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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.
| 1401. |
A 200 V varialbe frequency a.c. source is connected to a series combination the angular `C = 80 mu F` and `R = 40 Omega`. Calculate the angular frequency of the soucre to get maximum current in the cicuit, the current amplitude at resonance and power dissipated in the circuit. |
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Answer» Here, `E_(v) = 200 V, L = 5 H, C = 80 xx 10^(-6) F` `R = 40 Omega`, To get maximum current, `omega = (1)/(sqrt(LC)) = (1)/(sqrt(5 xx 80 xx 10^(-6))) = (1000)/(20) = 50 rad//s` `I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(R ) = (1.414 xx 200)/(40) = 7.07 A` `P = (E_(V)^(2))/(R ) = (200 xx 200)/(40) = 1000 W` |
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| 1402. |
In a series L-R circuit `(L=35 mH and R=11 Omega)`, a variable emf source `(V=V_(0) sin omega t)` of `V_(rms)=220V` and frequency 50 Hz is applied. Find the current amplitude in the circuit and phase of current with respect to voltage. Draw current-time graph on given graph `(pi=22/7)`. |
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Answer» Here, `L = 35 mH = 35 xx 10^(-3) H, R = 11 Omega` `E_(v) = 220 V, (omega)/(2 pi) = 50 Hz = v` `X_(L) = omega L = 2 pi xx 50 xx 35 xx 10^(-3) = 11 ohm` Impedance of the circuit `Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(11^(2) + 11^(2)) = 11 sqrt2 ohm` `E_(0) = sqrt2 E_(v) = sqrt2 xx 220 V` `I_(0) = (E_(0))/(Z) = (sqrt2 xx 220)/(11 sqrt2) 20 A` If `phi` is phase difference between current and voltage, then `tan phi (X_(L))/(R ) = (11)/(11) = 1` `phi = pi//4` or `45^(@)` |
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| 1403. |
We can reduce eddy currents in the core of transformerA. by increasing the number of turns in secondary coilB. by taking laminated coreC. by making step-down transformerD. by using a weak ac at high potential |
| Answer» Correct Answer - B | |
| 1404. |
If the current `30A` flowing in the primary coil is made zero in `0.1` sec. the emf induced in the secondary coil is `1.5` volt. The mutual inductance between the coil isA. `0.05H`B. `1.05H`C. `0.1H`D. `0.2H` |
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Answer» Correct Answer - A `e=M(di)/(dt)implies1.5=Mxx(30)/(0.1)impliesM=0.05H` |
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| 1405. |
A 50 Hz ac current of crest value 2 A flows mutual inductance between the primary and secondary be 0.25 H, the crest voltage induced in the secondary isA. 15.7 VB. 1.57 VC. 17.5 VD. 157 V |
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Answer» Correct Answer - D `e=M(dI)/(dt)` `=M(d)/(dt)I_(0)sinomegat=MI_(0)omegacos omegat` `therefore e_(0)=MI_(0)omega` `=0.25xx2xx2pixx50=157V` |
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| 1406. |
The mutual inductance of a pair of a coil is 0.75 H. if the current in primary coil changes from 0.5 A to 0A in 0.01 s. the average induced emf in secondary coil isA. 73.5 VB. 3.75 VC. 37.5 VD. 3.5 V |
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Answer» Correct Answer - C `e=M(dI)/(dt)` `=(0.75xx0.5)/(0.01)=37.5`V |
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| 1407. |
A certain current flows in the primary of an induction coil of mutual inductance 5 H. when the break occurs, an emf of 30,000 V is induced in the secondary. If the current in the primary falls uniformly to zero in `10^(-4)`s, find the current in the primary just before the break?A. 0.6 AB. 0.3 AC. 0.2 AD. 0.1 A |
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Answer» Correct Answer - A `e=M(dI)/(dt)` `dI=(edt)/(M)=(30000xx10^(-4))/(5)=0.6A` |
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| 1408. |
The mutual inductance between two coil is 2.5 H. If the current in one coil is changed at the rate `2.0 As^(-1)` , what will be the emf induced in the other coil? |
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Answer» Correct Answer - 5.0 V |
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| 1409. |
Flux passes through coil changes from `2xx10^(-3)Wb` to `3xx10^(-3)` Wb during 25s. The induced emf isA. 0.02 mVB. 0.03 mVC. 0.05 mVD. 0.04 mV |
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Answer» Correct Answer - D `e=(phi_(2)-phi_(1))/(dt)=(3xx10^(-3)-2xx10^(-3))/(25)` `e=0.04` mV |
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| 1410. |
A square loop of edge b having M turns is rotated with a uniform angular velocity `omega` about one of its diagonals which is kept fixed in a horizontal position. A uniform magnetic field `B_(0)` exists in the vertical direction. Find (i) the emf induced in the coil as a function of time t. (ii) the maximum emf induced. (iii) the average emf induced in the loop over a long period. (iv) if resistance of loop is R, amount of charge flown in time t = 0 to t = 2T. (v) heat produced in time t = 0 to t = 2T. |
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Answer» Initially, plane of loop is `_|_^(ar)` to the field. Let at any time, normal to loop makes an angle `theta` with magnetic field `theta=omegat`. The flux passing through loop, `phi_(B)=MB_(0)Scos theta=MB_(0)b^(2)cosomegat` `rArr" "e=-(dphi_(B))/(dt)=MB_(0)b^(2)omega sin omegat=e_(0)sinomegat,` where, `e_(0)=MB_(0)b^(2)omega` (ii) For emf to be maximum, then sin `omegat` should be equal to 1. `rArr" "e_(max)=e_(0)` (iii) Long period means one time period, i.e., `0rarrT` We know that average emf is given as `bar(e)=(int_(0)^(T)edt)/(int_(0)^(T)dt)=(1)/(T).e_(0)int_(0)^(T)sinomegatdt=0" "(becauseint_(0)^(T)sinomegatdt=0)` (iv) Using the relation, induced current, `i=(e)/(R)=(e_(0))/(R)sinomegat=i_(0)sinomegat` where, `i_(0)=(e_(0))/(R)`. We know q = idt So, for time `Orarr2T`, `q=int_(0)^(2T)"idt"=i_(0)int_(0)^(2T)sinomegatdt=0` (v) Heat (H) generated in a loop during time `0rarr2T`, is given as `H=int_(0)^(2T)(e^(2))/(R)dt=(e_(0)^(2))/(0)int_(0)^(2T)sin^(2)omegatdt` `=(e_(0)^(2))/(R).2.(T)/(2)=(e_(0)^(2))/(R).(2pi)/(omega)=(2pie_(0)^(2))/(omegaR)" "(becauseT=(2pi)/(omega))` |
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| 1411. |
Assertion A conducting loop is rotated in a uniform magnetic field with constant angular velocity `omega` as shown in figure. At time t = 0, plane of the loop is perpendicular to the magnetic field. Induced emf produced in the loop is maximum when plane of loop is parallel to magnetic field. Reason When plane of loop is parallel to magnetic field, then magnetic flux passing through the loop is zero. A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is ture. |
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Answer» Correct Answer - B `phi=BA cos omegat` As t = 0, `phi` = maximum `|e|=|(dphi)/(dt)|=B omega A sin omega t` `|e|_("max")` when `omegat=90^(@)`, i.e., loop is rotated `90^(@)`, or it is parallel to magnetic field. At this instant `phi=0` |
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| 1412. |
Asseration: Acceleration of a magnet falling through a long soleneoid decreases. Reason: the induced current produced in a circuit always flow in such direction that it opposes the change or the cause the produced it.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A The induced current in the ring opposes the motion of falling magnet. Therefore, the acceleration of the falling magnet. Therefore, the acceleration of the falling magnet will be less than that due to gravity. |
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| 1413. |
A metal disc of radius 200 cm is rotated at a constant angular speed of `rad s^(-1)` in a plane at right angles to an external field of magnetic induction `0.05 Wb m^(-2)`. Find the e.m.f. induced between the centre and a point on the rim. |
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Answer» Here, `r = 200 cm = 2 m, omega = 60 rad//s`. `B = 0.05 Wb m^(-2)` `e = BA v = BA ((omega)/(2 p)) = B (pi r^(2)) ((omega)/(2 pi))` `e = (1)/(2) B r^(2) omega = (1)/(2) xx 0.05 (2)^(2) xx 60 = 6 V` |
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| 1414. |
Asseration:Induced coil are made of copper. Reason:Induced current is more in wire having less resistance.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A The inductance coils made of copper will have very small ohmic resistance. Due to change in magnetic flux a large induced current will be produced in such an inductance, which will offer appreciable opposition to the flow of current. |
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| 1415. |
Statement-1 : The induced e.m.f. and current will be same in two identical loops of copper and aluminium, when rotated with same speed in the same magnetic field. Statement-2 : Induced e.m.f. is proportional to rate of change of magnetic field while induced current depends on resistance of wire.A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is ture. |
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Answer» Correct Answer - D Since, both the loops are identical (same area and number of turns) and moving with a same speed in same magnetic field. Therefore, same emf is induced in both the coils. But the induced current will be more in the copper loop as its resistnace wire be lesser as compared to that of the aluminium loop. |
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| 1416. |
If a straight conductor is moved with a uniform velocity at right angle to a uniform magneti field then e.m.f. induced in it isA. `e=-B//v`B. `e=B//v^(2)`C. `e=B//lv`D. `e=l//vB` |
| Answer» Correct Answer - A | |
| 1417. |
Asseration:The induced e.m.f. and curent will be same in two identical loops of copper and aluminium, when rotated with same speed in the same magnetic field. Reason:Induced e.m.f. is proportional to rate of change of magnetic field while induced current depends on resistance of wire.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D Since both the loops are idential (same area and number of turns) and moving with a same speed in same magnetic field. Therefore same emf is induced in both the coils. But the induced current will be more in the copper loop as its resistance will be lesser as compared to that of the aluminium loop. |
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| 1418. |
The value of current in 10 `Omega` resistor, when plug of key K is inserted in the adjoining figure is A. 0.5 AB. 0C. 0.3 AD. 0.2 A/s |
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Answer» Correct Answer - B Inductor offers zero resistance to dc. Thus, when plug key K is closed entire current flows through inductor only. Htus, current through `10Omega` resister is zero. |
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| 1419. |
In a coil, the current increases from 0 to 6 A in 0.4 s. If an induced of e.m.f of 15V is produced in the coil, then the coeffiecient of self-induction of the coil will beA. 0.5 HB. 0.75 HC. 1 HD. 1.5 H |
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Answer» Correct Answer - C `e=L(dI)/(dt) :. 15 = L xx 6/0.4 :. L = 1H`. |
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| 1420. |
The self induced emf in a coil when current charges in it is given by. |
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Answer» `e = - L (dI)/(dt)` where symbols have usual meaning. |
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| 1421. |
Two conducting circular loops of radii `R_1 and R_2` are placed in the same plane with their centres coincidingt. Find the mutual inductane between them assuming `R_2ltltR_1`.A. `R_(1)//R_(2)`B. `R_(2)//R_(1)`C. `R_(1)^(2)//R_(2)`D. `R_(2)^(2)//_(1)` |
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Answer» Correct Answer - D Mutual inductance between two coil in the same plane with their centres coinciding is giving by `M=(mu_(0))/(4pi)((2pi^(2)R_(2)^(2)N_(1)N_(2))/(R_(1)))`henry. |
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| 1422. |
A magnet is taken towards a conducting ring in such a way that a constant current of `10mA` is induced in it. The total resistance of the ring is `0.5Omega`. In `5s`, the magnetic flux through the ring changes byA. 0.25m WbB. 25mWbC. 50mWbD. 15mWb |
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Answer» Correct Answer - D |
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| 1423. |
Two thin concentric wires shpaed as circles with radii `a` and `b` lie in the same plane. Allowing for `a lt lt b`, find: (a) their mutual inductane, (b) the magnetic flux through the surface enclosed by the outside wire, when the inside wire carries a current `I`. |
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Answer» Correct Answer - `(mu_(0)pia^(2))/(2b); (b) (mu_(0)pi a^(2)I)/(2b)` |
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| 1424. |
Find the inductane of a unit length of a double line if the radius of each wire is `eta` times less than the distance between the axes of the wires. The field inside the wires is to be neglected, the permeability is assumed to be equal to unity throughout, and `eta gt gt 1`. |
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Answer» Correct Answer - `(mu_(0))/(pi)In(eta)` |
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| 1425. |
The induced e.m.f. is sometimes called back e.m.f. Why ? |
| Answer» This is because induced e.m.f. opposes the current due to the actual source of e.m.f. | |
| 1426. |
A 44 m H inductor is conneted to 220 V, 50 Hz a.c. supply. Determine the r.m.s. value of currents in the circuit. |
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Answer» Here, `L = 44 mH = 44 xx 10^(-3) H, E_(v) = 220 V, v = 50 Hz, I_(v) = ?` `X_(L) = omega L = 2 pi v L = 2 xx (22)/(7) xx 50 xx 44 xx 10^(-3) = 13.83 ohm` `I_(v) = (E_(v))/(X_(L)) = (220)/(13.83) = 15.9 A` |
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| 1427. |
Give expression for average value of a.c. voltage `V = V_(0) sin omega t` over interval `t = 0` to `t = (pi)/(omega)` |
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Answer» As `t = (pi)/(omega) = (1)/(2) . (2 pi)/(omega) = (1)/(2) T`, therefore, we wish to know average value of a.c. voltage over first half cycle `(0 rarr T//2)`. It is `E_(av) = (2 V_(0))/(pi)`. |
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| 1428. |
A coil has an inductane of 1 H. At what frequency will it have a reactance of 3142 ohm ? |
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Answer» Here, `L = 1 H, V = ?, X_(L) = 3142 ohm` From `X_(L) = omega L = 2 pi v L` `v = (X_(L))/(2 pi L) = (3142)/(2 xx 3.142 xx 1) = 500 Hz` |
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| 1429. |
A capacitor is `5 mu F` is conneted to 20 V supply through a resistance R. What should be the value of resistance so that condenser acquires 63.2% of its final charge in 10 millisecond ? |
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Answer» Here, `C = mu F = 5 xx 10^(-6) F, V = 20 v o l t, R = ?` To acqure 63.2% of its final charge, time required, `t = tau = RC = 10 xx 10^(-3) s` `R = (10 xx 10^(-3))/(C ) = (10^(-2))/(5 xx 10^(-6)) = 2 xx 10^(3) Omega` |
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| 1430. |
The voltage and current in an a.c. circuit are given by `E = 300 sin (omega t + pi//2)` and `I = 5 sin omega t`. What is power factor of the circuit and power dissipated in each cycle ? |
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Answer» Here, `E = 300 sin (omega + pi // 2), I = 5 sin omega t` `:. E_(v) = (300)/(sqrt 2)` and `I_(v) = (5)/(sqrt 2) A, phi = pi//2` Power factor` = cos phi = cos pi // 2 =` zero Average power dissipated/cycle `= E_(v) I_(v) cos phi = (300)/(sqrt 2) xx (5)/(sqrt 2) cos 90^(@) = 0` |
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| 1431. |
When an alternating emf, e=300 sin `100pit` volt is applied across a bulb, the peak value of current is found to be 2 A. the average power isA. 100 WB. 200 WC. 300 WD. 400 W |
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Answer» Correct Answer - C `overline(P)=(e_(0)I_(0))/(2)` `=(300xx2)/(2)=300W` |
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| 1432. |
An alternating e.m.f. `E=10sqrt(2) sin omega` volt is applied to a circuit containing a pure inductance L and a resistance and voltage acros the resistance is 6 v between the current and potential difference in this circuit isA. 10 VB. 8 VC. 6 VD. 12 V |
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Answer» Correct Answer - B `V_(rms) = (10sqrt(2))/(sqrt(2)) = 10 V` `:. 10^(2) = sqrt(V_(L)^(2)+V_(R)^(2)) :. 100 = 36 + V_(L)^(2) :. V_(L) = 8V`. |
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| 1433. |
A coil has an inductance of 1 henry. (a) At what frequency will it have a reactance of 3142 ohm? (b) What should be the capacity of a condenser which has the same reactanc at frequency ? |
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Answer» Here, `L = 1` henry, `v = ?` As `X_(L) = omega L = 2 pi v L` `:.v = (X_(L))/(2 pi L) = (3142)/(2 xx 3142 xx 1) = 500` hertz (b) `C = ? V = 500` hertz, `X_(C) = X_(L) = 3142 ohm` As `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `:. C = (1)/(2 pi v X_(C )) = (1)/(2 xx 3.142 xx 500 xx 3142)` `= (10^(-6))/((3.142)^(2))` farad `= 0.11 mu F` |
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| 1434. |
An alternating voltage is represented by `V=80 sin(100 pi t)cos(100 pi t)` volt, What is the peak voltage?A. 20 VB. 30 VC. 40 VD. 50 V |
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Answer» Correct Answer - C `V = 80 sin (100 pi t)cos (100 pi t)` `=80/2 xx 2 sin (100 pi t) cos (100 pi t)` Using `sin 2 theta = 2 sin (theta)/(2) cos (theta)/(2)`, we get `V = 40 sin [200 pi t]` Thus peak voltage `V_(max) or V_(0) = 40 V`. |
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| 1435. |
A coil of 0.01 henry inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current. |
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Answer» Here, `L = 0.01 H, R = 1 Omega, E_(v) = 200 V, v = 50 Hz, Z = ?, Delta t = ?` `X_(L) = omega L = 2 pi v L = 2 xx 3.14 xx 50 xx 0.01 = 3.14 Omega` `Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(1^(2) + (3.14)^(2)) = sqrt(10.86) = 3.3 ohm` `tan phi = (omega L)/(R ) = 3.14` `phi = tan^(-1) (3.14) = 72^(@) = 72 xx (pi)/(180) rad` Time lag, `Delta t = (phi)/(omega) = (72 xx pi)/(180 xx 2 pi xx 50) = (1)/(250) sec` |
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| 1436. |
A coil of 1 H and `100 Omega` resistance has a peak voltage of `5 sqrt2` volt 50 hz connected across it. Calculate the current through the coil and power absorbed. |
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Answer» Here, `L = 1 H, R = 1000 Omega, E_(0) = 5 sqrt 2 V, v = 50 hz, I_(v) = ? P = ?` `X_(L) = omega L = 2 pi v L = 2 pi xx 50 xx 1 = 1000 pi` `Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(100^(2) + (100 pi)^(2)) = 1.52 xx 10^(-2) A` `I_(v) = (E_(v))/(Z) = (E_(0))/(sqrt2 (Z)) = (5 sqrt 2)/(sqrt 2 xx 3.3 xx 10^(2)) = 1.52 xx 10^(-2) A` `P = E_(v) I_(v) cos phi = E_(v) I_(v) (R//Z)` `P = 5 xx 1.52 xx 10^(-2) xx (100)/(3.3 xx 10^(2)) = 2.29 xx 10^(-2)` watt |
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| 1437. |
At what frequency 1 Henry inductance offers the same impedance as `1 mu F` capacitor?A. `500/(2 pi) Hz`B. `500/(pi) Hz`C. `500pi Hz`D. `1000pi Hz` |
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Answer» Correct Answer - B `omega L = 1/(omega C) :. Omega^(2) = 1/(LC)` `:. F = 1/(2pisqrt(LC)) = 1/(2pisqrt(1xx10^(-6)))` `= (1000)/(2pi) = 500/(pi) Hz`. |
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| 1438. |
When a capacitance is connected in series with a series L-R, a.c. circuit, the total impedance of the circuitA. is increasedB. is decreasedC. does not changeD. is doubled |
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Answer» Correct Answer - B `Z = sqrt(R^(2)+(omega L - (1)/(omegaC)^(2))) = sqrt(R^(2)+L^(2)omega^(2)))`. |
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| 1439. |
A current of 6A flows through a coil when connected to a 24 volt d.c. supply. To get the same current with a 50 Hz a.c. supply, the voltage required is 30 V. What is the power factor of the coil?A. `2/5`B. `4/5`C. `5/6`D. `3/5` |
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Answer» Correct Answer - B `R = 24/6 = 4 Omega and Z = 30/6 = 5Omega` `:.` Power factor = `cos theta = R/Z = 4/5`. |
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| 1440. |
The phase angle between the current and voltage in an L-R circuit is `30^(@)`. What is the impendance of the circuit if the resistance in the circuit is `10sqrt(3) Omega`?A. `10 Omega`B. `15 Omega`C. `20 Omega`D. `30 Omega` |
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Answer» Correct Answer - C Power factor = `cos theta = cos 30^(@) = (sqrt3)/2` Power factor = `R/Z` `:. Z = R/(cos phi) = (10sqrt(3))/(sqrt(3)//2)= 20 Omega`. |
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| 1441. |
If the impedance of an L-C R circuit is `40 Omega`, then the admitance of the circuit will beA. A. 0.1 siemenB. B. 0.025 siemenC. C. 0.05 siemanD. D. 0.075 siemen |
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Answer» Correct Answer - B Admittance `K = 1/1 = 1/40 = 0.025` siemen. |
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| 1442. |
Choose the correct option: A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each:A. `(1)/(4)revolution`B. `(1)/(2)revolution`C. `1revolution`D. `2revolution` |
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Answer» Correct Answer - B This is the case of periodic `EMI` |
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| 1443. |
A transformer is an electrical device which is used for changing the a.c. voltage. It is based on the phenomenon of mutual induction. It can be shown that `(E_(s))/(E_(P)) = (I_(P))/(I_(s)) = (n_(s))/(n_(P)) = K` Where the symbols have their standard meaning. For a step up trasnformer,`K gt 1` and for a step down transformer, `K lt 1`. The above relations are on the assumption that efficiency to transformer is `100%` . In fact, efficiency `eta = ("outpu power")/("input power") = (E_(s) I_(s))/(E_(P) I_(P))` The number of turns in the primary and secondary coils of a transformer are 2000 and 50 respectively. The primary coil is connected to main supply of `120 V` and secondary to a night bulb `0.6 ohm`. The efficiency of transfomer is `80%` Voltage across the secondary of transformer isA. `125 V`B. `360 V`C. `40 V`D. `3 V` |
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Answer» Correct Answer - D Here, `n_(P) = 2000, n_(s) = 50, E_(P) = 120 V`, `R_(s) = 0.6 ohm eta = 80%` As `(E_(s))/(E_(P)) = (n_(s))/(n_(P))` `:. E_(s) = (n_(s))/(n_(P)) xx E_(P) = (50)/(2000) xx 120 = 3 V` |
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| 1444. |
A transformer is an electrical device which is used for changing the a.c. voltage. It is based on the phenomenon of mutual induction. It can be shown that `(E_(s))/(E_(P)) = (I_(P))/(I_(s)) = (n_(s))/(n_(P)) = K` Where the symbols have their standard meaning. For a step up trasnformer,`K gt 1` and for a step down transformer, `K lt 1`. The above relations are on the assumption that efficiency to transformer is `100%` . In fact, efficiency `eta = ("outpu power")/("input power") = (E_(s) I_(s))/(E_(P) I_(P))` The number of turns in the primary and secondary coils of a transformer are 2000 and 50 respectively. The primary coil is connected to main supply of `120 V` and secondary to a night bulb `0.6 ohm`. The efficiency of transfomer is `80%` Current in night bulb isA. `5 A`B. `3 A`C. `0.5 A`D. `0.6 A` |
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Answer» Correct Answer - A Current in night bulb is `I_(s) = (E_(s))/(R_(s)) = (3)/(0.6) = 5 A` |
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| 1445. |
A 20 V, 750 Hz source us connected to a series combination of `R = 100 Omega, C = 10 mu F` and `L = 0.1803 H`. Calculate the time in which resistance will get heated by `10^(@) C`, if thermal capacity of the material `= 2 J//.^(@)C` |
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Answer» Here, `E_(v) = 20 V, v = 750 Hz, R = 100 Omega, C mu F = 10^(-5) F, L = 0.1803 H, Delta theta = 10^(@)C, ms = 2 J//.^(@)C` `X_(L) = 2 pi v L = 2 xx (22)/(7) xx 750 xx 0.1803 Omega = 850 Omega, X_(C ) = (1)/(2 pi v C) = (1)/(2 xx (22)/(7) xx 750 xx 10 xx 10^(-6)) Omega = 21.2 Omega` `Z = sqrt(R^(2) + (X_(L) - X_(C))^(2)) = sqrt(100^(2) + (850 - 21.2)^(2)) = 834.8 Omega` `I_(v) = (E_(v))/(Z) = (20)/(834.8) = 0.02395` As `I_(v)^(2) Rt = (ms) Delta theta = 2 xx 10 = 20 :. t = 20//I_(v)^(2) R = (20)/((0.02395)^(2) xx 100) = 348.7 s` |
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| 1446. |
The phenomenon of generating current/e.m.f. in a circuit by changing magnetic field linked with the circuit is called electromagnetic induction (EMI). In our daily lives, we use (EMI) in many ways without even realiaing it. Read the above passage and answer the following question : (i) What is the most universal application of EMI ? (ii) How is the phenomenon being used to enhance the safety of all our important establishments ? |
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Answer» (i) The most universal application of the phenomenon of EMI is in the generation of electricity all over the world. Electricity is the eighth wonder of modern civilized world. It is just impossible to imagine the world without electricity as nothing works without electric power. And this electric power generation is based on hte phenomenon of electromagnetic induction. We are using electricity every moment without realizing the phenomenon of EMI, used in its production. (ii) All key estabishments like airports, railway stations, bus terminals, shopping malls, defence, scientific research areas etc. are provided with metal detectors at their main entrance. The visitors are to pass through the are of the detector and they are subjected to a series of magnetic field pulses. If the visitor has some metal object like knife, dagger, pistol etc. on his person, the currents induced in these objects create their own magnetic field echo, which is detected. The safety inspectors on duty can then check the visitor more thoroughly. |
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| 1447. |
Self-inductance of a coil is 50mH. A current of 1 A passing through the coil reduces to zero at steady rate in 0.1 s, the self-induced emf isA. 5 VB. 0.05 VC. 50 VD. 0.5 V |
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Answer» Correct Answer - D Here, `L=50xx10^(-3)H` `(dI)/(dt)=((1-0))/(0.1)=10` emf, `e=(L.dI)/(dt)=50xx10^(-3)xx10` `=50xx10^(-2)=0.5V` |
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| 1448. |
Current in a circuit falls form 0.5 A to 0.0 A in 0.1 s. If an average e.m.f. of 200 V is induced, give an estimate of the self inductance of the circuit ? |
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Answer» Here, `(dI)/(dt) = ((I_(2) - I_(1)))/(t) = (0.0 - 0.5)/(0.1) = - 50 As^(-1), e = 200 V, L = ?` As `|e| = L |(dI)/(dt)| :. L = (|e|)/(|dI // dt|) = (200)/(50) = 4 H` |
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| 1449. |
In a coil current falls from 5 A to 0 A in 0.2 s. If an average emf of 150 V is induced, then the self inductance of the coil isA. 4HB. 2 HC. 3 HD. 6 H |
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Answer» Correct Answer - D Here `(dI)/(dt)=(0-5)/(0.2)=-25A s^(-1),epsi=150V` `As |epsi|=|(dI)/(dt)|` `:. L=(|epsi|)/(|dI//dt|)=(250)/(25)=6H` |
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| 1450. |
The north pole of a magnet is brought near a coil. The induced current in the coil as seen by an observer on the side of magnet will be-A. in the clockwise directionB. in the anticlockwise directionC. initially in the clockwise and then anticlockwise directionD. initially in the anticlockwise and then clockwise direction |
| Answer» Correct Answer - B | |