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Physics NEET MCQ
Electro-Magentic Induction And Alternating Currents Mcq
Quiz 9
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Q.1
The core of any transformer is laminated so as to [ AIEEE 2003]
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a) reduce the energy loss due to eddy currents
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b) make it light weight
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c) make it robust and strong
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d) increases the secondary voltage
Explanation
Answer: (a)
Q.2
Alternating current can not be measured by D.C. ammeter because [ AIEEE 2004]
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a) average value of current for complete cycle is zero
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b) A.C. changes direction
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c) A.C. can not pass through D.C. Ammeter
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d) D.C. Ammeter will get damaged
Explanation
Answer: (a)
Q.3
In an LCR series a.c. circuit, the voltage across each of the components L, C and R is 50V. The voltage across the LC combination will be [ AIEEE 2004]
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a) 100V
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b) 50√2 V
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c) 50 V
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d) zero
Explanation
Since phase difference between L and C is π ∴ net voltage difference across LC = 50 - 50 = 0 Answer:(d)
Q.4
A coil having n turns and resistance RΩ is connected with a galvanometer of resistance 4RΩ This combination is moved in time t seconds from a magnetic field W1 weber to W2 weber. The induced current in the circuit is [ AIEEE 2004]
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a)
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b)
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c)
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d)
Explanation
E = - ndφ/dt = (W2 - W1) / t Total resistance is R + 4R = 5R< if i is the current then 5Ri = n(W2 - W1) / t i = n(W2 - W1) / 5Rt Answer: (b)
Q.5
In a uniform magnetic field of induction B a wire in the form of semicircle of radius r rotates about the diameter of the circle with an angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, the mean power generated per period of rotation is [ AIEEE 2004]
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a) (Bπrω)2 / 2R
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b) (Bπr2ω)2 / 8R
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c) (Bπr2ω) / 2R
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d) (Bπrω2)2 / 8R
Explanation
φ = BACosωt E = - dφ/dt = ωBAsinωt i = E/R = (ωBAsinωt) / R Pinst = i2R Answer: (b)
Q.6
In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to [ AIEEE 2004]
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a) L/2
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b) 2L
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c) 4L
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d) L/4
Explanation
For resonant frequency to remain same LC should be constant LC = L'×2C L' = L/2 Answer: (a)
Q.7
one conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B, l and V where l is the width of each tube will be [ AIEEE 2005]
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a) - Blv
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b) Blv
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c) 2Blv
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d) zero
Explanation
Area enclosed in decreasing relative velocity = v + v =2v ∴ E = Bl(2v) Answer: (c)
Q.8
A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is 0.2×10-4T, then the emf developed between the two ends of the conductor is [ AIEEE 2004]
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a) 5 mV
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b) 50 µV
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c) 5 µV
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d) 50 mV
Explanation
use the formula for induced emf E = Bvl / 2 Answer:(b)
Q.9
The self inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50Hz, it should be connected to a capacitance of [ AIEEE 2005]
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a) 8µF
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b) 4µD
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c) 2µF
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d) 1µF
Explanation
At resonance frequency power factor is one ∴ C = 1/Lω2 Answer: (d)
Q.10
The phase difference between the alternating current and emf is π/Which of the following cannot be the constituent of circuit [ AIEEE 2005]
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a) R,L
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b) C alone
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c) L alone
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d) L, C
Explanation
Phase difference for R-L circuit lies between ( 0, π/2) Answer: (a)
Q.11
A circuit has a resistance of 12Ω and an impedance of 15Ω. The power factor of the circuit will be [ AIEEE 2005]
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a) 0.4
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b) 0.8
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c) 0.125
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d) 1.25
Explanation
Power factor cosφ = R/Z Answer:(b)
Q.12
A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage 2V. The current reaches half of its steady state value in [ AIEEE 2005]
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a) 0.1 s
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b) 0.05s
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c) 0.3s
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d) 0.15s
Explanation
equation for current through inductor Answer: (a)
Q.13
Which of the following units denote the dimension ML2 / QWhere Q denotes electric charge? [ AIEEE 2006]
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a) Wb/m2
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b) Henry (H)
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c) H/m2
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d) Weber(Wb)
Explanation
Mutial inductance = φ/I = BA/I Answer: (b)
Q.14
The flux linked with coil at any instant 't' is given by φ = 10t2 - 50t +250 The induced emf at t = 3s is [ AIEEE 2006]
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a) -190 V
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b) -10 V
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c) 10 V
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d) 190 V
Explanation
Use E = -dφ/dt to get equation then substitute t = 3 Answer: (b)
Q.15
An inductor (L = 100 mH) a resistor ( R = 100Ω) and a battery (E = 100 V) are initially connected in series as shown in figure. After a long time the battery is disconnected after short circuiting the point A and B. the current in the circuit 1 ms after the short circuit is [ AIEEE 2006]
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a) 1/e A
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b) e A
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c) 0.1 A
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d) 1 A
Explanation
Current when discharging I = Io e-Rt/L Maximum current Io = E/R Answer:(a)
Q.16
In an a.c. circuit the voltage applied is E = Eosinωt. the resulting current in the circuit is I = Iosin( ωt - π/2). The power consumption in the circuit is given by [ AIEEE 2007]
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a) P = √2 EoIo
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b) P = (EoIo) / √2
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c) P = zero
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d) P = EoIo / 2
Explanation
phase difference between E and I is π/2. power factor is zero. Power consumed is zero Answer: (c)
Q.17
An ideal coil of 10H is connected in series with a resistance of 5Ω and a battery of 5V. 2 second after the connection is made, the current flowing in ammeter in the circuit is [ AIEEE 2007]
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a) (1 - e-1)
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b) (1 - e)
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c) e
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d) e-1
Explanation
Equation for current , when current is growing in RL circuit Answer: (a)
Q.18
two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and other 400 turns, their mutual inductance is [ AIEEE 2008]
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a) 2.4π ×10-5 H
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b) 4.8π ×10-4 H
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c) 4.8π ×10-5 H
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d) 2.4π ×10-4 H
Explanation
use formula Answer: (d)
Q.19
An inductor of inductance L = 400 mH and resistor of resistance R1 = 2Ω are connected to a battery of emf 12V as shown in figure. The internal resistance of the battery is negligible. The switch S is closed at t=The potential drop across L as a function of time is [ AIEEE 2009]
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a)
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b)
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c)
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d)
Explanation
potential across inductor and resistance R2 is E maximum current Io = E/ R2 Growth of current in LR2 branch when switch is closed is given by potential drop across L = rate of growth of current ×L (inductance) = Answer:(c)
Q.20
In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is [ AIEEE 2006]
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a) NABRω
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b) NAB
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c) NABR
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d) NABω
Explanation
Answer: (d)
Q.21
A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when
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a) Frequency of the AC source is decreased
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b) Number of turns in the coil is reduced
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c) A capacitance of reactance XC = XL is included in the same circuit
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d) An iron rod is inserted in the coil
Explanation
Brightness of bulb depends on the current. Increase in reactance decrease in brightness Reactant of coil is 2πfL thus when iron rod is inserted in the coil, value of inductance increases, thus option ‘d’ is correct. Answer:(d)
Q.22
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be
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a) p
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b)
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c)
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d)
Explanation
Power in A.C. is given p=I2Zcosθ For only resistance power θ=0 V2= pR…(i) For resistance and inductor in series combination impedance =Z Substituting value of V2 in above equation Answer:(b)
Q.23
A series R-C circuit is connected to an alternating voltage source. Consider two situations :- (a) When capacitor is air filled. (b) When capacitor is mica filled. Current through resistor is i and voltage acrosscapacitor is V then :-
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a) Va = Vb
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b) Va < Vb
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c) Va > Vb
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d) ia > ib
Explanation
When capacitor is filled with mica then capacitance C increases as Xc decreases so Va > Vb Answer:(c)
Q.24
An inductor 20 mH, a capacitor 50 µF and a resistor40Ωare connected inseries across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is :-
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a) 0.51 W
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b) 0.47 W
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c) 0.76 W
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d) 0.89 W
Explanation
Answer:(b)
Q.25
A small signal voltage V(t) = V0 sin ωt is applied across an ideal capacitor C :- ….[AIPMT 2015]
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a) Current I (t), lags voltage V(t) by 90°.
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b) Over a full cycle the capacitor C does not consume any energy from the voltage source.
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c) Current I (t) is in phase with voltage V(t).
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d) Current I (t) leads voltage V(t) by 180°.
Explanation
Power = Vrms .Irms cosθ as cosθ = 0 (Because θ = 90°) ∴ power consumed = 0 ( in one complete cycle) Answer:(b)
Q.26
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ?
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a) R = 15 Ω, L = 3.5 H, C = 30 µF
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b) R = 25 Ω, L = 1.5 H, C = 45 µF
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c) R = 20 Ω, L = 1.5 H, C = 35 µF
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d) R = 25 Ω, L = 2.5 H, C = 45 µF
Explanation
For better tuning, Q-factor must be high. R and C should be small and L should be high Answer:(a)
Q.27
The potential difference across the resistance, capacitance and inductance are 80V, 40V and 100V respectively in an L-C-R circuit. The power factor of this circuit is …[NEET II 2016]
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a) 0.8
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b) 1.0
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c) 0.4
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d) 0.5
Explanation
Power factor =cosθ [since opposite side =3 and adjacent side =4, thus hypo = 5] Answer:(a)
Q.28
A 100Ω resistance and a capacitor of 100Ω reactance are connected in series across a 220V source. When the capacitor is 50% charged, the peak value of the displacement current is …[ NEET II-2015]
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a) 4.4 A
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b) 11√2 A
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c) 2.2 A
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d) 11 A
Explanation
Asked about the peak value of the displacement current Peak voltage = 220√2 Answer:(c)
Q.29
Figure shows a circuit contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 mH each,and an ideal battery with emf ε = 18 V. The current'i' through the battery just after the switch closed is …. [ NEET 2017]
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a) 2 mA
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b) 0.2 A
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c) 2 A
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d) 0 ampere
Explanation
When switch is closed, inductors and capacitor will not conduct and no current flows through them, current will pass through middle resistance. Thus current will flow through middle resistance I = 18/9 = 2A Answer:(c)
Q.30
The supply voltage to a room is 120 V. The resistance of the lead wires is 6Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb ? .. [ IIT Mains 2013]
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a) zero Volt
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b) 2.9 Volt
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c) 13.3 Volt
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d) 10.3 Volt
Explanation
Resistance of bulb Resistance of heater When only bulb is on Voltage across Bub Total resistance = 240 +6 = 246Ω Current in bulb = 120/246 Voltage across bulb when heater is on Total resistance of parallel combination R’ Current from supply when heater is switched on Resistance of circuit = 48+6 = 54 Current = 120/54 Potential across combination of bulb and heater Answer:(d)
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