_0}{2}
  • \dfrac{I_0}{\sqrt{2}}
  • \dfrac{I_0}{\sqrt{3}}
  • \dfrac{I_0}{3}
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 :
  • 4.0\ \text{A}
  • 8.0\ \text{A}
  • \dfrac {20}{\sqrt {13}}\ \text{A}
  • 2.0\ \text{A}
Regarding the given circuit, the correct statement is:

75669.PNG
  • (V_a - V_b) is increasing with time
  • (V_a - V_b) is decreasing with time
  • (V_a - V_b) = 10V
  • (V_a - V_b) = zero
Calculate \omega and T.
214151.PNG
  • 6.28\times{10}^{-3}rad/s; {10}^{-3}s
  • 6.28\times{10}^{3}rad/s; {10}^{-3}s
  • 7.28\times{10}^{3}rad/s; {10}^{-3}s
  • 7.28\times{10}^{-3}rad/s; {10}^{-3}s
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.
  • \displaystyle (\pi/ \sqrt 3)H
  • \displaystyle (\sqrt 3/ \pi)H
  • \displaystyle (2/\pi)H
  • None of these
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 :
  • 2.4A
  • 1.8A
  • 1.5A
  • 1.2A
Identify the graph which correctly represents the variation of capacitive reactance X_C with frequency.
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

222192.JPG
  • 8V
  • 16V
  • 10V
  • Not possible to determine unless value of R and C are given.
The frequency of oscillation of current in the inductance is
120890.png
  • \dfrac {1}{3\sqrt {LC}}
  • \dfrac {1}{6\pi \sqrt {LC}}
  • \dfrac {1}{\sqrt {LC}}
  • \dfrac {1}{3\pi \sqrt {LC}}
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 :

222107.png
  • \displaystyle \frac{V_0}{\sqrt{\displaystyle \frac{1}{R^2}+\left(\omega L-\displaystyle\frac{1}{\omega C}\right)^2}}
  • \displaystyle V_0 \sqrt {1/ {R^2}+\left(\omega C-\displaystyle \frac{1}{\omega L}\right)^2 }
  • \displaystyle \frac{V_0}{\sqrt {R^2+\left(\omega L-\displaystyle \frac{1}{\omega C}\right)}^2}
  • None of these.
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?
  • \displaystyle \frac{1}{4\sqrt 2}A lagging in phase by \pi/4 with voltage.
  • \displaystyle \frac{1}{4\sqrt 2}A leading in phase by \pi/4 with voltage.
  • \displaystyle \frac{1}{\sqrt 2}A leading in phase by \pi/4 with voltage.
  • \displaystyle \frac{1}{4\sqrt 2}A leading in phase by \pi/2 with voltage.

A signal generator supplies a sine wave of 200V, 5kHz to the circuit as shown in the figure. Then choose the wrong statement:

219031.png
  • The current in the resistive branch is 0.2A.
  • The current in the capacitive branch is 0.126A.
  • Total line current is =0.283A.
  • Current in both the branches is same.
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
  • 1\space A
  • 1.5\space A
  • 2\space A
  • 2.5\space A
Resonance occurs in a series LCR circuit when the frequency of the applied emf is 1000\space Hz.
  • When frequency = 900\space Hz, then the current through the voltage source will be ahead of emf of the source
  • The impedance of the circuit is minimum at f = 1000\space Hz
  • At only resonance the voltages across L and C differ in phase by 180^{\small\circ}
  • If the value of C is doubled, resonance occurs at f = 2000\space 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?
  • \displaystyle\frac{5(\pi^2 - 5)}{\pi}\space \Omega
  • \displaystyle\frac{10(10 - \pi^2)}{\pi}\space \Omega
  • \displaystyle\frac{10(\pi^2 - 5)}{\pi}\space \Omega
  • \displaystyle\frac{5(10 - \pi^2)}{\pi}\space \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
  • 10^{-4}\space H
  • 10^{-6}\space H
  • 10^{-8}\space H
  • 10^{-10}\space H
Impedance of box P at the above frequency is 
  • 70\Omega
  • 77\Omega
  • 90\Omega
  • 100\Omega
The angular frequency for which he detects maximum current in the circuit is
  • 10^5/7 rad/s
  • 10^4 rad/s
  • 10^5 rad/s
  • 10^4/7 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
  • I_0 = 40\space A
  • V_0 = 9.8\space kV
  • V_1 = 9.8\space kV
  • V_1 = 19.6\space kV
0:0:1


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 12 Medical Physics Quiz Questions and Answers