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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.
| 1001. |
A conductor is moving with the velocity v in the magnetic field and induced current is I. If the velocity of conductor becomes double, the induced current will beA. 0.5 IB. `sqrt(2)I`C. `2I`D. `4I` |
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Answer» Correct Answer - C Indued emf is proportional to B. thus, induced current is also proportional to B. |
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| 1002. |
An alternating emf, e=200 sin `omegat` is applied to a lamp, whose filament has a resistance of `1000Omega`. The rms valve of current isA. 0.5AB. 1.4AC. 0.14AD. 0.04A |
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Answer» Correct Answer - C `I_(rms)=(e_(rms))/(R)` `=(e_(0))/(sqrt(2)R)` `=(200)/(sqrt(2)xx1000)=0.1414A` |
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| 1003. |
The alternating emf of `e=e_(0)` sin `omegat` is applied across capacitor C. the current through the circuit is given byA. `I=I_(0)sinomegat`B. `I=I_(0)sin(omegat+(pi)/(2))`C. `I=I_(0)sin(omegat-(pi)/(2))`D. `I=I_(0)sin(omegat-pi)` |
| Answer» Correct Answer - B | |
| 1004. |
The phase difference between the voltage and the current in an AC circuit is `pi//4`. If the frequency is `50Hz` then this phase difference will be equivalent to a time ofA. 0.05 sB. 2.5 millisecondC. 25 millisecondD. 0.25 s |
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Answer» Correct Answer - B Phase diff. = `(pi)/4 T = 1/n = 1/50 s` `:.` Time T correspond to a phase diff. of `2 pi` `:.` for a phase diff. of `(pi)/4`. `t = T/8 = 1/(50 xx 8) = 1/400 s :. t= 2.5` millisecond. |
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| 1005. |
When a charged condenser is allowed to discharge through inductor the electrical oscillations are produced called asA. LC oscillationsB. RC oscillationsC. RL oscillationsD. none of these |
| Answer» Correct Answer - A | |
| 1006. |
In `L-C` oscillationsA. time period of oscillation is `(2pi)/sqrt(LC)`B. maximum curret in circuit is `q_0/sqrt(LC)`C. maximum rate of change of current in circuit is `q_0/(LC)`D. maximum potential difference across the inductor is `q_0/(2C)` Here `q_0` is maximum charge on capacitor. |
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Answer» Correct Answer - B::C `i_(max)=omegaq_0=(1/sqrt(LC))q_0` `((di)/(dt))_(max)=omega^2q_0=(1/(LC))q_0` |
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| 1007. |
Which of the following statements is/are correct?A. Self-induced emf may tend to decrease the currentB. Self-induced emf may tend to increse the currentC. Self-induced emf triesto keep the current constnatD. None of these |
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Answer» Correct Answer - A::B::C::D |
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| 1008. |
In the circuit shown the key `(K)` is closed at `t=0`, the current through the key at the instant `t=10^-3 ln, 2` is A. 2AB. 8AC. 4AD. Zero |
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Answer» Correct Answer - C |
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| 1009. |
In the circuit shown in figure, circuit is closed at time `t=0`. At time `t=ln(2)` second A. rate of energy supplied by the battery is 16 J/sB. rate of heat dissipated across resistance is 8J/sC. rate of heat dissipated across resistance is 16 J/sD. `V_a-V_b=4V` |
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Answer» Correct Answer - A::B::D `tau_L=L/R=2/2=1s` `t 1/2=(ln 2) tau_L=(ln 2) s` Hence the given time is half time. `:. i=i_0/2=(8//2)/2=2A` Rate of energy supplied by battery `=Ei=8xx2=16J//s` `P_R=i^2R=(2)^2(2)=8J//s` `V_a-V_b=E-iR=9-2xx2=4V` |
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| 1010. |
In the circuit shown the key `(K)` is closed at `t=0`, the current through the key at the instant `t=10^-3 ln, 2` is A. `2A`B. `8A`C. `4A`D. zero |
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Answer» Correct Answer - A `tau_L=L/R=0.01/10=10^-3s` `tau_C=CR=(0.1xx10^-3)(10)=10^-3s` `(i_0)_L=20/10=2A` `(i_0)_C=20/10=2A` The given time is the half life time of both the circuits. `:. i_L=i_C=2/2=1A` or total current is `2A`. |
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| 1011. |
In the circuit shown in figure, circuit is closed at time `t=0`. At time `t=ln(2)` second A. Rate of energy supplied by the battery is 16J/sB. Rate of heat dissipated across resistance is 8J/sC. Rate of heat dissipated across resistance is 16J/sD. `V_(a)-V_(b)=4V` |
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Answer» Correct Answer - A::C |
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| 1012. |
The potential differences `V` and the current `i` flowing through an instrument in an `AC` circuit of frequency `f` are given by `V=5 cos omega t` and `I=2 sin omega t` amperes (where `omega=2 pi f`). The power dissipated in the instrument isA. ZeroB. 10WC. 5WD. 2.5W |
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Answer» Correct Answer - D |
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| 1013. |
For an LCR series circuit with an aac source of angular frequency `omega`.A. Circuit will be capacitive if `omegagt(1)/sqrt(LC)`B. Circuit will be inductive if `omega=(1)/sqrt(LC)`C. Power factor of circuit will by unity if `omegaL=(1)/(omegaC)`D. Current will be leading if `omegagt(1)/sqrt(LC)` reactance equals inductive reactance. |
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Answer» Correct Answer - C::D |
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| 1014. |
For an LCR series circuit with an aac source of angular frequency `omega`.A. Circuit will be capacitive if `omegagt(1)/sqrt(LC)`B. Circuit will be inductive if `omega=(1)/sqrt(LC)`C. Power factor of circuit will by unity of capacitive reactance equals inductive reactance.D. Current will be leading voltage if `omegagt(1)/sqrt(LC)` |
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Answer» Correct Answer - C |
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| 1015. |
A coil having an inductance `L` and a resistance `R` is connected to a battery of `emf epsilon`. Find the time elapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches half its maximum value and (c) the magnetic field energy stored in the circuit reaches half its maximum value.A. `R/L ln [sqrt2/(sqrt-1)]`B. `L/Rln [(sqrt2-1)/sqrt2]`C. `L/R ln (sqrt2/(sqrt2-1))`D. `R/L ln [(sqrt2-1)/sqrt2]` |
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Answer» Correct Answer - C `1/2Li^2=1/2[1/2Li_0^2]` `:. i=i_0/sqrt2=i_0(1-e^(-t//tau_L))` `e^(t//tau_L)=1-1/sqrt2=(sqrt(2)-1)/(sqrt(2))` `:. t/tau_L=ln (sqrt2/(sqrt2-1))` `or t=tau_L ln (sqrt2/(sqrt2-1))` `=L/R ln ((sqrt(2))/(sqrt(2)-1))` |
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| 1016. |
An inductor coil stores 32 J of magnetic field energy and dissiopates energy as heat at the rate of 320 W when a current of 4 A is passed through it. Find the time constant of the circuit when this coil is joined across on ideal battery. |
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Answer» Magnetic field energy ` = 32 = (1)/(2) LI^(2) = (1)/(2) L (4)^(2)` `:. L = 4 H` Power dissipated as heat, `P = I^(2) R` `320 = 4^(2) R R = 20 Omega` Time constant `tau = (L)/(R ) = (4)/(20) = 0.2 s` |
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| 1017. |
A hundred turns of insulated copper wire are wrapped around an iron cylinder of area `1xx10^(-3) m^(2)` and are connected to a resistor. The total resistance in the circuit is `10` ohms. If the longitudinal magnetic induction in the iron changes from `1` weber `m^(-2)`, in one direction to `1` weber `m^(-2)` in the opposite direction, how much charge flows through the circuitA. `2xx10^(-2)`CB. `2xx10^(-3)`CC. `2xx10^(-4)`CD. `2xx10^(-5)`C |
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Answer» Correct Answer - A |
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| 1018. |
A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is `n` and the cross sectional area of the coil is `A`. When the coil turns through `180^(@)` about its diameter, the charge flowing through the coil is `Q`. the total resistance of the circuit is `R`. what is the magnitude of the magnetic induction?A. `(QR)/(nA)`B. `(2QR)/(nA)`C. `(Qn)/(2RA)`D. `(QR)/(2nA)` |
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Answer» Correct Answer - D |
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| 1019. |
A conductuing loop having a capacitor is moving outward from the magnetic field. Which plate of the capacitor will be positive? A. Plate AB. Plate BC. Plate A and Plate B bothD. None |
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Answer» Correct Answer - A |
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| 1020. |
The figure shows four wire loops, with edge length of either `L` or `2L`. All four loops will move through a region of uniform magnetic field `vecB` (directed out of the page) at the same constant velocity . Rank the four loops according to the maximum magnitude of the e.m.f. induced as they move through the field, greatest first A. `(e_(c) = e_(d)) lt (e_(a) = e_(b))`B. `(e_(c) = e_(d)) gt (e_(a) = e_(b))`C. `e_(c) gt e_(d) gt e_(b) gt e_(a)`D. `e_(c) lt e_(d) lt e_(b) lt e_(a)` |
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Answer» Correct Answer - b |
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| 1021. |
The magnetic field in the cylindrical region shown in figure increase at a constant rate of `20mT//sec`. Each side of the square loop `ABCD` has a length of `1 cm` and resistance of `4Omega`. Find the current in the wire `AB` if the switch `S` is closed A. `1.25xx10^(-7)`A( anticlockwise)B. `1.25xx10^(-7)`A ( clockwise)C. `2.5xx10^(-7)` A ( anticlockwise)D. `2.5xx10^(-7)`A(clockwise) |
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Answer» Correct Answer - A |
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| 1022. |
A plane rectangular loop is placed in a magnetic field. The emf induced in the loop due to this field is `epsilon_(1)` whose maximum value is `epsilon_(im)`. The loop was pulled out of the magnetic field at a variable velocity. Assume that `vec(B)` is uniform and constant `epsilon_(1)` is plotted against t as shown in the graph. Which of the following are/is correct statement(s): A. `epsi_(im)` is independent of rate of removal of coil from the field.B. The total charge that passes through any point of the loop in the process of complete removal of the loop does not depend on velocity of removal.C. The total area under the curve (`epsi_(i)`vs t) is independent of rate of removal of coil from the field.D. The area under the curve is dependent on the rate of removal of the coil. |
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Answer» Correct Answer - B::C |
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| 1023. |
A hollow cylinder made of material of resistivity `rho` has length a and wall thickness `d(l gt gt a gt gt d)`. A current I flows through the cylinder in tangential direction and is uniformly distributed along its length. (a) Find the emf developed along the circumference of the cylinder if the current changes at a rate `(dl)/(dt) = beta` (b) Assume that no external source is present and the current at time t = 0 is `I_(0)`. The current decays with time. Write I as a function of time. |
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Answer» Correct Answer - (a) `(mu_(0)pia^(2)beta)/(l)` (b) `I = I_(0)e^(-(2rhot)/(mu_(0)ad))` |
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| 1024. |
A `1.5 mu F` capacitor is charged to 57 volt.The charging battery is then disconnected and a 12 mH coil is connected across the capacitor so that is zero, what is the maximum value of current in the coil ? |
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Answer» As `R = 0`, therefore, `(1)/(2) LI^(2) = (1)/(2) CV^(2)` `:. I = V sqrt((C )/(L)) = 57 sqrt((1.5 xx 10^(-6))/(12 xx 10^(-3)))` `= 63.72 xx 10^(2) A = 637.2 mA` |
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| 1025. |
A `100 mu F` capacitor is charged to a potentail of 50 V. The battery is then disconnected to it. Calculate maximum current in the coil and frequency of LC oscillations |
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Answer» Here, `L = 100 mu H = 100 xx 10^(-6) H = 10^(-4) H` `C = 0.01 mu F = 10^(-8) F, v = ? , lambda = ?` `v = (1)/(2 pi sqrt(LC)) = (1)/(2 pi sqrt(10^(-4) xx 10^(-8))) = (10^(6))/(2 pi)` ` = 1.592 xx 10^(5) hz = 159.2 kHz` `lambda = (upsilon)/(v) = (3 xx 10^(8))/(1.592 xx 10^(5)) = 1.884 xx 10^(3) m` If we assume that there is no loss of energy, then the energy oscillation will continue for infinite time. |
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| 1026. |
Calculate the mutual inductance between two coils when the currents change from 2 to 5 A in 0.25s in one coil, induces an emf of 12 mV in the other. |
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Answer» Correct Answer - 1 mH |
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| 1027. |
Calculate the mutual inductance between two coils when a current 2A changes to 6A in and 0.2 s and induces an emf of 20mV in secondary coil. |
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Answer» We know that emf is related to the mutual inductance by `e=M(Deltai)/(Deltat)` Substituting the given values, `20xx10^(-3)=(M(6-2))/(0.2)=(Mxx4)/(0.2)=20M` `rArr" "M=10^(-3)H="1 mH"` |
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| 1028. |
A small town with a demand of 1 MW at 220 V is situated 20 km away from an electric plant generating power at 440 V. The resistance of two line wires carrying is `0.5 Omega` per km. Two step down transformers are available : 4000 V - 220 V and 100000 V - 220 V at a substationin the town for supply of power. Which one will you prefer and why? |
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Answer» Here, power required `= 1 MW = 10^(6) W = 10^(3) kW` Total resistance of two line wires `R = 2xx 20 xx 0.5 = 20 ohm`. (a) If power is supplied throuhg `4000 V - 220 V` transformer `E_(v) = 4000 V, I_(v) = (P)/(E_(v)) = (10^(6)W)/(4000 V) = 250 A` Line power loss `=I_(v)^(2) = (250)^(2) xx 20 = 1250000 W = 1250 kW` If there is no other power loss, then the essential plant supply `= (1000 + 1250) kW = 2250 kW` voltage drop on the line `= I_(v) xx R = 250 xx 20 = 5000 V :.` voltage for transmission `= 4000 + 5000 = 9000 V` As power is generated at 440 V, the step up transformer needed at the plant is 440 V - 9000 V Power loss in the process `= (1250)/(2250) xx100% = 55.5%` (b) If is supplied through 100,000 V - 220 V transformer `E_(v) = 1000,000 V` `I_(v) = (P)/(E_(v)) = (10^(6)W)/(100,000 V) = 10 A` Line power loss `= I_(v)^(2) = 10^(2) xx 20 = 2000 W = 2 kW` If there is no other power loss, then the essential plant supply `= (100 + 2 ) kW = 1002 kW` voltage drop on the line `= I_(v) xx R = 10 xx 20 = 200 V` `:.` Voltage for transmission `= (100000 + 200) V = 100200 V` As power is generated at 440 V, the step up transformer needed at the plant is `440 V - 100200 V` |
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| 1029. |
In circuit shown in figure-5.149 if switches `S_(1) " and " S_(2)` are closed at t=0, find the oscillation frequency of charge in this circuit. Neglect mutual induction between inductors. |
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Answer» Correct Answer - `(1)/(6pi sqrt(LC))` |
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| 1030. |
A small circular loop is suspended from an insulating thread. Another coaxial circular loop carrying a current `I` and having radius much larger than the first loop starts moving towards the smaller loop. The smaler loop will A. Be attracted towards the bigger loopB. Be repelled by the bigger loopC. Expreince no forceD. All of the above |
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Answer» Correct Answer - D |
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| 1031. |
An alternating current generator has an internal resistance `R_(g)` and an internal reactance `X_(g)`. It is used to supply power to a passive load consisting of a resistance `R_(g)` and a rectance `X_(L)`. For maximum power to be delivered from the generator to the load, the value of `X_(L)` is equal toA. zeroB. `X_(g)`C. `-X_(g)`D. `R_(g)` |
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Answer» Correct Answer - C For delivering maximum power from the generator to the passive load, total reactance must vanish, i.e., `X_(L) + X_(g) = 0` or `X_(L) = - X_(g)`. Choice (c ) is correct. |
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| 1032. |
A coil takes a current of 2.0 ampere and 200 watt power from an alternating current source of 220 volt, 50 hertz. Calculate resistance and inductance of the coil. |
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Answer» `R = P//I^(2)` because power is disspated only due to ohmic resistance. Use `Z = V//I sqrt(R^(2) + X_(L)^(2))` |
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| 1033. |
Equation of emf of a generator is `V = 282 sin 100 pi t` volt. Internal resistance of generator is `2000 Omega`. It is connected as shown in Fig. Find the frequency of generator and impedance of circuit. |
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Answer» Here, `V = 282 sin 100 pi t, R = 2000 Omega` `L = (40)/(pi) H` and `C = (10)/(pi) mu F . V = ? , Z = ?` The standard form of alternating emf is `E = E_(0) sin omega t`. Compare it with `V = 282 sin 100 pi t`, to get `E_(0) = 282 V, omega = 2 pi v = 100 pi`, `v = (100)/(2) = 50 Hz` `X_(L) = omega L = 2 pi v xx (4)/(pi)` `= 2 xx 50 xx 40 = 4000 ohm` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `(1)/(2 pi xx 50 xx (10)/(pi) xx 10^(-6)) = 1000 Omega` `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` `= sqrt((2000)^(2) + (4000 + 1000)^(2))` `= 10^(3) sqrt( 4 + 9) = 3.6 xx 10^(3) Omega` |
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| 1034. |
In an A.C. circuit the instantaneous values of e.m.f. And current are given by `E=100 sin (200 t)` volt and `I=2 sin(200 t +(pi)/3)` ampere. The average power consumed is the circuit isA. 200 WB. 100 WC. 50 WD. 25 W |
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Answer» Correct Answer - C `P_(av) = (E_0)/(sqrt2) xx (I_0)/(sqrt2) xx cos theta = (100 xx 2)/(2) xx 1/2 = 50 W`. |
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| 1035. |
The e.m.f. In an A.C. circuit at any instant is `E=200 sin[100 pit+(pi)/6]` volt. The time when the voltage become maximum for the first time isA. `1/10 s`B. `1/100 s`C. `1/200 s`D. `1/300 s` |
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Answer» Correct Answer - D `E = 200 sin [100 pi t + (pi)/6]` will be maximum, when `sin[100 pi t+(pi)/6] = 1 = sin(pi)/2` `:. 100 pi t +(pi)/6 = (pi)/2` `:. 100 pi t = (pi)/2 - (pi)/6 = (pi)3 :. T = 1/300 s`. |
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| 1036. |
What are phasa lines and neutral line in respect of a generator ? |
| Answer» In a polyphase generator, one end of each coil is brought to a common point through shaft of the generator. The line wire from this points is called neutral line. The line wires from other ends of different coils through their slip rings and brushes are called phase lines. | |
| 1037. |
In an ac generator, the number of turns in the coil is 2000 and the area of coil is `0.2 m^(2)`. If the coil is rotated at an angular frequency of `400 S^(-1)` in a magnetic field of `0.5 T`, then what is the peak value of induced emf ?A. `40,000 V`B. `60,000 V`C. `80,000 V`D. `20,000 V` |
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Answer» Correct Answer - C Here, `N = 200, A = 0.2 m^(2), B = 0.5 T, omega = 400 s^(-1)` Peak value of induced e.m.f. `E_(0) = NAB omega` `= 2000 xx 0.2 xx .5 xx 400 = 80,000 V` |
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| 1038. |
The e.m.f. of a.c. Generator is given by `E = 200 sin (100 pi t+ pi//3)` where E is in volt and t in sec. Which is the correct option from the following?A. The peak value of the e.m.f. is `200sqrt(2)V`B. At time t=0, the plane of the coil is perpendicular to the fieldC. At times t=0, the plane of the armature makes an angle of `60^(@)` with the magnetic fieldD. The frequency of rotation of the armature is 50 Hz |
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Answer» Correct Answer - D a,b and c are wrong because (i) the peak value of the e.m.f. = 200V (ii) at time `t=0, E=E_(0) sin (pi//3)` or `alpha = (pi)/(3)` i.e. the plane of the armature makes an angle of 60^(@) with the magnetic field. The angles are not measured with the plane of the coil. However angular frequency `omega t = 100 pi t` `:. 2 pi n =100 pi :. n = 50 Hz` `:.` (d) is the correct option. |
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| 1039. |
The coil of an a.c. generator has 100 turns, each of cross sectional area `2 m^(2)`. It is rotating at a constant angular speed of 30 radians`//`s, in a uniform magnetic field of `2 xx 10^(-2)`T. What is the maximum power dissipated in the circuit, if the resistance of the circuit including that of the coil is `600 Omega`?A. 6 WB. 9 WC. 12 WD. 24 W |
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Answer» Correct Answer - C `N = 1000,A = 2 m^(2), omega = 30 rad//s, T = 2 xx 10^(-2)T`, `R = 600 Omega` Maximum power dissipated in the circuit, `P_(max) = E_(rms) xx I_(rms) = (E_0)/(sqrt2) xx (I_0)/(sqrt2) = (E_(0)I_(0))/(2)` But `I_(0) = (E_0)/R` `:. P_(max) = (E_(0) cdot E_(0))/(R xx 2) = (E_(0)^(2))/(2R) but E_(0) = NAB omega` `:. P_(max) = ((NABomega)^(2))/(2R)` `=((100 xx 2 xx 2 xx 10^(-2) xx 30)^(2))/(2 xx 600)` `:. P_(max) = (120 xx 120)/(2 xx 600) = 12 W`. |
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| 1040. |
In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency `(omega)` in a magnetic field B. The maximum value of emf generated in the coils isA. NABRB. `omega NAB`C. `omega NABR`D. NAB |
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Answer» Correct Answer - B The induced emf `E = nAB omega sin omega t` It is maximum when `sin omega t = 1` `:. E_(max) = nAB omega`. |
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| 1041. |
A circuit element is placed in a closed box. At time I= 0, constant current generator supplying a current of IA, is connected across the box. Potential difference across the box varies according to graph shown in figure-5.248. The element in the box is: A. Resistance of `2Omega`B. Battery of emf `6V`C. Inductance of 2HD. Capacitors of 3F |
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Answer» Correct Answer - C |
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| 1042. |
An a.c. generator consists of a coil of 1000 turns and area `2m^(2)`. The coil is rotating in a transverse uniform magnetic field of 0.2 T at an angular speed of `60 rad//s`. What is the maximum current drawn from the generator if resistance of the circuit including that of the coil is `6000 Omega`?A. 2.5 AB. `3 A`C. `4 A`D. `5 A` |
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Answer» Correct Answer - C `N = 1000, A = 2m^(2), B = 0.2 T and omega = 60 rad//s and R = 6000 Omega` `E_(0) = NAB omega and I_(0) = (E_0)/(R)` `:. I_(0) = (NABomega)/(R) = (1000 xx 2 xx 0.2 xx 60)/(6000) = 4A`. |
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| 1043. |
What is meant by form factor of an a.c. generator ? |
| Answer» Form factor `= (I_(v))/(I_(av)) = (I_(0)//sqrt2)/(2 I_(0)//pi) = (pi)/(2 sqrt 2) = 1.1` | |
| 1044. |
Consider the LCR circuit shown in Fig. Find the net current I and the phse of i. show that `i= (upsilon)/(Z)` . Find the impedence Z for this circuit. |
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Answer» In Fig. let i. be the total current form the source. It is divided into two parts : `i_(1)` through R and `i_(2)` through series combination of C and L, such that `i= i_(1) + i_(2)`. As `R i_(1) = upsilon_(m) sin omega t, i_(1) = (upsilon_(m) sin omega t)/(R )` If `q_(2)` is charge on C at time t, then for series combination of C and L, `(q_(2))/(C ) + (dq_(2)^(2))/(dt^(2)) = upsilon_(m) omega t` Let `q_(2) = q_(m) sin (omega t + phi)` `:. (dq_(2))/(dt) = q_(m) omega cos (omega t + phi)` and `(d^(2) q_(2))/(dt) = - q_(m) omega^(2) sin (omega t + phi)` Putting in (ii), we get `q_(m) [(1)/(C ) = L omega^(2)] sin (omega t + phi) - upsilon_(m) sin omega t` If `phi = 0` and `((1)/(C ) - L omega^(2)) gt 0`, then `q_(m) = (upsilon_(m))/(((1)/(C ) - L omega^(2)))` From (iii), `i_(2) = (dq_(2))/(dt) = omega q_(m) cos (omega t + phi)` , using (iv), `i_(2) = (omega upsilon_(m) cos (omega t + phi))/((1)/(C ) - L omega^(2))` Taking `phi = 0`, `i_(2) = (upsilon_(m) cos (omega t))/(((1)/(omega C) - L omega))` From (i) and (v), we find that `i_(1)` and `i_(2)` are out of phase. Now, `i_(1) + i_(2) = (upsilon_(m) sin omega t)/(R ) + (upsilon_(m) cos omega t)/(((1)/(omega C) - L omega))` Put `(upsilon_(m))/(R ) = A = C cos phi` and `(upsilon_(m))/(((1)/(omega C) - L omega)) = B = C sin phi` `:. i_(1) + i_(2) = C cos phi sin omeg t + C sin phi cos omega t = C sin (omega t = phi)` Where `C = sqrt(A^(2) + B^(2))` and `phi = tan^(-1) ((B)/(A))` `C = [(upsilon_(m)^(2))/(R^(2)) + (upsilon_(m)^(2))/((1)/(omega C) - L omega)^(2) ]^(1//2)` and `phi (tan^(-1)) (R )/(((1)/(omega C) - L omega))` Hence, `i= i_(1) + i_(2) = [(upsilon_(m)^(2))/(R^(2)) + (upsilon_(m)^(2))/(((1)/(omega C) - L omega)^(2))]^(1//2) sin (omega t + phi)` or `(i)/(upsilon_(m)) = (1)/(Z) = [(1)/(R^(2)) = (1)/(((1)/(omega C) - L omega)^(2))]^(1//2)` This is the expression for impedance Z of the circuit. |
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| 1045. |
In series LCR circuit `R = 18 Omega` and impedence is `33 Omega`. An Vrms voltage `220 V` is applied across the circuit . The true power consumed in AC circuit isA. 220 WB. 400 WC. 600 WD. 800 W |
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Answer» Correct Answer - D Power consumed in a circuit `P = e_(rms) xx I_(rms) xx cos phi` `= e_(rms) xx (e_(rms))/(Z) xx R/Z` `= 220 xx 220/33 xx 18/33` `= (20 xx 20 xx 18)/(3 xx 3) =800 W` .....Option (d) |
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| 1046. |
The flux in a closed circuit of resistance `20 Omega` varies with time according to the equation `phi = 6t^(2)-5t+1`. What is the induced current at time `t = 0.25` second?A. 0.5 AB. 0.4 AC. 0.1 AD. 0.2 A |
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Answer» Correct Answer - C `phi=6t^2-5t+1 e = -(d phi)/(dt)=-[12t-5]` at time `t = 0.25 s, e = -[12 xx 0.25 - 5] = 2V` `:. E=e/R=2/20 = 0.1A`. |
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| 1047. |
The magnetic flux linked with a coil is given by `phi=5t^(2)+3t+2` What is the e.m.f. Induced in the coil in the third second?A. 5 VB. 10 VC. 15 VD. 20 V |
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Answer» Correct Answer - B `e = (d phi)/(dt) = d/(dt)(5t^(2)+3t+2)=10t+3` When `t=2 sec, e_(1)=20+3=23 V` When `t=3 sec, e_(2)=30+3=33` `:.` e.m.f. induced in the third second = 33 - 23 = 10V. |
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| 1048. |
The magnetic flux `(phi)` linked with a coil depends on time `t` as `phi=at^(n)` , where a and `n` are constants. The emf induced in coil is `e` . (i) If `0ltnltI,e=0` (ii) If `0ltnltI,een0` and `|e|` decreases with time (iii) If `ngtI,e` is constant (iv) If `nltI,|e|` increases with timeA. `(i),(iii)`B. `(ii), (iii), (iv)`C. `(i), (ii)`D. `(ii), (iv)` |
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Answer» Correct Answer - D `phi=at^(n)` `|e|=ant^(n-1)` For `0ltnlt1` , `|e|=(an)/(t^(1-n))` As `n` increases from `0` to `1` , `|e|` decreases For `ngt1` , `|e|=ant^(n-1)` , `|e|` increases |
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| 1049. |
The sum and difference of self-inductances of two coils are `13 mH` and `5 mH` respectively. What is the maximum value of mutual inductance (im milli henry) of the two coil ? |
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Answer» Correct Answer - 6 Here, `L_(1) + L_(2) = 12 mH, L_(1) = L_(2) mH , M = ?` Adding, we get `2 L_(1) = 18, L_(1) = 9 mH` `:. L_(2) = 13 = L_(1) = 13 - 9 = 4 mH` Max value of mutual inductance of two coils, `M = sqrt(L_(1) L_(2)) = sqrt(9 xx 4) = 6 mH` |
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| 1050. |
The sum and difference of self-inductances of two coils are `13 mH` and `5 mH` respectively. What is the maximum value of mutual inductance (im milli henry) of the two coil ?A. 6HB. 5HC. `sqrt65H`D. 18H |
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Answer» Correct Answer - A `L_(1) + L_(2)= 13` `L_(1) - L_(2)= 5` |
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