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Physics NEET MCQ
Electro-Magentic Induction And Alternating Currents Mcq
Quiz 2
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Q.1
The primary winding of a transformer has 500 turns where as its secondary has 5000 turns. The primary is connected to an a.c. supply of 20V, 50 Hz. The secondary will have an output of ... [ CBSE-PMT 1997]
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a) 2V, 5Hz
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b) 200V, 500Hz
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c) 2V, 50Hz
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d) 200V, 50Hz
Explanation
Vs / Vp = Ns / Np Es /20= 5000 / 500 Es = 200V Frequency don't change Answer: (d)
Q.2
A step-up transformer operates on a 230 V line and supplies a load of 2 amp. The ratio of the primary and secondary windings is 1:The current in the primary is ... [ CBSE-PMT 1997]
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a) 25A
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b) 50A
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c) 15A
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d) 12.5A
Explanation
Ip / Is = Ns / Np Ip/2 = 25/1 Ip = 50A Answer:(b)
Q.3
A transistor oscillating using a resonant circuit with an inductor L ( of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be... [ CBSE-PMT 2006]
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a) 8f
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b) f/2√2
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c) f/2
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d) f/4
Explanation
We know that frequency of electric oscillation in L.C. circuit is Now L' = 2L and C' = 4C then f' = f/√8 f' = f/2√2 Answer: (b)
Q.4
In the circuit as shown in figure, the bulb will become suddenly bright if ... [ CBSE-PMT 1989]
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a) contact is made or broken
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b) contact is made
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c) contact is broken
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d) wont become bright at all
Explanation
When circuit is broken, the induced e.m.f. is largest, Answer: (c)
Q.5
A coil of 40H inductance is connected in series with a resistance of 8Ω and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is ... [ CBSE-PMT 2004]
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a) 20 sec
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b) 5 sec
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c) 1/5 sec
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d) 40 sec
Explanation
Time constant for LR circuit is τ = L/R τ = 40/8 = 5 sec Answer: (b)
Q.6
The primary and secondary coil of a transformer have 50 and 1500 turns respectively. If the magnetic flux φ linked with primary coil is given by φ = φo + 4t, where φ is in weber, t is time in seconds and φo is a constant, the output voltage across the secondary coil is .. [ CBSE-PMT 2007]
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a) 120 volts
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b) 220 volts
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c) 30 volts
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d) 90 volts
Explanation
W e know that E = dφ / dt ∴ E = 4V ∴ Vp=4 V Since Vp / Vs = Np / Ns on substituting the values in above equation we get Vs = 120V Answer:(a)
Q.7
A transformer is used to light a 100W and 110V lamp from 220V mains. If the main current is 0.5 amp., the efficiency of transformer is approximately ... [ CBSE-PMT 2007]
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a) 50%
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b) 90%
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c) 10%
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d) 30%
Explanation
Input power Pin = VI Pin = 220 × 0.5 = 110 Efficiency η = (Power output/ power input )× 100 η = 100 ×100 / 100 = 90.9% Answer: (b)
Q.8
A 220 V and 50 Hz is supplied to LCR circuit. Voltage across Capacitor and inductor is 300 V each what is the Voltage and current across resistance 100Ω
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a) 150V, 2.2A
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b) 220V, 2.20A
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c) 220V, 2.0A
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d) 100V, 2.0A
Explanation
As VL = VC = 300 V resonance will take place ∴ VR = 220 V Current I = V/R = 220 /100 = 2.2 A Answer: (b)
Q.9
An inductance L having a resistance R is connected to an alternating source of angular frequency ω . The quality factor Q of the inductance is ... [ CBSE-PMT 2000]
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a) R/ ωL
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b) ( ωL / R)2
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c) ( R / ωL)1/2
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d) ωL / R
Explanation
Quality factor = Potential drop across capacitor or inductor / Potential drop across R Quality factor = IωL / IR Quality factor = ωL / R Answer: (d)
Q.10
A coil of inductive reactance 31Ω has a resistance of 8Ω. It is placed in series with a condenser of capacitive reactance 25Ω. The combination is connected to an a.c. source of 110V. The power factor of the circuit is ... [ CBSE-PMT 2006]
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a) 0.64
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b) 0.80
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c) 0.33
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d) 0.56
Explanation
Power factorcosφ Answer:(b)
Q.11
In a series resonant circuit, having L,C,R as its elements, the resonant current is 'i'. The power dissipated in the circuit at resonance is ... [ CBSE-PMT 2002]
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a) i2 ωC
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b) zero
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c) i2ωL
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d) i2R
Explanation
At resonance XL = XC thus circuit is resistive Thus power dissipated = i2R Answer: (d)
Q.12
The rms value of potential difference V shown in figure is ... [ CBSE-PMT 2011]
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a) V0
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b) V0 / √2
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c) V0 / 2
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d) V0 / √3
Explanation
Answer: (b)
Q.13
Which quantity is increased in step-down transformer? [ MPPMT 1998]
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a) current
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b) Voltage
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c) Power
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d) Frequency
Explanation
Answer: (a)
Q.14
The current passing through a choke coil of 5 henry is decreeing at the rate of 2amp/sec. The e.m.f. developed across the coil is [ MPPPMT 1999]
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a) 10 Volts
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b) -10 Volts
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c) 2.5 Volts
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d) -2.5 Volts
Explanation
From the formula E = - L( dI/dt) given dI/dt = -2 amp E = -5(-2) = 10 V Answer:(a)
Q.15
Quantity which remains unchanged in a transformer is [ MPPMT 1998]
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a) Voltage
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b) Current
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c) Frequency
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d) none of the above
Explanation
Answer: (c)
Q.16
Lenz's law is a consequence of the law of conservation of [ MPPMT 1999]
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a) charge
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b) mass
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c) momentum
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d) energy
Explanation
Answer: (d)
Q.17
Two coils X and Y are placed in a circuit such that a current changes by 2 amp per sec. in coil X and the magnetic flux change of the 0.4 weber occurs in Y. The value of mutual inductance of the coil and its unit are [ MPPMT 1990]
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a) 0.2 henry
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b) 5 henry
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c) 0.8 henry
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d) 0.2 weber
Explanation
dφ /dt = M (dI/dt) 0.4 = M(2) M = 0.4/2 = 0.2 henry Answer:(a)
Q.18
Q47) The current in a self inductance L = 40mH is to be increased uniformly from 1A to 11A in 4 milliseconds. The induced e.m.f produced in L during this process will be [ CBSE 1990]
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a) 100 V
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b) 0.4 V
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c) 440 V
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d) 40 V
Explanation
Use formula E = -L( dI/dt) Answer: (a)
Q.19
A straight conductor of length 0.4m is moved with a speed of 7m/s perpendicular to a magnetic field of intensity 0.9 Wb/mThe induced emf across the conductor is [ CBSE 1995]
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a) 5.04 V
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b) 1.26V
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c) 2.52V
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d) 25.2V
Explanation
Use formula E = Blv Answer: (c)
Q.20
A coil of wire of a certain radius has 600 turns and self-inductance of 108mH. The self-inductance of a second similar coil of 500 turns will be [ MPPMT 1990]
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a) 74 mH
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b) 75mH
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c) 76mH
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d) 77mH
Explanation
From the formula for self inductance ∴ L ∝ N2 L' / L = (500/600)2 L' = 108(500/600)2 L' = 75 mH Answer: (b)
Q.21
The back emf in a DC motor is maximum when [ Raj.PMT 1997]
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a) the motor has pick up maximum speed
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b) the motor has just started moving
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c) the speed of motor is still on the increase
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d) the motor has just been switched off
Explanation
Answer: (a)
Q.22
A resistance coil is held horizontally and a magnet is allowed to fall vertically through it. Then the acceleration of the magnet is: [ MPPET 1999]
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a) equal to g
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b) less than g
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c) more than g
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d) sometimes less and some times more than g
Explanation
direction of induced current in coil will be such that it will oppose the cause of it. Thus magnetic field produced in coil will oppose motion of magnet thus acceleration will be less than g Answer: (b)
Q.23
With the decrease of current in the primary coil from 2 amp to zero value in 0.01 second, the emf generated in the secondary coil is 1000V. The mutual inductance of the two coils is [ MPPMT 1989]
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a) 1.25 henry
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b) 2.5 henry
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c) 5.0 henry
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d) 10.0 henry
Explanation
here dI/dt = -2/0.010 = -200 amp/s use formula E - -L ( dL/dt) Answer:(c)
Q.24
The core used in transformers and other electromagnetic devices is laminated to [ MPPMT 1995]
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a) increase the magnetic field
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b) increase the level of the magnetic saturation of the core
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c) reduce the residual magnetism in the core
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d) reduce eddy current losses in the core
Explanation
Answer: (d)
Q.25
The armature current in a D.C. motor is maximum when the motor has [ MPPET 1995]
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a) picked up maximum speed
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b) just started
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c) intermediate speed
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d) just been switched off
Explanation
Answer: (b)
Q.26
An ideally efficient transformer has primary power input of 10kW. The secondary current when the transformer is on load is 25 amperes. If the primary to secondary turns ratio is 8:1, then the potential difference applied to the primary coil is [ CMPT 1998]
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a)
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b)
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c)
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d)
Explanation
Input power = Output power 104 = EsIs Es = 104 / 25 Now Es / Ep = ns / np Es / Ep = 1 / 8 Ep = 8× Es Ep = (8×104) / 25 Answer: (b)
Q.27
A solenoid has n turns. Its coefficient of an inductance L varies with n as [ CBSE 1998]
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a) L ∝ n
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b) L ∝ n2
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c) L ∝ n-1
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d) L ∝ n
Explanation
Answer:(b)
Q.28
An emf of 5V is produced by a self-inductance when the current changes at a steady rate from 3 to 2 amp in 1 millisecond. The value of self-inductance is [ CPMT 1993]
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a) zero
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b) 500 henries
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c) 5 henries
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d) 5 milihenries
Explanation
Answer: (d)
Q.29
A loop of area 0.1m2 rotates with a speed of 60 rev/s with the axis of rotation perpendicular to a magnetic field B = 0.4T. If there are 100 turns in the loop, the maximum voltage induced in the loop is [ PMT 1995]
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a) 15.07 V
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b) 150.7 V
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c) 1507 V
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d) 250 V
Explanation
Emax = NABω ω = 120π Emax = 100×0.1×0.4×120π Emax ≈ 1507V Answer: (c)
Q.30
The current of LR circuit is reduced to half. What will be energy stored in it [ CPMT 1999]
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a) Four times
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b) Two times
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c) 1/2
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d) One fourth
Explanation
Energy stored U = ½ Li2, when i reduced to half energy redued to one fourth Answer: (d)
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