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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.
| 301. |
When a current of 2 A is passed through a coil of 100 turns , flux associated with it is `5 xx 10^(-5) `Wb. Find the self inductance of the coil.A. `4 xx 10^(-2) H`B. `2.5 xx 10^(-3) H`C. `10^(-3) H`D. `4 xx 10^(-3) H` |
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Answer» Correct Answer - B The flux associated with the coil = `n phi and n phi = LI` `:. L = (n phi)/(I) = (100 xx 5 xx 10^(-5))/(2) = 2.5 xx 10^(-3)H`. |
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| 302. |
Magnetic flux in a circuite containing a coil of resistance `2Omega`change from `2.0Wb` to `10 Wb` in `0.2 sec`. The charge passed through the coil in this time isA. 5.0 CB. 4.0 CC. 1.0 CD. 0.8 C |
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Answer» Correct Answer - B As, charge, `DeltaQ=(Deltaphi)/(R)=((10-2))/(2)=4C` |
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| 303. |
Magnetic flux in a circuite containing a coil of resistance `2Omega`change from `2.0Wb` to `10 Wb` in `0.2 sec`. The charge passed through the coil in this time isA. `5.0` coulombB. `4.0` coulombC. `1.0` coulombD. `0.8` coulomb |
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Answer» Correct Answer - B `DeltaQ=(Deltavarphi)/(R )=((10-2))/(2)=4C` |
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| 304. |
Magnetic flux in a circuite containing a coil of resistance `2Omega`change from `2.0Wb` to `10 Wb` in `0.2 sec`. The charge passed through the coil in this time isA. `0.8 C`B. `1.0 C`C. `5.0 C`D. `4.0 C` |
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Answer» Correct Answer - D `q=(N)/(R )(Deltavarphi)=(1)/(2)xx(10-2)=4C` |
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| 305. |
Two metal bars are fixed vertically and are connected on the top by a capacitor `C`. A sliding conductor of length land mass m slides with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement `x(t)` of the conductor as a function of time t. |
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Answer» Correct Answer - B::C Let `v` be the velocity at some instant. Then motional emf `V=Bvl` Charge stored in capacitor `q=CV=(CBl)v` Current in the wire `=(dq)/(dt)=(CBl)(dv)/(dt)` Magnetic force `F_m=ilB=(CB^2l^2)(dv)/(dt)`(upwards) `:.` Net force `F_("net") = mg - F_(m)` or `m=(dv)/(dt) = mg -(CB^(2)l^(2)) (dv)/(dt)` `:. (dv)/(dt) =`acceleration, `a=(mg)/(m+CB^2l^2)` Since, `a =`constant `:. x=1/2at^2=(mg t^2)/(2(m+CB^2l^2))` |
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| 306. |
A circuit `ABCD` is held perpendicular to the uinform magnetic field of `B = 5 xx 10^(-2)T` extending over the region `PQRS` and directed into the plane of the paper. The cicuit is moving out of the field at a uin form speed of `0.2 m s^(-1)` for `1.5 s`. During this time, the current in the `5 Omega` resistor is A. (a) `0.6mA from B to C`B. (b) `0.9mA from B to C`C. ( c) `0.9mA from C to B`D. (d) `0.6mA from C to B` |
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Answer» Correct Answer - A (a) `I = (Blv)/( R)` `I = (5 xx 10^(-2) xx 0.3 xx 0.2)/(5)A = 0.6 mA`. Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from `B to C` through `5 Omega`. |
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| 307. |
a conducting wire os mass `m` slides down two smooth conducting bars, set at an angle `theta` to the horizontal as shown in . The separatin between the bars is `l`. The system is located in the magnetic field `B`, perpendicular to the plane of the sliding wire and bars. The constant velocity of the wire is A. (a) `(mgR sin theta)/(B^(2)l^(2))`B. (b) `(mgR sin theta)/(Bl^(3))`C. ( c) `(mgR theta)/(B^(2)l^(5))`D. ( d) `(mgR sin theta)/(Bl^(4))` |
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Answer» Correct Answer - A (a) Component of weight along the inclined plane `= mg sin theta` Again, `F = Bil = B(Nlv)/(R )l = (B^(2)l^(2)v)/(R )` Now, `(B^(2)l^(2)v)/(R ) = mg sin theta` or `v = (mg R sin theta)/(B^(2)l^(2))` |
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| 308. |
An elasticized conducting band is around a spherical ballon . Its plane through the center of the balloon. A uniform nmagnetic field of magnitude `0.04T` is directed perpendicular to the plane of the band. Air is let out of the balloon at `100 cm^(3) s^(-1)` at an instant when the radius of the balloon is `10 cm`. The induced emf in the band is A. (a) `15mu v`B. (b) `25mu v`C. ( c) `10mu v`D. (d) `20mu v` |
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Answer» Correct Answer - D (d) Volume of the balloon at any instant, when rdius is `r`, `V = (4)/(3)pir^(3)` Time rate of change of volume, `(dV)/(dt) = 4pir^(2)(dr)/(dt)` Time rate of change of radius of balloon, `(dr)/(dt) = (1)/(4pir^(2))(dV)/(dt)` Flux through rubber band at the given instant, `phi = B(pir^(2))` `Induced emf = -(dphi)/(dt) = -(d)/(dt)(Bpir^(2)) = -2pirB(dr)/(dt)` `= -2pirB.((1)/(4pir^(2))(dV)/(dt)) = -(B)/(2r)(dV)/(dt)` As volume of the balloon is decreasing, `(dV)/(dt)` is negative. `E_(induced) = -((0.04))/(2 xx 10 xx 10^(-2)) xx (-100 xx 10^(-6)) = 20mu V` |
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| 309. |
A sereis R-C circuit is connected to AC voltage source. Consider two cases, (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current `I_(R)` through the resistor and voltage `V_(c)` across the capacitor are compared in the two cases. Which of the following is/ are true?A. `I_(R )^(B) gt I_(R )^(B)`B. `I_(R )^(B) = I_(R )^(B)`C. `V_(A)^(C ) gt V_(c )^(B)`D. `V_(A)^(C ) lt V_(c )^(B)` |
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Answer» Correct Answer - C For circuit `A`, Impedance, `Z_(A) = sqrt(R^(2) + (1)/(omega^(2) C^(2)))` Current in circuit, `I_(A)^(2) = (V)/(sqrt(R^(2) + (1)/(omega^(2) C^(2)))` Pot. Diff. across `C`, `V_(C )^(A) = I_(R)^(A) xx (1)/(omega C) = (V)/(sqrt((R omega C)^(2) + 1)` For circuit `B`, impedance ` Z_(B) = sqrt(R^(2) + (1)/((4 omega)^(2) C^(2)))` Current in circuit, `I_(B)^(B) = (V)/(sqrt(R^(2) + (1)/((4 omega)^(2) C^(2)))` Pot. diff. across `C`, `V_(C )^(B) = I_(R )^(B) xx (1)/( 4 omega C) = (V)/(sqrt((4 R omega C)^(2) + 1))` We conclude, form (i) and (iii) , `I_(R )^(B) gt I_(R )^(A)` From (ii) and (iv), `V_(C )^(A) gt V_(C )^(B)` choice (c ) is correct |
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| 310. |
A sereis R-C circuit is connected to AC voltage source. Consider two cases, (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current `I_(R)` through the resistor and voltage `V_(c)` across the capacitor are compared in the two cases. Which of the following is/ are true?A. `I_(R)^(A)gtI_(R)^(B)`B. `I_(R)^(A)ltI_(R)^(B)`C. `V_(A)^(C)gtV_(B)^(C)`D. `V_(A)^(C)gtV_(B)^(C)` |
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Answer» Correct Answer - A::C We know that `Z=sqrt(r^(2)+((1)/(WC))^(2))` The capacitance in case B is four times the capacitance in case A Impedance in case B is less then that of case A `(Z_(B)ltZ_(A))` Now `I=V/Z` `:. (I_(R)^(A))lt(I_(R)^(B))`. Option (a) is correct. `:. (V_(R)^(A))lt(V_(R)^(A))` `:. (V_(C)^(A))lt(V_(C)^(A))` if V is the applied potential potential difference access series R-C circuit then `V=sqrt((V_(R)^(2))+(V_(C)^(2)))`] (c) is the correct option . |
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| 311. |
A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is `L`. A conducting massless rod of resistance `R` can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass `m` tied to the other end of the string hangs vertically. A constant magnetic field `B` exists perpendicular to the table. If the system is released from rest, calculate a. the terminal velocity achieved by the rod and b. The acceleration of the mass of the instant when the velocity of the rod is half the terminal velocity. |
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Answer» a. Let `v` be the velocity of the wire (as well as block )at any instant of time `t`. Motional emf `e=BvL` Motional current `i=e/r=(BvL)/L` and magnetic force on the wire `F_m=iLB=(vB^2L^2)/R` Net force on the system at this moment will be `F_("net")=mg-F_m=mg-(vB^2L^2)/R` `ma=mg-(vB^2L^2)/R` `a=(vB^2L^2)/(mR)` Velocity will acquire its terminal value i.e. `v=v_T` when `F_("net")` or acceleration `a` of the particle becomes zero. Thus, `0=g-(v_TB^2L^2)/(mR)` or `v_T=(mgR)/(B^2L^2)` b. when `v=v_T/2=(mgR)/(2B^2L^2)` Then from Eqn (i) acceleratin of the block `a=g-((mgR)/(2B^2L^2))((B^2L^2)/(mR))=g-g/2` or `a=g/2` |
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| 312. |
The phase difference between the current and voltage of `LCR` circuit in series combination at resonance isA. zeroB. `pi//4`C. `pi//2`D. `pi` |
| Answer» Correct Answer - A | |
| 313. |
A conducting square loop of side L and resistance R moves in its plance with a uniform velocity v perpendicular to one of its sides. A magnetic induction B, constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere. The current induced in the loop is A. `BLv//R colckwise`B. `BLv//R anticlockwise`C. `2BLv//R anticlockwise`D. zero |
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Answer» Correct Answer - D Since the rate of change of magnetic flux is zero, hence there will be no net induced emf and hence no current flowing in the loop. |
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| 314. |
An AC voltage source of variable angular frequency `(omega)` and fixed amplitude `V_(0)` is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When `(omega)` is increasedA. the bulb glows dimmerB. the bulb glows brighterC. total impendance of the circuit is unchangedD. total impendance of the circuit increases |
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Answer» Correct Answer - B `(I_(rms))=(V_(rms))/(sqrt(R^(2)+(1/(omega C))^(2)))` As `(omega)` increses, `(I_(max))` increases. Therefore, the blub glows brighter. |
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| 315. |
An e.m.f. `E=4cos(1000t)`volt is applied to an `LR` circuit of inductance `3mH` and resistance `4ohm`. The amplitude of current in the circuit isA. 1.0AB. `4/7 A`C. `0.8 A`D. `4/(sqrt(7))A` |
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Answer» Correct Answer - C ` E = 4 cos (1000t)` `:. E_(0) = 4 V and omega = 1000` `:.` inductive reactance `X_(L) = omega L = 1000 xx 3 xx 10^(-3) = 3 Omega` `:.` Impedance `Z = sqrt(R^(2)+X_(L)^(2)) = sqrt(4^(2)+3^(2))=5` `I_(0) = (V_0)/(Z) = 4/5 = 0.8 A`. |
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| 316. |
L, C and R represent the physical quantities, inductance, capacitance and resistance respectively. The combination(s) which have the dimensions of frequency areA. `1//RC`B. `R//L`C. `1//sqrt(LC)`D. C//L` |
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Answer» Correct Answer - A::B::C `1/(RC), R//L and 1//(sqrt(LC))` have the dimensions of frequency. |
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| 317. |
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. `sqrt2`B. `(1)/(sqrt2)`C. 2D. 1 |
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Answer» Correct Answer - A In series `LR` circuit, Power factor, `cos phi_(1) = (R )/(Z)` `= (R )/(sqrt(R^(2) +X_(L)^(2))) = (R )/(sqrt(R^(2) + 9 R^(2))` ` = (R )/(R sqrt(10)) = (1)/(sqrt10)` When capacitor with `X_(C ) = R` is added in series, new power factor, `phi_(2) = (R )/(Z) = (R )/(sqrt(R^(2) + (X_(L) - X_(C ))^(2)))` `cos phi_(2) = (R )/(sqrt(R^(2) + (3 R - R)^(2))) = (R )/(R sqrt5)` `:. (cos phi_(2))/(cos phi_(1)) = (1)/(sqrt5) sqrt(10)/(1) = sqrt2` |
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| 318. |
L, C and R represent the physical quantities, inductance, capacitance and resistance respectively. The combination(s) which have the dimensions of frequency areA. `L/C`B. `(LC)^(2)`C. `(LC)^(-1//2)`D. `((LC)/(R))^(1//2)` |
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Answer» Correct Answer - C The resonant frequency, `f=(1)/(2pisqrt(LC)) = 1/(2pi(LC)^(1//2))` `:. F = 1/(2pi)(LC)^(-1//2)` `:. (LC)^(-1//2)` will have the dimensions of frequency as `1/(2pi)` is dimensionless. |
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| 319. |
Suppose the emf of the battery, the circuit shown varies with time `t` so the current is given by `i(t)=3+5t`, where `i` is in amperes & `t` is in seconds. Take `R=4Omega, L=6H` & find an expression for the battery emf as function of time. |
| Answer» Correct Answer - B::D | |
| 320. |
The current in a coil of self-inductance `4H` is given by `i=4sint^(2)` . Find the amount of energy spent during the period when the current changes from `0` to `4A` . |
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Answer» When the current in the coil is variable, small work done in time `dt` `dW=Pdt=eidt=(L(di)/(dt))idt=Lidi` `W=Lint_(0)^(4)idi` `=4|(i^(2))/(2)|_(0)^(4)=32J` |
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| 321. |
A copper ring is tied to a string and suspended vertically. On bringing a magnet towards the coil, as shown in the figure- A. the ring will move away from magnetB. the ring will move towards the magnetC. the ring will remain stationaryD. none of these |
| Answer» Correct Answer - A | |
| 322. |
A transformer whose efficiency is 90%, draws 5 A when 200 V is applied to its primary coil. If output is drawn at 300 V, what is the current in secondary coil ? If number of turns in primary coil is 500, what is the number of turns in secondary coil? |
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Answer» Here,`eta = 90%, I_(P) = 5 A, E_(P) = 200 V` `I_(s) = ? E_(s) = 300 V, N_(P) = 500, N_(s) = ?` As `eta = (E_(s) I_(s))/(E_(P) I_(P))` `:. I_(s) = (eta E_(P) I_(P))/(E_(s)) = (90)/(100) xx (200 xx 5)/(300) = 3 A` From `(E_(s))/(E_(P)) = (N_(s))/(N_(P))` `N_(s) = (E_(s))/(E_(P)) xx N_(P) = (300)/(200) xx 500 = 750` |
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| 323. |
A transformer is used to reduce the main supply of 220V to 22V. If the currents in the primary and secondary are 2A and 15 A respectively, then the efficiency of the transformer isA. 0.65B. 0.75C. 0.8D. 0.9 |
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Answer» Correct Answer - B `eta = (V_(S)I_(S))/(V_(P)I_(P)) xx 100 = 22/220 xx 15/2 xx 100 = 75%`. |
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| 324. |
A transformer is employed to reduce `220V` to `11V`. The primary draws a current of `5A` and the secondary `90A`. The efficiency of the transformer isA. `20%`B. `40%`C. `70%`D. `90%` |
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Answer» Correct Answer - D `eta=(V_(s)i_(s))/(V_(p)i_(p))xx100=(11xx90)/(220xx5)xx100=90%` |
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| 325. |
The primary winding of a transformer has `100` turns and its secondary winding has `200V` turns. The primary is connected to an ac supply of `120V` and the current flowing in it is `10A`. The voltage and the current in the secondary areA. `240 V,5A`B. `240 V,10A`C. `60V,20A`D. `120 V,20A` |
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Answer» Correct Answer - A `(N_(s))/(N_(p))=(V_(s))/(V_(p))implies(200)/(100)=(V_(s))/(120)impliesV_(s)=240V` Also `(V_(s))/(V_(p))=(i_(p))/(i_(s))implies(240)/(120)=(10)/(i_(s))implies5A` |
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| 326. |
A step up transformer connected to a `220V AC` line is to supply `22 kV` for a neon sign in secondary circuit. In primary circuit a fuse wire is connected which is to blow when the current in the secondary circuit exceeds `10mA`. The turn ratio of the transformer isA. 50B. 100C. 150D. 200 |
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Answer» Correct Answer - B `K=(n_(S))/(n_(P))=(e_(S))/(e_(P))=(22xx10^(3))/(220)=100` |
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| 327. |
In a step up transformers, `220V` is converted into `200V`. The number of turns in primary coil is `600`. What is the number of turns in the secondary coil?A. `60`B. `600`C. `6000`D. `100` |
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Answer» Correct Answer - C `(N_(s))/(N_(p))=(V_(s))/(V-(p))implies(N_(s))/(600)=(2200)/(220)impliesN_(s)=6000` |
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| 328. |
The turn ratio of a transformers is given as `2:3`. If the current through the primary coil is `3 A`, thus calculate the current through load resistanceA. `1A`B. `4.5A`C. `2A`D. 1.5A` |
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Answer» Correct Answer - C As `(I_(p))/(I_(s))=(n_(p))/(n_(s))implies(3)/(I_(s))=(3)/(2)impliesI_(2)=2A`. |
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| 329. |
A transformer with efficiency `80%` works at `4 kW` and `100 V`. If the secondary voltage is `200 V`, then the primary and secondary currents are respectivelyA. `40A,16A`B. `16A,40A`C. `20A,40A`D. `40A,20A` |
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Answer» Correct Answer - A `eta=("outepur power")/("input power")=(E_(s)I_(s))/(E_(p)I_(p))implies(80)/(100)=(200xxI_(s))/(4xx10^(3))` `impliesI_(s)=(80)/(100)xx(4xx1000)/(200)=16A` Also,`E_(p)I_(p)=4KWimpliesI_(p)=(4xx10^(3))/(100)=40A` |
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| 330. |
A transformer with efficiency `80%` works at `4 kW` and `100 V`. If the secondary voltage is `200 V`, then the primary and secondary currents are respectivelyA. 16A, 40AB. 40A, 16AC. 30A, 45AD. 50A, 30A |
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Answer» Correct Answer - B `P_(i) = 4 xx 10^(3) W` `:. I_(P) = (P_i)/(v) = (4 xx 10^3)/(100) = 40 A` (Primary) `eta = (P_o)/(P_i) :. 8/10 = (P_o)/(4 xx 10^3) :. P_(o) = 32 xx 10^(2)` `:. I_(S) = (P_o)/(V_S) = (32 xx 10^(2))/(200) = 16 A` (Secondary). |
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| 331. |
A transformer with efficiency `80%` works at `4 kW` and `100 V`. If the secondary voltage is `200 V`, then the primary and secondary currents are respectively |
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Answer» `eta = 80%, E_(P) = 100 V, E_(P) I_(P) = 4 kW = 4000 W` `E_(s) = 240 V, I_(P) = ?, I_(s) = ?` `I_(P) = (E_(P) I_(P))/(E_(P)) = (4 xx 1000)/(100) = 40 A` To calculate `I_(s)` use `eta = (E_(s) I_(s))/(E_(P) I_(P))` |
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| 332. |
A transformer has 500 turns in the primary and 1000 turns in its secondary. The primary voltage is 200 V and load in secondary is 100 ohm. Calculate the current in primary assuming it to ideal transformer. |
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Answer» Here, `n_(p) = 500, n_(s) = 1000` `E_(p) = 200 V, R_(s) = 100 Omega, I_(p) = ?` In an ideal transformer, `(E_(s))/(E_(p)) = (n_(s))/(n_(p)) = (1000)/(500) = 2`, `E_(s) = 2 E_(p) = 2 xx 200 = 400 V` Also, `(I_(p))/(I_(s)) = (n_(s))/(n_(p)) = 2`, `I_(p) = 2 I_(s) = 2 ((E_(s))/(R_(s))) = (2 xx 400)/(100) = 8 A` |
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| 333. |
In an ideal transformer, number of turns in primary and secondary are 200 and 1000 respectively. If the power input to the primary is 10 kW at 200 V, calculate (i) output voltage (ii) primary current. |
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Answer» Here, `n_(P) = 200, n_(s) = 1000`, `P_(i) = 10 kW = 10000 W, E_(P) = 200 V` `E_(s) = ?, I_(P) = ?` As `(E_(s))/(E_(P)) = (n_(s))/(n_(P)) = (1000)/(200) = 5` Also, `E_(P) I_(P) = P_(i) = 1000`, `I_(P) = (10000)/(E_(P)) = (10000)/(200) = 50 A` |
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| 334. |
The primary of a transformer has 200 turns and secondary has 1000 turns. The power output from secondary at 1000 V is 9 kW. Calculate primary voltage and heat loss in primary. Take resistance of primary coil `0.2 Omega` and efficiency of transformer = 90%. |
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Answer» Here,`x_(P) = 200, n_(s) = 1000 , E_(s) = 1000 V`, `P = 9 kW = 9000 W` `E_(P) = ?, H = ?, R_(P) = 0.2 Omega, eta = 90%` As `(E_(s))/(E_(P)) = (n_(s))/(n_(P)) = (1000)/(200) = 5` `:. E_(P) = (E_(s))/(5) = (1000)/(5) = 200 V` From `eta = (P_(0))/(P_(i)) , P_(i) = (P_(0))/(n) = 10 kW` `:. I_(P) = (1000)/(200) = 50 A` Energy loss in primary `= I_(P)^(2) xx R_(P) = (50)^(2) xx 0.2 = 500 W` |
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| 335. |
A light bulb is rated at 100 W for a 220 V supply. Find (a) the resistance of the bulb. (b) the peak voltage of the source (c ) the rms current through the bulb. |
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Answer» Here, `P = 100 W, E_(v) = 220 V`, `R = ? E_(0) = ? I_(v) = ?` From, `P = (E_(v)^(2))/(R ) , R = (E_(v)^(2))/(R ) = (220 xx 220)/(100) = 484 Omega` `E_(0) = sqrt2 E_(v) = 1.414 xx 220 = 311 V` `I_(v) = (P)/(E_(v)) = (100)/(220) = 0.45 A` |
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| 336. |
In a series resonant LCR circuit the voltage across R is 100 volts and R = `1 k(Omega) with C =2(mu)F`. The resonant frequency `(omega)` is `200 rad//s`. At resonance the voltage across L isA. 100 VB. 150 VC. 200 VD. 250 V |
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Answer» Correct Answer - D At resnonace, `omegaL = 1/(omegaC) and Z = R` Current through the circuit `I = (V_R)/(R) = 100/1000 = 0.1A` `:.` Voltage across L is given by `V_(L) = IX_(L) = I omega L` but `omega L = 1/(omega C)` `:. V_(L) = 1/(omega C)` `=(0.1)/(200 xx 2 xx 10^(-6)) = 1000/4 = 250V`. |
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| 337. |
In a series RC circuit, R = 30 ohm, `C = 0.25 mu F`, `V = 100 V, omega = 10000 rad//s`. Find the current in the circuit and calculate the voltage across the resistor and capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox. |
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Answer» Here, `R = 30 Omega, C = 0.25 mu F, E_(v) = 100 V`, `omega = 10000 rad//s, I_(v) = ?` `X_(C ) = (1)/(omega C) = (1)/(10000 xx 0.25 xx 10^(-6)) = 400 Omega` `Z = sqrt(R^(2) + X_(C )^(2)) = sqrt(30^(2) + (400)^(2)) = 401.1 Omega` `I_(v) = (E_(v))/(Z) = (100)/(401.1) ~= 0.25 A` `V_(R ) = I_(v) X_(R ) = 0.25 xx 30 = 7.5 V` `V_(C ) = I_(V) X_(C ) = 0.25 xx 400 V` `V_(R ) + V_(C ) = 7.5 + 100 = 107.5 V`, which is more than source voltage = 100 V .This is because `V_(R )` and `V_(C )` are not in the same phase. |
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| 338. |
A 40 ohm resistor, 3 mH inductor and `2 mu F` capacitor are connected in series to a 110 V, 5000 Hz a.c. source. Calculate the value of current in the circuit. |
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Answer» Here, `R = 40 Omega, L = 3 mH = 3 xx 10^(-3) H`, `C = 2 mu F = 2 xx 10^(-6) F, E_(v) = 110 V, v = 5000 Hz` `X_(L) = omega L = 2 pi L = 2 xx 3.14 xx 5000 xx 3 xx 10^(-3)` `= 94.2 Omega` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `= (1)/(2 xx 3.14 xx 5000 xx 2 xx 10^(-6)) = 15.92 Omega` `Z = sqrt(R^(23) + (X_(L) - X_(C ))^(2))` `= sqrt(40^(2) + (94.2 - 15.92)^(2)) = 87.9 Omega` `I_(v) = (E_(v))/(Z) = (110)/(87.9) = 1.25 A` |
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| 339. |
A resistor of 50 ohm, an inductor of `(20//pi)` H and a capacitor of `(5//pi) mu F` are connected in series to an a.c. source 230 V, 50 Hz. Find the current in the circuit. |
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Answer» Here, `R = 50 ohm` `L = (20)/(pi) H, C = (5)/(pi) mu F = (5xx 10^(-6))/(pi) F` `E_(v) = 230 V, v = 50 Hz, I_(v) = ?` `X_(L) = omega L = 2 pi v L = 2 pi v ((20)/(pi))` ` = 40 xx 50 = 2000 Omega` `X_(C ) = (1)/(omega C) = (1)/(2 pi v c) = (1 pi)/(2 pi v xx 5 xx 10^(-6))` `= (1)/(10 xx 50) = 2000 Omega` ` Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` `= sqrt(50^(2) + (2000 - 2000)^(2)) = 50 Omega` `I_(v) = (E_(v))/(Z) = (230)/(50) = 4.6 A` |
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| 340. |
An inductor L, capacitor of `20 mu F` and a resistor of `10 Omega` are connected in series with 220 V, 50 Hz a.c. supply. If current is in phase with the voltage, calculate the inductance. What is the value of current in the circuit ? |
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Answer» Here, `C = 20 mu F = 20 xx 10^(-6) F, R = 10 Omega` `E_(v) = 220 V, v = 50 Hz, L = ? I_(v) = ?` As current is in phase with the voltage, `X_(L) = X_(C )` `omega L = (1)/(omega C)`, `L = (1)/(omega^(2) C ) = (1)/((2 pi v )^(2) C)` `L = (1)/((2 pi xx 50)^(2) xx 20 xx 10^(-6)) = 0.506 H` ` Z = sqrt(R^(2) + (x_(L) - X_(C ))^(2)) = R = 10 Omega` `I_(v) = (E_(v))/(Z) = (220)/(10) = 22 A` |
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| 341. |
A `1 mu F` capacitor is connected to a 220 V, 50 Hz a.c. source. Find the virtual value of current through the circuit if a resistor of `100 Omega` is connected in series with the capacitor. |
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Answer» Here, `C = 1 mu F = 10 ^(-6) F, E_(v) = 220 V, v = 50 Hz` `R = 100 Omega, I_(v) = ?` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C ) = (1)/(2 xx 3.14 xx 50 xx 10^(-6))` ` = 3.185 xx 10^(-3) ohm` `Z = sqrt(R^(2) + X_(C )^(2)) = sqrt(100^(2) + (3.185 xx 10^(3))^(2))` ` = 3186.6 ohm` `I_(v) = (E_(v))/(Z) = (200)/(3186.6) A = 0.069 A` |
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| 342. |
A resistance of `40 Omega` is connected to an source of 220 v, 50 Hz. Find the (i) rms current (ii) maximum instantaneous current in resistor. |
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Answer» Here, `R = 40 Omegam E_(v) = 220 V, v = 50 Hz`, (i) `I_(v) = (E_(v))/( R) = (220)/(40) = 5.5 A` `I_(v) = sqrt 2 I_(v) = 1.414 xx 5.5 = 7.78 A` If alternating current is given by `I = I_(0) sin omega t`, then `I_(0) = I_(0) sin omega t_(1) = 1` or `omega t_(1) = (pi)/(2)` and `I_(v) = (I_(0))/(sqrt 2) = I_(0) sin omega t_(2)`, which implies `omega t_(2) (pi)/(2) + (pi)/(4)` `:. omega (t_(2) - t_(1)) = (pi)/(2) + (pi)/(4) - (pi)/(2) = (pi)/(4)` `t_(2) - t_(1) = (pi)/(4 omega) = (pi)/(4 xx 2 pi v) = (1)/(8 v) = (1)/(8 xx 50)` `= 2.5 xx 10^(-3) s = 2.5 ms` |
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| 343. |
An electrical device draws 2 kW power form AC mains [voltage 223 V (rms) `= sqrt(50,000) V`]. The current differs (lags) in phase by `phi (tan = (-3)/(4))` as compared to voltage. Find (i) R, (ii) `X_(C ) - X_(L)`, and (iii) `I_(M)`. Another device has twice the values for R, `X_(C )` and `X_(L)`. How are the answers affected ? |
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Answer» Here, `P = 2 kW = 2000 W. E_(v) = sqrt(50000) = 223 V, tan phi = - (3)/(4) , R = ? , X_(C ) - X_(L) = ?, I_(M) = I_(0) = ?` From `P = (E_(upsilon)^(2))/(Z) , Z = (E_(upsilon)^(2))/(P) = (50000)/(2000) = 25 Omega` Now, `tan phi = (X_(C ) - X_(L))/(R ) = - (3)/(4) :. X_(C ) - X_(L) = - (3)/(4) R` From `Z^(2) = R^(2) + (X_(C ) - X_(L))^(2)` `(25)^(2) = R^(2) + (-(3)/(4) R)^(2) = (25)/(16) R^(2)` `R = (4)/(5) xx (25) = 20 ohm` From (i), `X_(C ) - X_(L) = - (3)/(4) xx 20 = - 15 Omega` Now, `I_(upsilon) = (E_(upsilon))/(Z) = (223)/(25) ~= 9 A` `I_(0) = sqrt(2) I_(upsilon) = sqrt2 xx 9 A = 12.6 A` When R, `X_(C ), X_(L)` are all doubled, `tan phi` does not change , Z is doubled, current is halved and power drawn is halved. |
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| 344. |
An are lamp requires a direct current of 10A at 80V to function. If it is connected to a 220V(rms), 50 Hz AC supply, the series inductor needed for it to work is close to:A. 0.044HB. 0.065HC. 80HD. 0.08H |
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Answer» Correct Answer - B here, `i=(e)/(sqrt(R^(2)+X_(L)^(2)))=(e)/(sqrt(R^(2)+omega^(2)L^(2)))=(e)/(sqrt(R^(2)+4 pi^(2)v^(2)L^(2)))` `10=(220)/(sqrt(64xx4 pi ^(2)(50^(2)L` [`:. R=v//(i) = 80/18]` pm solving we get ltbr. `L=0.065h`. |
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| 345. |
An AC voltage is given as `e=e_(1) sin omega t`. Find the RMS value of this voltage. |
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Answer» Correct Answer - `sqrt((e_(1)^(2)+e_(2)^(2))/(2))` |
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| 346. |
A `15.0 mu F` capacitor is connected to a 220 V, 50 Hz source. Find the capacitative reactance and the current (rms and peak) in the circuit. If the frequency is doubled. What happens to the capactive reactance and current ? |
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Answer» Here, `C = 15.0 mu F = 15 xx 10^(-6) F` `E_(v) = 220 V, v = 50 Hz, X_(C ) = ?` `I_(v) = ?, I_(0) = ?` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C) = (1)/(2 xx (22)/(7) xx 50 xx 10^(-6))` `= (7 xx 10^(4))/(22 xx 15) = 212 ohm` `I_(v) = (E_(v))/(X_(C )) = (220)/(212) = 1.04 A`, `I_(0) = sqrt 2 I_(v) = 1.414 xx 1.04 = 1.47 A` When frequency is doubled, `X_(C)` is halved and current is doubled. |
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| 347. |
A long straight wire is parallel to one edge of a rectangular loop as shown in figure-5.310. if the current in the long wire varies with time as `I=I_(0)e^(-tr)`, what will be the induced emf in the loop at t=`tau`? A. `(mu_(0)bI)/(pitau)In((d+a)/(d))`B. `(mu_(0)bI)/(2pitau)In((d+a)/(d))`C. `(2mu_(0)bI)/(pitau)In("(d+a)/(d))`D. `(mu_(0)bI)/(pitau)In((d)/(d+a))` |
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Answer» Correct Answer - C |
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| 348. |
In a wire direct current `i_(1)` and an AC current `i_(2)=i_(20) sin omega`t is superposed. Find the RMS value of current in wire. |
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Answer» Correct Answer - `sqrt(i_(1)^(2)+(i_(20)^(2))/(2))` |
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| 349. |
Two coils have a mutual inductance `0.005 H`. The current changes in the first coil according to equation `I=I_(0)sin omegat`, where `I_(0)=10A` and `omega=100pi`radian//sec`. The maximum value of e.m.f. in the second coil isA. `2piV`B. `5piV`C. `piV`D. `4piV` |
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Answer» Correct Answer - B |
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| 350. |
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))/2`B. `P = (E_(0)I_(0))/(sqrt2)`C. P=0D. `P=sqrt(2) E_(0)I_(0)` |
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Answer» Correct Answer - C There is phase difference of `pi//2` between E and I. `:.` Power factor `cos phi = cos pi//2 =0` or Average power = `(E_0)/(sqrt2)(I_0)/(sqrt2) xx cos ((pi)/(2)) =0`. |
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