MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Electro-Magentic Induction And Alternating Currents Mcq
Quiz 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
In an LCR series circuit, the voltage across each of the components. L, C, and R is 50 V. The voltage across the LC combination will be....[ AFMC 2004]
0%
a) 50V
0%
b) 0V
0%
c) 50√2 V
0%
d) 100 V
Explanation
Voltage across LC combination is given by Answer: (b)
Q.2
What is the maximum value of inductance L for which the current is maximum in a LCR circuit with C = 10µF and ω = 1000 s-1 [ CBSE-PMT 2007]
0%
a) 1 mH
0%
b) cannot be calculated unless R is known
0%
c) 10 mH
0%
d) 100 mH
Explanation
At resonance Impedance becomes only resistive at resonance ω = 1/ (LC) 1/2 L = 1/ ω2 C L = 1/ ( 106 × 10 ×10-6 ) L = 0.1 H L = 100 mH Answer: (d)
Q.3
The time constant of C-R circuit is [ CBSE-PMT 1992]
0%
a) 1 / CR
0%
b) C/R
0%
c) CR
0%
d) R/C
Explanation
Answer:(c)
Q.4
In a LCR series circuit, the potential difference between the terminals of the inductance is 60V, between the terminals of the capacitor is 30V and across resistance is 40V. Then supply voltage will be equal to .... [ AFMC 2008]
0%
a) 50 V
0%
b) 70V
0%
c) 130V
0%
d) 10V
Explanation
In case of inductor voltage leads the current and for capacitance voltage lags the current hence we can not add them algebraically. To find the resultant voltage we have to use phasor diagram Answer: (a)
Q.5
In A.C. circuit in which inductance and capacitance are joined in series, current is found to be maximum when the value of inductance is 0.5H and the value of capacitance is 8µF. The angular frequency of applied alternating voltage will be .. [ AFMC 2010]
0%
a) 400 Hz
0%
b) 5000Hz
0%
c) 2×105 Hz
0%
d) 500 Hz
Explanation
At resonance frequency current is maximum so ω = 1/ √(LC) By substituting values we get ω = 500Hz Answer:(d)
Q.6
In a series LCR circuit, resonance occurring at 105Hz. At that time, the potential difference across the 100Ω resistance is 40V while the potential difference across the pure inductor is 30v. The inductance L of the inductor is equal to [ AFMC 2011]
0%
a) 10-3 H
0%
b) 7.5× 10-4 H
0%
c) 5.0× 10-4 H
0%
d) 1.2×10-4 H
Explanation
At resonance imaginary part of current is zero Now V=IR Thus I = 40/100 = 0.4A XL = V L / I XL = 30/0.4 = 75Ω L = XL / ω L = 7.5× 10-4 Answer: (b)
Q.7
An emf of 15 Volt is applied in a circuit containing 5H inductance and 10Ω resistance . The ratio of current at=∞ and t=1 sec is ...[ AFMC 2002]
0%
a)
0%
b)
0%
c) 1 - e
0%
d) e-1
Explanation
For growth of current in LR circuit is given by formula CaseI: for t = ∞, I = I 0 CaseII: for t=1 sec Ratio of current at t=∞ and t =1 sec is Answer: (a)
Q.8
The reactance of coil when used in an A.C. power supply ( 220volts, 50 cycles/sec) is 50ohms. The inductance of the coil is nearly [ AFMC 2001]
0%
a) 0.16 henry
0%
b) 0.22 henry
0%
c) 2.2 henry
0%
d) 1.6 henry
Explanation
Reactance of coil = ωL = 2πfL = 50 L = 50/ 2πf = 0.16 henry Answer: (a)
Q.9
If a capacitance C is connected in series with an inductor of inductance L, then the angular frequency will be.. [ AFMC 1998]
0%
a)
0%
b)
0%
c) LC
0%
d)
Explanation
At resonance angular frequency is as per option (a) Answer:(a)
Q.10
Which of the following is not transducer? [ AFMC 2004]
0%
a) Loudspeaker
0%
b) Amplifier
0%
c) Microphone
0%
d) All of these
Explanation
A transducer is a device which produce electrical signal from other form of energy and vise a versa Amplifier is a device which amplifies in put signal hence not transducer Answer: (b)
Q.11
A choke is preferred to a resistance for limiting current in AC circuit because [ AFMC 2008]
0%
a) choke is cheap
0%
b) there is no wastage of power
0%
c) choke is compact in size
0%
d) choke is a good absorber of heat
Explanation
Choke is a device having high inductance and negligible resistance. It is used to control current in ac circuits Answer: (b)
Q.12
A.C supply gives 30 V r.m.s. which passes through a 10Ω resistance. The power dissipated in it is :
0%
a) 90√2 W
0%
b) 90 W
0%
c) 45√2 W
0%
d) 45 W
Explanation
Power P = V2 / R P = 302 / 10 = 90W Answer: (b)
Q.13
The value of inductance which should be connected in series with capacitance of 0.5µF, resistance of 10Ω and an A.C. source of 50cps so that the power factor of the circuit is unit, will be ..[ AFMC 210]
0%
a) 20.28 H
0%
b) 10.13 H
0%
c) 15.26 H
0%
d) 25.28 H
Explanation
At resonance A.C circuit have power factor unity ω= 1 / √(LC) L = 1 / (ω2 C) here ω = 2πf C = 0.5µF R = 10Ω, f = 50cps by substituting values in equation for L we get L =20.41H Answer:(a)
Q.14
A condenser of capacity 1µF and resistance 0.5 MΩ are connected in series with D.C. supply of 2V. The time constant of the circuit is : [ AFMC 2010]
0%
a) 0.25 s
0%
b) 0.5s
0%
c) 1 s
0%
d) 2 s
Explanation
time constant τ = RC substituting R = 0.5×106 Ω C = 1 × 10-6F In above equation we get τ = 0.5sec Answer: (b)
Q.15
The core of transformer is laminated because... [ CBSE-PMT 2006]
0%
a) the weight of the transformer may be reduced
0%
b) rusting of the core may be prevented
0%
c) ratio of voltage in primary and secondary may be increases
0%
d) energy losses due to eddy currents may be minimised
Explanation
When there is change of flux in the core of transformer due to change in current around it, eddy current is produced. The direction of this current is opposite to the current which produces it, so it will reduce the main current. We laminate the core so that flux reduced resulting in the reduced production of eddy current. Answer: (d)
Q.16
A capacitor has capacitance C and reactance X. If capacitance and frequency become double, then reactance will be.. [ CBSE-PMT 2001]
0%
a) 4X
0%
b) X/2
0%
c) X/4
0%
d) 2X
Explanation
Reactance X = 1 / 2πfC Thus when frequency and capacitance is doubled New reactance will be X/4 Answer: (c)
Q.17
A 220 volts input is supplied to a transformer. The output circuit draws a current of 2.0amp at 440 volts . If the efficiency of the transformer is 80, the current drawn by the primary windings of the transformer is
0%
a) 3.6 amp
0%
b) 2.8 amp
0%
c) 2.5 amp
0%
d) 5.0 amp
Explanation
Equation for transformer Here Primary voltage V1 = 220 V Secondary voltage V2 = 440 V Secondary current I2 = 2.0A efficiency n = 0.8 on substituting above values in equation and on solving I1 = 5 A Answer:(d)
Q.18
In an a.c circuit an alternating voltage e = 200√2 sin 100t volts is connected to a capacitor of capacity 1 µF. The r.m.s. value of the current in the circuit is ... [ CBSE-PMT 2011]
0%
a) 10 mA
0%
b) 100 mA
0%
c) 200 mA
0%
d) 20 mA
Explanation
Vr.m.s = Vmax / √2 Vr.m.s = 200√2 / √2 = 200 V Reactive capacitance XC =1/ ωC XC = 1/100× 10-6= 104 Ir.m.s = Vr.m.s / XC Irms = 200/104 = 20×10-3 = 20 mA Answer: (d)
Q.19
In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by e = Eo sin ωt i = Io sin (ωt - φ) Average power is given by [CBSE-PMT 2008]
0%
a) EoIo / 2
0%
b) (Eo Io/2) sin φ
0%
c) (Eo Io/2) cos φ
0%
d) EoIo
Explanation
The average power in the circuit over one cycle of a.c. is given by P ave. = er.m.s ir.m.s cos φ Answer: (c)
Q.20
In an A.C circuit with voltage V and current I the power dissipated is .. [ CBSE-PMT 1997]
0%
a) VI/√ 2
0%
b) VI/2
0%
c) depends on the phase difference between V and I
0%
d) VI
Explanation
Power dissipated depends upon phase difference Power P = ErmsIrmscosθ Answer: (c)
Q.21
Energy in a current carrying coil is stored in the form of ... [ CBSE-PMT 1988]
0%
a) electric field
0%
b) magnetic field
0%
c) dielectric strength
0%
d) heat
Explanation
Energy density of for Inductor = B2/2µ Answer:(b)
Q.22
A coil of resistance 30Ω and inductive resistance 20Ω at 50Hz frequency. If an a.c. source of 200 volt, 100Hz is connected across the coil, the current in the coil will be ... [ CBSE-PMT 2011]
0%
a) 4.0A
0%
b) 8.0A
0%
c) 20/ √13 A
0%
d) 2.0A
Explanation
Inductive resistance XL = 2π×f×L 20 = 2π × 50 × L ∴ L = 1/ 5π Now for frequency 100 Hz Inductive reactance XL = 2π ×100× ( 1/5π ) XL = 40 Ω Now impedance of L-R circuit =Z = [ R2 + (XL)2 ] 1/2 Z = [ 302 + 402 ] 1/2 Z = 50 Ω Now current I = V/Z I = 200 /50 = 4A Answer: (a)
Q.23
An a.c. voltage is applied to a resistance R and inductor L in series. If R and the inductive reactance both are equal to 3Ω, the phase difference between the applied voltage and current in the circuit is .. [ CBSE-PMT 2011]
0%
a) π/6
0%
b) π /4
0%
c) π/2
0%
d) zero
Explanation
The phase difference Φ is given by tanΦ = XL / R tanΦ = 3/3 = 1 ∴ Φ = π /4 Answer: (b)
Q.24
In LCR circuit the capacitance is changed from C to 4C. For the same resonant frequency, the inductance should be changed from L to [ BHU 1999]
0%
a) 2L
0%
b) L/2
0%
c) L/4
0%
d) 4L
Explanation
Since resonant frequency is same L1C1 = L2C2 LC = 4C×L2 L/4 = L2 Answer: (c)
Q.25
An LCR series circuit is connected to a source alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of ... [ CBSE-PMT 1994]
0%
a) π
0%
b) π /2
0%
c) π /4
0%
d) 0
Explanation
At resonance circuit behaves as if it contains only resistance. So phase difference is zero. Answer: (d)
Q.26
Eddy current are produced when... [ CBSE-PMT 1988]
0%
a) a metal is kept in varying magnetic field
0%
b) a metal is kept in steady magnetic field
0%
c) a circular coil is placed in a magnetic field
0%
d) through a circular coil, current is passed
Explanation
Eddy current is produced when a metal is kept in a varying magnetic field Answer: (a)
Q.27
In an a.c. circuit, the r.m.s. value of current, ir.m.s. is related to the pick current, i0 by the relation.. [ CBSE-PMT 1994]
0%
a) Ir.m.s. = √2 I0
0%
b) Ir.m.s. = π I0
0%
c) Ir.m.s. = I0 / π
0%
d) Ir.m.s. = I0 / √2
Explanation
Answer: (d)
Q.28
In a region of uniform magnetic induction B= 10-2 tesla, a circular coil of radius 30cm and resistance π2Ω is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of coil. If the coil rotates at 200rpm the amplitude of the alternating current induced in the coil is ... [ CBSE-PMT 1988]
0%
a) 4π2 mA
0%
b) 30 mA
0%
c) 6 mA
0%
d) 200 mA
Explanation
Maximum Induced voltage E = nABω Maximum current I = E/I = nABω / R -- eq(1) Here A is area of coil = πr2 = π × (0.3)2 ω = 2πf = 2π (200/60) Substituting values in eq(1) we get I = 6mA Answer:(c)
Q.29
In an experiment, 200 V a.c. is supplied at the ends of an LCR circuit. The circuit consists of an inductive reactance XL = 50Ω, capacitive reactance XC = 50 Ω and ohmic resistance = 10 Ω. The impedance of the circuit is ... [ CBSE-PMT 1996]
0%
a) 10 Ω
0%
b) 20 Ω
0%
c) 30 Ω
0%
d) 40 Ω
Explanation
Impedance of the LCR circuit Answer: (a)
Q.30
Power dissipated in an LCR series circuit connected to an a.c. source of e.m.f. E is ... [ CBSE-PMT 2009]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Power dissipated P = I2R But I = E/Z here Z is impedance P = (E/Z)2R Impedance Answer: (c)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page