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InterviewSolution
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.
| 201. |
For L-R circuit, the time constant is equal toA. twice the ratio of the energy stored in the magnetic field to the ratio of dissipation of energy in the resistanceB. the ratio of the energy stored in the magnetic field to the ratio of dissipation of energy in the resistanceC. half of the ratio of the energy stored in the magnetic field to the ratio of dissipation of energy in the resistanceD. square of the ratio of the energy stored in the magnetic field to the ratio of dissipation of energy in the resistance |
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Answer» Correct Answer - A |
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| 202. |
In a series LR circuit, `X_L=R` and the power factor of the circuit is `P_1`.When a capacitor with capacitance C such that `X_C=X_L` is put in series , the power factor becomes `P_2`.Find out `P_1//P_2`. |
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Answer» In a series LR circuit, power factor `(P_1)=R/Z` Here, Z =impedance = `sqrt(R^2+X_L^2)` Here, `Z=sqrt(R^2+R^2)=sqrt2R` `thereforeP_1=1/sqrt2` In a series LCR circuit power factor `(P_2)=R/Z` where,`Z=sqrt(R^2+(X_L-X_C)^2)=R` `thereforeP_2=1` Hence,`P_1/P_2=1/sqrt2` |
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| 203. |
L, C and R represent the physical quantities, inductance, capacitance and resistance respectively. The combination(s) which have the dimensions of frequency areA. `LC`B. `(LC)-1//2`C. `(L/C)^(-1//2)`D. `C/L` |
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Answer» Correct Answer - B Frequency `=1/(2pisqrt(LC))` so the combination which represents dimension of frequency is `1/(sqrt(LC))=(LC)^(-1//2)` |
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| 204. |
In series `LR` circuit, `X_(L) = 3 R`. Now a capacitor with `X_(C ) = R` is added in series. The ratio of new to old power factorA. `sqrt3`B. 2C. `(1)/(sqrt2)`D. `sqrt2` |
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Answer» Correct Answer - C |
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| 205. |
Same current is flowing in two alternating circuits. The first circuit contains only inductances and the other contains only a capacitor, if the frequency of the e.m.f of AC is increased, the effect on the value of the current will beA. increased in the first circuit an decreased in the otherB. increased in both circuitsC. decreased in both the circuitsD. decreased in the first circuit and increased in the other |
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Answer» Correct Answer - D |
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| 206. |
A light bulb is rated at 100W for a 220 V supply. Find The rms current through the bulb. |
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Answer» Since, P = I V `I = (P)/(V)=(100 W)/(220 V)=0.450 A`. |
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| 207. |
A `0.2 k Omega` resistor and `15 mu F` capacitor are connected in series to a 220 V, 50 Hz ac source. The impadance of the circuit isA. `250 Omega`B. `268 Omega`C. `29.15 Omega`D. `291.5 Omega` |
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Answer» Correct Answer - D Here, `R=0.2 k Omega = 200 Omega` `C =15 mu F = 15xx10^(-6)F, V_("rms")=220 V, upsilon = 50 Hz` Capacitive reactance, `X_(C )=(1)/(2pi upsilon C)=(1)/(2xx3.14xx50xx15xx10^(-6))=212 Omega` The impedance of the Rccircuit is `Z = sqrt(R^(2)+X_(C )^(2))=sqrt((200)^(2)+(212)^(2))=291.5 Omega` |
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| 208. |
A `100 Omega` resistor is connected to a 220 V. 50 Hz ac supply. What is the net power consumed over a full cycle ? |
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Answer» Given resistance `R = 100 Omega` `V_("rms")=220 V` Frequency `f = 50 Hz` Net power consumed in full cycle `P = V_("rms")xx I_("rms")` `= 220xx2.2` `= 484 W` |
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| 209. |
A `100 Omega` resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?A. `1.56 A`B. `1.56 mA`C. `2.2 A`D. `2.2 mA` |
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Answer» Correct Answer - C Here, `R = 100 Omega V_("rms") = 220 V, upsilon = 50 Hz` `therefore I_("rms")=(V_("rms"))/(R )=(220)/(100)=2.2A` |
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| 210. |
A `100 Omega` resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? |
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Answer» Resistance of the resistor, `R=100Omega` Supply voltage, `V=220V` (a) The rms value of current in the circuit is given as: `I=(V)/(R)` `=(220)/(100)= 2.20A` (b) The net power consumed over a full cycle is given as: P=VI `=220xx2.2=484W` |
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| 211. |
A fully charged capacitor C with initial charge `q_(0)` is connected to a coil of self inductance L at t=0. The time at which the energy is stored equally between the electric and the magnetic fields isA. `(pi)/(4)sqrt(LC)`B. `2pi sqrt(LC)`C. `sqrt(LC)`D. `pi sqrt(LC)` |
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Answer» Correct Answer - A As `omega^(2)=(1)/(LC)` or `omega=(1)/(sqrt(LC))` Maximum energy stored in capacitor `=(1)/(2)(Q_(0)^(2))/(C )` Let at any instant t, the energy be stored equally between electric and magnetic field. Then energy stored in electric field at instant t is `(1)/(2)(Q^(2))/(C )=(1)/(2)[(1)/(2)(Q_(0)^(@))/(C )]` or `Q^(2)=(Q_(0)^(2))/(2)` or `Q=(Q_(0))/(sqrt(2)) rArr Q_(0) cos omega t = (Q_(0))/(sqrt(2))` or `omega t=(pi)/(4)` or `t=(pi)/(4omega) = (pi)/(4xx(1//sqrt(LC)))=(pi sqrt(LC))/(4)` |
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| 212. |
A fully charged capacitor C with initial charge `q_(0)` is connected to a coil of self inductance L at t=0. The time at which the energy is stored equally between the electric and the magnetic fields isA. `(pi)/(4) sqrt(LC)`B. `2 pi sqrt(LC)`C. `sqrt(LC)`D. `pi sqrt(LC)` |
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Answer» Correct Answer - 1 As initially charge is maximum `q = q_(0) cos omega t` `implies i = (dq)/(dt) = - omega q_(0) sin omega t` Given `(1)/(2) Li^(2) = (q^(2))/(2C)` `implies (1)/(2) (omega q_(0) sin omega t)^(2) = ((q_(0) cos omega t)^(2))/(2C)` But, `omega = (1)/(sqrt(LC)) implies tan omega t = 1` `omega t = (pi)/(4) implies t = (pi)/(4 omega) = (pi)/(4) sqrt(LC)` |
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| 213. |
A voltage , `E=60 sin(314t)`, is applied across a resistor of `20 Omega`. What will be the reading of `I_(rms)`. (a) in an ac ammeter? (b) in an ordinary moving coil ammeter in series with the resistor? |
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Answer» Given that `E=60 sin (314 t)` (a) An ac ammeter will read the rms value `E_(rms)=(E_0)/(sqrt(2))=(60)/(sqrt(2))=42.2V` `I_(rms)=(E_(rms))/(R )=(42.4)/(20)=2.12A` Therefore, ac ammeter will read 2.12A. (b) An ordinary moving coil ammeter will read the average value of alternating current. since the average value of ac is zero. This meter will give zero reading. |
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| 214. |
In an a.c. Circuit the voltage applied is `E=E_(0) sin (omega)t`. The resulting current in the circuit is `I=I_(0)sin((omega)t-(pi/2))`. The power consumption in the circuit is given byA. `P=(E_(0)I_(0))/(sqrt(2))`B. `P=sqrt(2) E_(0) I_(0)`C. `P=(E_(0)I_(0))/2`D. `P=0` |
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Answer» Correct Answer - D Phase angle `varphi=90^(@)`, so power `P=Vi cos varphi=0` |
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| 215. |
In the given LC circuit,if initially capacitor C has charge Q on it and 2C has charge 2Q. The polar ar as shown in figure. Then after closing switch S and t=0 A. energy will get equally distributed in both the capacitors ust after closing the switchB. initial rate of groqth of current in inductor will be 2Q/3CLC. maximum energy in the inductor will be `3Q^(2)//2C`D. none of these |
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Answer» Correct Answer - C |
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| 216. |
An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall beA. 10mAB. 20mAC. 40mAD. 80mA |
| Answer» Correct Answer - B | |
| 217. |
An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall beA. 10 mAB. 20 mAC. 40 mAD. 80 mA |
| Answer» Correct Answer - b | |
| 218. |
An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall beA. `10 mA`B. `20 mA`C. `40 mA`D. `80 mA` |
| Answer» Correct Answer - B | |
| 219. |
In an AC circuit , an alternating voltage e = 200``sin 100t V is connected to a capacitor of capacity `1mu F` . The rms value of the current in the circuit isA. 100 mAB. 200 mAC. 20 mAD. 10 mA |
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Answer» Correct Answer - C Given, e = 200`sqrt 2` sin 100t and C = 1`mu`F `underset(rms)(E)` = 200 V `underset(C )(X)` = 1/`omega`C = 1/`1 xx 10^(-6) xx 100` = `10^(4)Omega` `underset(rms)(i)` = `underset(rms)(E)`/`underset(C )(X)` `adr` `underset(rms)(i)` =`200 / 10^(4)` = `2 xx 10^(-2)` A = 20 mA |
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| 220. |
A device X is connected across an ac source of voltage `V=V_0sinomegat`. The current through X is given as `I=I_0sin(omegat+pi/2)`. Identify the device X and write the expression for its reactance. |
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Answer» Device X is a capacitor `therefore` Capacitive reactance , `X_C=1/(omegaC)=1/(2pifC)` |
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| 221. |
A small signal voltage V(t) =`V_0sinomegat` is applied across an ideal capacitor C.A. over a full cycle, the capacitor C does not consume any energy from the voltage sourceB. current I(t) is in phase with voltageV(t)C. current I(t) leads voltage V(t) by `180^@`D. current I(t) lags voltage V(t) by`90^@` |
| Answer» Correct Answer - A | |
| 222. |
A circuit contains resistance `R` and an inductance `L` in series. An alternating voltage `V=V_0sinomegat` is applied across it. The currents in `R` and `L` respectively will be .A. `I_(R)=I_(O)cosomegat, I_(L) = I_(0) cosomegat`B. `I_(R) = -I_(0) sin omegat, I_(L)cosomegat`C. `I_(R)=I_(0)sin omegat, I_(L) = -I_(0)cosomegat`D. none of the above |
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Answer» Correct Answer - D `I_(R) = I_(L)=I` Buy they have a phase difference `0ltphi lt 90^(@)`, with applied voltage. |
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| 223. |
The value of an ac voltage at time `0letle(pi)/omega` is given by `V=V_0sinomegat` and at time `pi/omegaletle(2pi)/omega` is given by `V=-V_0sinomegat` . The average value of V for a complete cycle isA. `V_0/sqrt2`B. `(2/pi)V_0`C. `V_0/2`D. zero |
| Answer» Correct Answer - B | |
| 224. |
The voltage over a cycle varies as `V=V_(0)sin omega t` for `0 le t le (pi)/(omega)` `=-V_(0)sin omega t` for `(pi)/(omega)le t le (2pi)/(omega)` The average value of the voltage one cycle isA. `(V_(0))/(sqrt2)`B. `((2)/(pi))V_(0)`C. `((2)/(pi))V_(0)`D. ZERO |
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Answer» Correct Answer - B |
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| 225. |
The voltage over a cycle varies as `V=V_(0)sin omega t` for `0 le t le (pi)/(omega)` `=-V_(0)sin omega t` for `(pi)/(omega)le t le (2pi)/(omega)` The average value of the voltage one cycle isA. `(V_(0))/(sqrt(2))`B. `(V_(0))/(2)`C. zeroD. `(2V_(0))/(pi)` |
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Answer» Correct Answer - D The average value of the voltage is `V_(av)=(int_(0)^(2pi//omega)Vdt)/(int_(0)^(2pi//omega)dt)=(int_(0)^(pi//omega)V_(0)sin omega dt + int_(pi//omega)^(2pi//omega)(-V_(0)sin omega t)dt)/((2pi)/(omega))` `=(omega)/(2pi)[|(-V_(0)cos omega t)/(omega)|_(0)^(pi//omega) + |(V_(0)cos omega t)/(omega)|_(pi//omega)^(2pi//omega)]=(2V_(0))/(pi)` |
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| 226. |
In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab) "at" t=5s`A. 50 VB. 35.5 VC. 46.5 VD. 40.2 V |
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Answer» Correct Answer - C |
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| 227. |
In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab)=(V_(a)-V_(b)) at t=3s`A. `-24.5 V`B. `18.5 V`C. `-25.5 V`D. `22.5 V` |
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Answer» Correct Answer - D |
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| 228. |
In the above question size element will raise the power factor to unity?A. an inductor should be placed in seriesB. a capacitor should be placed in seriesC. a resistance should be placed in seriesD. an inductor or a resistance should be placed in series |
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Answer» Correct Answer - D |
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| 229. |
A coil of inductance `0.50H` and resistance `100 Omega` is connected to a `240V`, `50Hz` ac supply. What are the maximum current in the coil and the time lag between voltage maximum and current maximum? |
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Answer» Given, Inductance L = 0.50 H Resistance `R = 100 Omega` `V_("rms")=240V, f=50Hz` Using the formula of time lag, `t=(phi)/(omega)` `tan phi = (X_(L))/(omega)=(omega L)/(R )=(2pi fL)/(R )= (2xx3.14xx50xx0.50)/(100)` `phi = tan^(-1)(1.571)=57.5` `phi=(57.5)/(180)pi` rod Time lag `t=(phi)/(omega)=(57.5pi)/(180xx2pi f)` `= (57.5)/(180xx2xx50)` `= 3.19xx10^(-3)S` Thus, the time lag between the voltage maximum and the current maximum is `3.19xx10^(-3)S`. |
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| 230. |
A `100(mu)F` capacitor in series with a `40 Omega` resistor is connected to a 110V, 60 HZ supply. What is the time lag between current maximum and voltage maximum? |
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Answer» Correct Answer - `3.24A; 1.55xx10^(-3)s` Impedance of circuit is `Z=sqrt((40)^(2)+(1)/((120 pi xx 100 xx10^(-6))^(2)) =48 Omega` Maximum current: `I_(0) =(E_0)/(Z) =(110 sqrt(2))/(48) = 3.24 A` [ we have `E=E_(0) sin omega t and I=(I_0) sin (omega t + phi]` `tan(phi)=(X_C)/(R )=(1)/(120 pi (100xx10^(-6)40)` `implies phi =33.5^(@)=0.186 pi` requried time lag: `(t_0)=(phi)/(omega)=(0.186 pi)/(2 pi 60) =1.55xx10^(-3)s`. |
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| 231. |
A coil of inductance 0.50H and resistanace `100 Omega` is connected to a 240V, 50Hz ac supply. What are the maximum current in the coil and the time lag between boltage maximum and current maximum? |
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Answer» Correct Answer - `1.82A; 3.2 xx10^(-3)s` Impedance of the coil: `Z=sqrt(R^(2)+(2 pi f l)^(2)) =sqrt(100^(2)+4 pi^(2) (50)^(2)(0.5)^(2))=186.2 Omega` Maximum current: `I_(0) =(E_0)/(Z) =(240 sqrt(2))/(186.2) = 1.82A` `tan phi - (X_(C )-X_(L))/(R ) =(0-2 pi (50)(0.5))/(100) = -(pi)/(2)` `implies phi = - tan^(-1) ((pi)/(2)) =-27.51^(@) =-1.003 rad ` Let emf be maximum at `(t_1)`, then `(omega t_(1))=(pi)/(2)` [ we have `E=E_(0) sin omega t and I=(I_0) sin (omega t + phi]` Let current be maximum at `(t_2)`, then` (omega t_(2))+phi =(pi)/(2)` form above we have `omega(t_(2)-t_(1)) =-phi` Time lag between voltage maximum and current maximum : `(t_2)-(t_1) =-(phi)/(omega) = (1.003)/(2 pi f ) =3.2 xx10^(-3)s`. |
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| 232. |
A 120V, 620W lamps is run froma 240V, 50Hz mains supply sing a capacitor connected connected in series with the lamp and supply. What is the teoretical value of the capacitor required to operate the lamp at its normal rating?A. `3.8muF`B. `6.6muF`C. `0.7muF`D. `13.3muF` |
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Answer» Correct Answer - C |
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| 233. |
Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil. |
| Answer» Correct Answer - `25 mJ,5 mJ` | |
| 234. |
A cylindrical space of radius R is filled with a uniform magnetic induction parallel to the axis of the cylinder. If B charges at a constant rate, the graph showin the variation of induced electric field with distance r from the axis of cylinder is B. C. D. |
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Answer» Correct Answer - A |
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| 235. |
In the above problem if `m=1kg` and teminal velocity attained by its is 4m/s after falling a height of 1m, the energy dissipated as heat till then is `(g=10m//s^(2))`A. 10JB. 2JC. epsilonJD. 12J |
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Answer» Correct Answer - B |
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| 236. |
A conducting rod AB of length `l=1m` is moving at a velocity `v_A=4m//s` making an angle `30^@` with its length. A uniform magnetic field `B=2T` exists in a direction perpendicular to the plane of motion. Then A. `V_(A)-V_(B)=8V`B. `V_(A)-V_(B)=4V`C. `V_(B)-V_(A)=8V`D. `V_(B)-V_(A)=4V` |
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Answer» Correct Answer - B |
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| 237. |
A square coil ABCD lying in `x-y` plane with its centre at origin. `A` long straight wire passing through origin carries a current `i=2t` in negative z-direction. The induced current in the coil is A. clockwiseB. anticlockwiseC. altenatingD. zero |
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Answer» Correct Answer - D |
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| 238. |
A current of `2A` is increasing at a rate of `4A//s` through a coil of inductance `2H`. The energy stored in the inductor per unit time isA. 2J/sB. 1J/sC. 16J/sD. 4J/s |
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Answer» Correct Answer - C |
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| 239. |
Two coil A and B have coefficient of mutual inductance M=2H. The magnetic flux passing through coil A charges by 4 Weber in 10 seconds due to the change in current in B. ThenA. change in current in B in this time interval is 0.5AB. the change in current in B in this time interval is 2AC. the change in current in B in this time interval is 8AD. a change in current of 1A in coil A will produces change in the flux passing through B by 4Wb |
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Answer» Correct Answer - B |
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| 240. |
Two circular coils A and B are facing each other in shown figure. The current I through A can be alterned A. there will be repuision between A and B if I is increasedB. there will be attraction between A and B if I is increasedC. there will be neither between A and B if I is increasedD. attraction of repusion between A and B depend on the direction of current, It does not depending wheather the current is increased or decreased |
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Answer» Correct Answer - A |
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| 241. |
Two identical coaxial circular loops carry a current `i` each circulating int the same direction. If the loops approch each other the current inA. the current in each loop will decreasesB. the current in each loop will increasesC. the current in each loop will remain the sameD. the current in one tlop will increase and in the other loop will decrease |
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Answer» Correct Answer - A |
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| 242. |
Two coils are at fixed location: When coil 1 has no corrent and the current in coil 2 increase at the rate of `15.0 A s^(-1)`, the emf in coil 1 is `25 mV`, when coil 2 has no current and coil 1 has a current of `3.6 A`, the flux linkange in coil 2 isA. 16mWbB. 10mWbC. 4.00mWbD. 6.00mWb |
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Answer» Correct Answer - D |
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| 243. |
The dimensions of magnetic flux areA. `"["MLT^(-3)A^(2)"]"`B. `"["ML^(2)A^(-1)"]"`C. `"["ML^(2)T^(2)A"]"`D. `"["ML^(2)TA^(-1)"]"` |
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Answer» Correct Answer - B |
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| 244. |
An inductor coil stores U energy when i` current is passed through it and dissipates energy at the rate of P. The time constant of the circuit, when this coil is connected across a battery of zero internal resistance isA. `(4U)/(P)`B. `(U)/(P)`C. `(2U)/(P)`D. `(2P)/(U)` |
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Answer» Correct Answer - C |
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| 245. |
A constant voltage at different frequencies is applied across a capacitance. C as shown in the figure. Which of the following graphs correctly depicts the varitaion of current with frequency? B. C. D. |
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Answer» Correct Answer - B For capacitivie circuits, `X_(C) = 1/(omegaC)` `I=1/(LC) = 1/(sqrt(10^(-3) xx 10 xx 10^(-6))` `omega = 1/(sqrt(10)^(-8))=10^(4)` `therefore X_(L) = omegaL=10^(4)xx 10^(-3)= 10 Omega` |
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| 246. |
In the non-resonant circuit, what will be the nature of the circuit for frequencies heigher than the resonant frequency?A. ResistiveB. CapacitiveC. InductanceD. None of the above |
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Answer» Correct Answer - B In non resonant circuits impendance `Z=1/(sqrt(1/(R^(2))+(omegaC-1/(omegaL))^(2))`, with rise in frequency `Z` decreases i.e., current increases so circuit behaves as capacitive circuit. |
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| 247. |
A constant voltage at different frequencies is applied across a capacitance `C` as shown in the figure. Which of the following graphs correctly depicts the variation of current with frequency A. B. C. D. |
| Answer» Correct Answer - b | |
| 248. |
A constant voltage at different frequencies is applied across a capacitance `C` as shown in the figure. Which of the following graphs correctly depicts the variation of current with frequency A. B. C. D. |
| Answer» Correct Answer - B | |
| 249. |
In series `L-C-R` resonant circuit, to increase the resonant frequencyA. `L` will have to be increasedB. `C` will have to be increasedC. `LC` will have to be decreasedD. `LC` will have to be increased |
| Answer» Correct Answer - 3 | |
| 250. |
In the non-resonant circuit, what will be the nature of the circuit for frequencies heigher than the resonant frequency?A. resistiveB. capacitiveC. inductiveD. both 1 and 2 |
| Answer» Correct Answer - 3 | |