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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 801. |
A 1000 hertz a.c. is flowing in a 14 mH coil. Find its reactance. |
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Answer» Here, `v = 100 hz, L = 4 mH = 14 xx 10^(-3) H` `X_(L) = omega L = 2 pi v L pi xx 100 xx 14 xx 10^(-3) = 8.8 ohm` |
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| 802. |
Shows a square loop having 100 turns, an area of `2.5 xx10^(-3) m^2` and a resistance of `100 Omega`. The magnetic field has a magnitude B =0.40 T. Find the work done in pulling the loop out of the field, slowly and uniformly is 1.0 s. |
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Answer» Correct Answer - `10^(-6)`J |
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| 803. |
Can we use 15 Hz ac for lighting purpose? |
| Answer» Yes, we can use `15 c//s`.a.c. for lighting purpose. The fluctuations in current will be so rapid `(30 times//sec)` that the bulbb will appear glowing continuously due to persistence of vision. | |
| 804. |
A 100 mHcoil carries a current of 1A. Energy stored in its magetic field isA. 0.5 JB. 0.1 JC. 0.05 JD. 1 J |
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Answer» Correct Answer - C `E = 1/2 LI^(2) = 1/2 xx 100 xx 10^(-3) xx 1^(2) = 0.05 J`. |
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| 805. |
Eddy currents are developed, whenA. conductor is kept in changing magnetic fieldB. conductor is kept in steady magnetic fieldC. conductor is kept in electric fieldD. none of these |
| Answer» Correct Answer - A | |
| 806. |
Eddy currents do not produceA. dampingB. heatingC. sparkingD. loss of energy |
| Answer» Correct Answer - C | |
| 807. |
Describe the principle, construction and working of an AC generator.A. magaetic effect of currentB. heating effect of currentC. chemical effect of currentD. electromagnetic induction |
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Answer» Correct Answer - D The working of a generator in based on the phenomen of electromagnetic induction. |
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| 808. |
In the preceding question, what will be the reading of a.c. voltmeter ? |
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Answer» A.C. voltmeter in the circuit will read effective `value//virtual` value of alternating e.m.f. `E_(v) = (E_(0))/(sqrt2) = (314)/(1.414) = 222` volt |
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| 809. |
In a series LCR circuit, `V_(L) = V_(C ) != V_(R )`. What is the value of power factor ? |
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Answer» When `V_(L) = V_(C ) , X_(L) = X_(C ) , Z = R, cos phi (R )/(Z) = 1`. i.e., power factor is unity. |
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| 810. |
What is the maximum value of power factor ? When does it occur ? |
| Answer» One. It occurs in a non-inductive circuit, i.e., in a circuit containing R only. | |
| 811. |
What is the average power consumed in a circuit consisting of resistanceless inductiance ? |
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Answer» Zero, because through resistance less inductance, `phi = 90^(@)` and `P = E_(v) I_(v) cos 90^(@) = 0`. |
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| 812. |
A long wire carries a current of 4.00 A. Find the energy stored in the magnetic field inside a volume of `1. mm^3` at a distance of 10.0 cm from the wire. |
| Answer» Correct Answer - A::B::D | |
| 813. |
In the circuit shown find (a)the power drawn from the cell, (b)the power consumed by the resistor which is converted into heat and ( c)the power given of the inductor. |
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Answer» Correct Answer - A::B::C `P_(cell)=epsilon=5xx1xx5 J//sec.=5 "watt".` `P_(res.)=I^(2)R=(1)^(2)xx3=3"watt"`. `P=5-3=2 "watt"`. |
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| 814. |
Figure shows a part of a circuit.Find the rate of change as shown. |
| Answer» Correct Answer - A::B::C::D | |
| 815. |
A `100 W 200 V` bulb is connected to a `160 V` power supply. The power consumption would beA. 64 WB. 80 WC. 100 WD. 125 W |
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Answer» Correct Answer - A `R_(b)=(V_(b)^(2))/(P)=(200xx200)/(100)=400Omega` `I=(e)/(R)=(160)/(400)=0.4A` |
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| 816. |
A solenoid is connedted to a source of constant EMF for a long time. A sof iron piece is inserted into it. Then :A. Self-inductance of the solenoid gets increasedB. Flux linked with the solenoid increases , hence steady state current gets decreased.C. Energy stored in the solenoid increases, hence steady state current gets decreased.D. Magnetic moment of the solenoid gets increased. |
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Answer» Correct Answer - A::B::C::D |
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| 817. |
STATEMENT-1 A vertical iron rod has a coil of wire wound over it at the bottom end.An alternating current flows in the coil.The rod goes through a conducting ring as shown in the figure.The ring can float at a certain height above the coil. because STATEMENT-2 In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True,Statement-2 is False.D. Statement-1 is False ,Statement-2 isTrue. |
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Answer» Correct Answer - A Due to induce current in coil, force between two coil is generated. |
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| 818. |
The average value of alternating current over a complete cycle is |
| Answer» Correct Answer - A | |
| 819. |
The working of dynamo is based on principle ofA. Electromagnetic inductionB. Conversion of energy into electricityC. Magneting effects of currentD. Heating effects of current |
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Answer» Correct Answer - A Rotation of magnet in the dynamo creates the varible flux which in turn produces the induced current. |
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| 820. |
Alternating current can not be measured by D.C. Ammeter becauseA. AC is virtualB. AC changes its directionC. AC can bot pass through DC ammeterD. average value of AC for complete cycle is zero |
| Answer» Correct Answer - D | |
| 821. |
With increase in frequency of an `AC` supply, the inductive reactance:A. increasesB. decreasesC. remains constantD. decreases sharply |
| Answer» Correct Answer - A | |
| 822. |
Current in the circuit is wattless, ifA. current is alternatingB. resistance in the circuit is zeroC. inductance in the circuit is zeroD. resistance and inductance both are zero |
| Answer» Correct Answer - B | |
| 823. |
With an increase in the frequency of an AC supply the capacitive reactanceA. increasesB. decreasesC. remains constantD. decreases sharply |
| Answer» Correct Answer - B | |
| 824. |
In an `AC` circuit containing only capacitance the currentA. current leads voltage by `pi`B. current is in phase with voltageC. current leads voltage by `pi//2`D. current lags voltage by `pi//2` |
| Answer» Correct Answer - C | |
| 825. |
Wattless current is obtainedA. when resistance is zeroB. when current in minimumC. when inductance is zeroD. when current is alternating current |
| Answer» Correct Answer - A | |
| 826. |
Resistance of a resistor is independent onA. lengthB. area of cross sectionC. temperatureD. frequency |
| Answer» Correct Answer - D | |
| 827. |
Production of induced emf in a coil due to the changes of current in the same coil isA. self inductionB. mutual inductionC. dynamoD. none of these |
| Answer» Correct Answer - A | |
| 828. |
The coefficient of self inductance of a coil isA. `L=e(dI)/(dt)`B. `L=e(dI)/(dr)`C. `L=-(dI)/(dr)`D. `L=-e(dt)/(dI)` |
| Answer» Correct Answer - D | |
| 829. |
When a rate of change of current in a circuit is unity, the induced emf is equal toA. thickness of the coilB. number of turns in the coilC. coefficient of self inductionD. total flux linked with the coil |
| Answer» Correct Answer - C | |
| 830. |
When a rate of change of current in a circuit is unity, the induced emf is equal toA. thickness of coilB. number of turns in coilC. coefficinet of self inductanceD. total flux linked with coil |
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Answer» Correct Answer - C As , `epsi=L(dI)/(dt)` When `(dI)/(dt)=1` `:. Epsi=L` |
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| 831. |
In the given branch `AB` of as circuit a current `I=(10t+5)A` is flowing, where `t` is time in second . At `t=0`, the potential difference between points `A` and `B(V_A-V_B)` is A. `15V`B. `-5V`C. `-15V`D. `5V` |
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Answer» Correct Answer - A `I=(10t+5)A` `(dI)/(dt)=10A//s=`constant At `t=0, I=5A` Now, `V_A-3xx5-1x10+10=V_A` `:. V_A-V_B=15V` |
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| 832. |
In an a.c. circuit, there is no power consumption in an ideal inductor. Explain. |
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Answer» Average power/cycle in an a.c circuit is `P = E_(v) I_(v) cos phi` In an ideal inductor,which has no ohmic resistance, phase angle `phi = 90^(@)` `:. P = E_(v) I_(v) cos 90^(@) = ` Zero. Infact, power supplied during growth of current is retrieved during decay of current through ideal inductor. |
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| 833. |
Can a.c. source be connected to a circuit and yet deliver no power to it ? If so, under what circumstance ? |
| Answer» Yes, this would happen when phase difference between alternating voltage and alternating current is `90^(@)`. It can happen when the circuit contians pure L or pure C. | |
| 834. |
In an A.C. circuit, `E=100 sin (500 t)` volt and `I=1000 sin(500 t +(pi)/3) mA`. The power dissipated in the circuit isA. 10 WB. 100 WC. 50 WD. 25 W |
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Answer» Correct Answer - D Power = `E_(rms) xx I_(rms) xx cos theta` `=(100)/(sqrt(2)) xx (1000 xx 10^(-3))/(sqrt(2)) xx cos 60^(@) = 25 W`. |
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| 835. |
The inductance of a coil is 10 H. What is the ratio of its reactance when it is connected first to an A.C. source and then to a D.C. source?A. 10B. 0C. `oo`D. `0.5` |
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Answer» Correct Answer - C For A.C., Reactance `(X_L) = omega L for D.C., f =0`, `:. Omega = 2pif = 0 :. Ratio = (omega L)/0) = oo`. |
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| 836. |
A step voltage `V_(0)` is applied to a series combination of `R` and `C` as shown Fig. Then A. after sufficienty long time `V_(R ) = 0`B. as time passes, `V_(R )` decreases as `(1//t)`C. after `1 ms, V_(C ) = 6.3` volt (approximately)D. initially, current through `R` is `10 mA`. |
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Answer» Correct Answer - A::C::D (a) After sufficently long time, since the charging current drops to zero, voltage drop across resistance is zero. (c ) Time constant ` = RC = K Omega xx 1 mu F = 1 ms`. So, accoridng to the definition of time constant, voltage across capacitor would be about `63%` of maximum voltage, i.e., `6.3 V` after 1 ms. (d) Initital current `= (10 V )/(1 KOmega) = 10 mA` |
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| 837. |
A coil of inductance 1.0 H and resistance 100 Omega is connected to a battery of emf 12 V. Find the energy stored in the magnetic field associated with the coil at an instant 10 ms after the circuit is switched on. |
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Answer» Time constant of circuiti is `tau = (L)/(R ) =(1.0)/(100) = 10^(-2)s` As `I = I_(0) (1 - e^(-t//tau)) = (E _(0))/(R ) (1 - e^(-t//tau`) `:. I = (12)/(100) (1 - e^(-10^(-2_(//10^(-2)))))` ` = 0.12 (1 - (1)/(e)) = 0.0758 A` Energy store `= (1)/(2) LI^(2) = (1)/(2) xx 1.0 (0.0758)^(2)` `= 2.87 xx 10^(-3) J` Maximum electric intnesity the capacitor can withstand = 10% dielectric strength i.e., `E = (10)/(100) xx 3.0 xx 10^(6) = 3 xx 10^(5) V//,` `d = 1 mm = 10^(-3) m` |
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| 838. |
When a current flows in the coil of a transformer, then why does its core become hot ? |
| Answer» Wher alternating current flows through the coil of a transformer, its core get magnetised and demagnetised repeatedly. The energy spent in magnetising the core is not returned fully in demagnetisation. This energy left in the core appears in the form of heat. | |
| 839. |
The core of a transformer is made of a material having norrow hysterisis loop. Why ? |
| Answer» A core material with narrow hysterisis loop involves much less energy loss (hysterisis loss) due to repeated magnetisation and demagnetisation of the core. | |
| 840. |
A `0.18 mu` F capacitor is first charged and then discharged through a high resistance. If it takes 0.5 sec for the chage to reduce to one forth of its initial value, find the value of the resistance. Given `log_(e) 4 = 1.386` |
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Answer» Here, `C = 1.08 mu F = 0.18 xx 10^(-6) F, t = 0.5 s, (q)/(q_(0)) = (1)/(4) = 0.25, R = ?` During dischanging, `q = q_(0) e^(-t//RC) :. e^(-t//RC) = (q)/(q_(0)) = (1)/(4)` `(-t)/(RC) log_(e) e = log_(e) 4 , (-t)/(RC) xx 1 = 0 - 1.386` `R = (t)/(1.386 C) = (0.5)/(1.368 xx 0.18 xx 10^(-6)) = 2 xx 10^(6) Omega = 2 M Omega` |
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| 841. |
In a d.c. motor, do we use slip ring or split ring arrangement and why? |
| Answer» In d.c. motor, we use split rings. This helps us in keeping same, the direction of rotational torque on the coil. | |
| 842. |
The one of a transformer is laminated, why ? |
| Answer» Lamination of core is required to reduce energy losses due to eddy currents. | |
| 843. |
The reactance of 2 H inductance in an ac circuit of frequency 50 Hz isA. `628Omega`B. `826Omega`C. `286Omega`D. `862Omega` |
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Answer» Correct Answer - A `X_(L)=2pifL` `=2xx3.14xx50xx2=628Omega` |
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| 844. |
At Curie point, a ferromagnetic material transforms into:A. DiamagneticB. ParamagneticC. FerromagneticD. Antiferromagnetic |
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Answer» Correct Answer - A |
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| 845. |
A parallel plate capacitor with plate area A and plate separation d, is charged by a constant current I. Consider a plane surface of area `A//2` parallel to the plates and situated symmetrically between the plates. Determine the displacement current through this area.A. IB. `I/2`C. `2I`D. `I/3` |
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Answer» Correct Answer - B Let q be the charge on the capacitor plates at any instant and `sigma` be the surface charge density. Then the electric field between the plates, `E = (sigma)/(epsilon_0) = (q)/(A epsilon_(0)) "where" sigma = q/A` `:.` Flux through the area `A/2 = E xx Area` `phi_(E) = q/(A epsilon_(0)) xx A/2 = q/(2 epsilon_0)` `:.` The displacement current `=I_(D) = epsilon_(0) (d phi_(E))/(dt) = epsilon_(0)1/(2 epsilon_(0)) xx (dq)/(dt)` `:. I_(D) = I/2` where `I = (dq)/(dt)` = the rate at which the charge flows through the conducting wire. |
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| 846. |
A choke of 0.5 H, a capacitor of `15 mu F` and resistance of 100 ohm are connected in series across 200 V, 50 Hz main. Find current in the circuit and power factor of the circuit. |
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Answer» Here, `L = 0.5 H, C = 15 xx 10^(-6) F, R = 100 ohm` `E_(v) = 200 V, v = 50 Hz, I_(v) = ?` power factor? `X_(L) = omega L = 2 pi v L = 2 pi xx 50 xx 0.5 = 157 Omega` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C) = (1)/(2 pi xx 50 xx 15 xx 10^(-6))` `= 2.12 xx 10^(2) ohm = 212 Omega` `Z = sqrt(R^(2) + (X_(C ) - X_(L))^(2))` `= sqrt(100^(2) + (212 - 1578)^(2))` `= sqrt(100^(2) + 55^(2)) = 114.127 Omega` `I_(v) = (E_(v))/(Z) = (202)/(114.127) = 1.752 A` Power factor, `cos phi = (R )/(Z) = (100)/(114.127) = 0.876` |
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| 847. |
Which graph gives the correct relation between Z and f for the given R-C circuit? A. B. C. D. |
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Answer» Correct Answer - C `Z=sqrt(R^(2)+X_(C)^(2))=sqrt(R^(2)+((1)/(2pifC))^(2))` Thus, as f increases Z decreases. |
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| 848. |
Keeping the source of frequency equal to the resonating frequency of the series LCR circuit, if the three elements L, C and R in are arranged in parallel , show that the total current in the parallel LCR circuit is a minimum at this frequency. Obtain the r.m.s. value of current in each brach of the circuit for the elements and source specified in for this frequency. |
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Answer» If Z is the effecitve impedance of the parallel LCR circuits shown in Fig then in vector form, `(1)/(vec Z) = (1)/(vec R) + ((1)/(vec X_(L)) - (1)/(vec X_(C )))` As reactance `(X_(C ) - X_(L))` is perpendicular to ohmic resistance R, therefore, in magnitude, we may write `(1)/(Z) = sqrt((1)/(R^(2)) + ((1)/(omega L) - omega C)^(2))` At `omega = omega_(r) = (1)/(sqrt(LC)) , (1)/(Z)` = minimum. `:.` Z is maximum. `:.` Total current will be minimum. Through L, `I_(v) = (E_(v))/(omega L) = (230)/(50 xx 5) = 0.92 A` Through C, `I_(v) = (E_(v))/(1 // omega C) = omega C E_(v) = 5 xx (80 xx 10^(-6)) xx 230 = 0.92 A` Through R, `I_(v) = (E_(v))/(R ) = (230)/(40) = 5.75 A` As currents in L and C are equal, but `180^(@)` out of phase, therefore, they cancel out. `:.` Total current `= I_(v) =` current through `R = 5.75 A` |
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| 849. |
In the circuit shown in the figure initialy the switch in position 1 for a long time, then suddenly at `t=0` the switch is shifted to position 2. It is required that a constant current should flow in the circuit, the value of resistance `R` in the circuit A. should be decreased at constant rateeB. should be increased at a constant rateC. should be maintained constantD. not possible |
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Answer» Correct Answer - D In decay of current though `L-R` circuit, current can not remain constant. |
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| 850. |
In figure, the switch is in the position 1 for long time, then the switch is shifted to position 2 at `t=0`. At this instant the value of `i_1` and `i_2` are A. `E/R,0`B. `E/R,(-E)/R`C. `E/(2R),(-E)/(2R)`D. None of these |
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Answer» Correct Answer - B At `t=0, i=E//R` Now, this current will decay in closed loop in anti clockwise direction. So, `|i_2|=i_2=E//R` in upward or opposite direction. Hence, `i_2=-E/R` |
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