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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.
| 851. |
In the circuit shown in the figure initialy the switch in position 1 for a long time, then suddenly at `t=0` the switch is shifted to position 2. It is required that a constant current should flow in the circuit, the value of resistance `R` in the circuit A. Should be decreased at a constant rateB. Should be increased at a cosntant rateC. should be maintained constantD. Not possible |
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Answer» Correct Answer - A |
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| 852. |
An inductor acts as a conductor for d.c. why ? |
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Answer» The inductive reactance is given by `X_(L) = omega L = 2 pi v L` For d.c., `v = 0, :. X_(L) = 0` i.e.., inductor offers no resistance to d.c. Hence for d.c. an inductor acts as a conductor. |
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| 853. |
Why connot we use a.c. for electrolysis? |
| Answer» For electrolysis, we require fixed cathode and fixed anode. In a.c., the direction of current is changing periodically. | |
| 854. |
A series LCR circuit is made by taking `R = 100 Omega, L = (2)/(pi)` and `C = (100)/(pi) mu F`. The series combination is connected across an a.c. source of 220 V, 50 Hz. Calculate impedance of the circuit and peak value of current flowing n the circuit. What is power factor of the circuit ? Compare it with one at resonance frequency. |
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Answer» Here, `R = 100 Omega`, `L = (2)/(pi) H, C = (100)/(pi) xx 10^(-6) F`, `E_(v) = 220 V, v = 50 Hz, Z = ? I_(0) = ? Cos phi = ?` `X_(L) = omega L = 2 pi L = 2 pi xx 50 xx (2)/(pi) = 200 Omega` `X_(C ) = (1)/(omega C) = (1)/(2 pi v xx (100)/(pi) xx 10^(-6)) = 100 Omega` `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` `= sqrt(100^(2) + (200 - 100)^(2)) = 141.4 Omega` `I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(Z) = (sqrt2 xx 200)/(141.4) = 2.2 A` Power factor `= cos phi = (R )/(Z) = (100)/(100 sqrt2) = (1)/(sqrt 2)` At resonance, `Z = R , cos phi = (R )/(R ) = 1` |
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| 855. |
Assertion : A glowing bulb becomes dim when an iron bar is put in the inductor in the a.c. circuit Reason : Resistance of the circuit increases.A. AB. BC. CD. D |
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Answer» Correct Answer - A On introducing an iron bar in the inductor, inductive reactance `X_(L) = oemga L` increases. Therefore impedance of the circuit `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` , increases, current through the bulb decreases and it becomes dim. |
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| 856. |
For an a.c. can ever: (a) r.m.s value to equal to peak value ? (b) average value be equal to peak value ? (c ) r.m.s value be equal to average value (d) all the three values be equal ? |
| Answer» Yes, when the a.c. is a square wave, r.m.s value, average value and peak value of a.c. are equal Choiec (d) is correct. | |
| 857. |
Explain : voltages across L and C in series are `180^(@)` out of phase, while for L and C in parallel, current in L and C are `180^(@)` out of phase. |
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Answer» When L and C in series, for a given a.c., voltage acorss L leads `pi//2`.and voltage across C lags by `pi//2`. Therefore, voltages acorss L and C differ in phase by `pi` When L and C are in parallel, for a given voltage, current in L lag by `pi//2` and current in C leads by `pi//2`. Therefor, currents in L and C differ in phase by `pi`. |
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| 858. |
An inductor is connected in series with a bulb to an a.c source. What happens to brighness of bulb when number of turns in the inductor is reduced ? |
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Answer» Current through the bulb `I_(v) = (E_(v))/(Z) = (E_(V))/(sqrt(R^(2) + X_(L)^(2)))`, where `X_(L) = omega L` As `L prop N^(2)`, therefore, on decreasing number of turns, L decreases , Z decreases, `I_(v)` increases. Hence brightness of bulb increases. |
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| 859. |
Distinguish between alternating current and direct current by giving two points. |
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Answer» (i) Alternating current is the current which shows periodic variation in its value in its value with time. Direct current is that current whose magnitude and direction do not change. (ii) The frequency of alternating current has some finite value. The frequency of direct current is zero. |
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| 860. |
The resistance of a coil for direct current is `10 Omega`. An alternating current is sent through it. Will its resistance remain the same ? |
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Answer» No, the resistance of the coil will not remain the same. It will increases to `Z = sqrt(R^(2) + (omega L)^(2))`, which is obviously more. When potential difference is constant, heat produced/sec in inversely proportional to resistance. Hence heat produced/sec in case of a.c. will be less compared to heat produced/sec in case of d.c. |
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| 861. |
A Capacitor blocks d.c. and allows a.c. Why ? |
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Answer» The capacitative reactance is `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` For d.c., `v = 0 , :. X_(C ) = oo` i.e., capacitor offers infinite resistance to d.c. and hence block it. For a.c., `v != 0`, `:. X_(C ) != oo`, but it has only a finite value Therefore, a.c. can pass through the capacitor. |
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| 862. |
The powerr factor in CR circuit is given byA. `(R)/(omegaC)`B. `(1)/(omegaCR)`C. `(R)/(sqrt(R^(2)+((1)/(omegaC))^(2)))`D. `(R)/(R+(1)/(omegaC))` |
| Answer» Correct Answer - C | |
| 863. |
For minimum dissipation energy in the circuit, the power factor should beA. largeB. smallC. moderateD. can not say |
| Answer» Correct Answer - A | |
| 864. |
The applied emf lags behind the current by an angle of `45^(@)` in the circuit which containsA. resistance onlyB. resistance and inductanceC. capacitance onlyD. capacitance and resistance |
| Answer» Correct Answer - D | |
| 865. |
An ac ammeter is used to measure currnet in a circuit. When a given direct current passes through the circuit. The ac ammeter reads 3 A. When another alternating current passes through the circuit, the ac ammeter reads 4A. Then find the reading of this ammeter (inA), if dc and ac flow through the circuit simultaneously.A. 7AB. 5AC. 3AD. 1A |
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Answer» Correct Answer - B A.c. ammeter works on the heating effect of electric current. `:.` Heat produced in the ammter of resistor R per sec due to d.c. of `3A = ((I^(2)R)/(J))=((3)^(2)R)/(J) = (9R)/(J)` and that due to a.c. of `4 A = ((4)^(2)R)/(J) = (16R)/(J)` `:.` Total heat prodcued`//`sec = `(9R)/J + (16R)/J = (25R)/(J)` If is the equivalent rms current, then `(I^(2)R)/(J) = (25R)/(J)` `:. I^(2)=25 or I=5A`. |
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| 866. |
A square loop of wire with side length 10 cm is placed at angle of `45^(@)` with a magnetic field that changes uniformly from 0.1 T to zero in 0.7s. The induced current in the loop (its resistance is `1Omega`) isA. 1.0 mAB. 2.5 mAC. 3.5 mAD. 4.0 mA |
| Answer» Correct Answer - A | |
| 867. |
There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of the paper. ThenA. current is induced in the loop in the anticlockwise directionB. current is induced in the loop in the clockwise directionC. Altrenating current is induced in the loopD. no current is induced in the loop |
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Answer» Correct Answer - A The area of loop is increasing |
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| 868. |
There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of the paper. ThenA. Current is induced in the loop in anticlockwise directionB. Current is induced in the loop in clockwise directionC. AC is induced in the loopD. No current induced in the loop |
| Answer» Correct Answer - A | |
| 869. |
Two coils P and S have a mutual inductance of `5 xx 10^(-3)H`. If the current in the primary is `I=10 sin(100 pi t)`, then the maximum value of the e.m.f. induced in S isA. 6.82 VB. 12.56 VC. 15.70 VD. 3.14 V |
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Answer» Correct Answer - C `I_(P) = I_(0) sin omega t :. (dI_(P))/(dt) = I_(0) omega cos omega t` `:. E_(S) = M(dI_(P))/(dt) = MI omega cos omega t` `:. Max E_(S) = 5 xx 10^(-3) xx 10 xx 100 pi xx 1` `=5 pi = 15.70 V`. |
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| 870. |
What is the SI unit of mutual inductance of two coils ? |
| Answer» 1 Henry, is SI unit of mutual inductance of two coils. | |
| 871. |
If the current increases from zero to 1A in 0.1 s in a coil of 5 mH then magnitude of induced emf will beA. 5 VB. 0.5 VC. 0.05 VD. 0.005 V |
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Answer» Correct Answer - C `e=L(dI)/(dt)` `=(5xx10^(-3)xx1)/(0.1)=0.05V` |
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| 872. |
If an emf of 1 volt is induced in the coil due to the change of current of 1A/s then the inductance isA. a henryB. a faradC. a ohmD. none of these |
| Answer» Correct Answer - A | |
| 873. |
The coils in resistance boxes are made from doubled insulated wire to nullify the effect ofA. heatingB. magnetismC. pressureD. self induced emf |
| Answer» Correct Answer - D | |
| 874. |
What are the factors on which self inductance of a coil depend ? |
| Answer» (i) Number of turns (ii) area of cross section (iii) permeability of core material. | |
| 875. |
The core of any transformaer is laminated so as toA. Eddy currentsB. HysteresisC. Resistance in windingD. None of these |
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Answer» Correct Answer - A Circulation of eddy currents is prevented by use of laminated core. |
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| 876. |
Which of the following does not depend upon the magnetic effect of some sortA. moving coil glavanometer B. hot wire ammeterC. dynamoD. electric motor |
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Answer» Correct Answer - B Hot wire ammeter is not based on the phenomenon of electromagnetic induction. |
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| 877. |
The self inductance of a coil does not depend uponA. currentB. timeC. induced voltageD. resistance of coil |
| Answer» Correct Answer - D | |
| 878. |
The self inductance of a coil does not depend uponA. the diameter of the coilB. the length of the coilC. the resistance of the wire of coilD. the number of turns in the coil |
| Answer» Correct Answer - C | |
| 879. |
Which of the following does not depend upon the magnetic effect of some sortA. moving coil glavanometerB. hot wire ammeterC. dynamoD. electric motor |
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Answer» Correct Answer - B Hot wire ammeter is not based on the phenomenon of electromagnetic induction. |
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| 880. |
In the circuit shown in figure switch `S` is closed at time `t=0`. The charge which passes through the battery in one time constant is A. `(eR^(2)E)/(L)`B. `E((L)/(R ))`C. `(EL)/(eR^(2))`D. `(eL)/(ER)` |
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Answer» Correct Answer - C The current at time `T` is given by `i=i_(0)(1-e^(-t//tau))` where `i_(0)=(E)/(R )` and `tau=(L)/(R )` So `dq=I dt=i_(0)(1-e^(-t//tau))dt` `Q=intdQ=in_(0)^(R )i_(0)(1-e^(t//tau))dt=i0{[in_(0)^(R )dt-int_(0)^(R )e^(-t//tau)dt]}` `=(i_(0)tau)/(e)impliesQ=((E)/(R ).(L)/(R))/(e)=(EL)/(eR^(2))impliesQ=(EL)/(ER^(2))` |
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| 881. |
A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B = 0.2 T . The normal to the plane of the coil makes an angle of `60^(@)` with the direction of the magnetic field. The magnetic flux linked with the coil isA. `5xx10^(-3)`WbB. `2.5xx10^(-3)`WbC. `3.5xx10^(-3)`WbD. `4.5xx10^(-3)` Wb |
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Answer» Correct Answer - A `phi=nABcos60` `=5xx10^(-3)Wb` |
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| 882. |
The magnetic field in a coil of 100 turns and 40 square cm area is increased from 1 Tesla to 6 Tesla in 2 second. The magnetic field is perpendicular to the coil. The e.m.f. generated in it isA. 0.5 VB. 1 VC. 1.5 VD. 2 V |
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Answer» Correct Answer - B `e=nA(dB)/(dt)=1V` |
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| 883. |
A field of 0.2 tesla acts acts at right angles to a coil of area `100 cm ^(2)m` with 50 turns. The coil is removed from the field in 0.1s . Find the emf induced . |
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Answer» Correct Answer - 1V |
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| 884. |
STATEMENT - 1 : The magnetic flux through a closed surface is zero. STATEMENT - 2 : When a rod oriented along magnetic north and magnetic south direction falls under gravity. Potential difference developed across its ends is zero. STATEMENT - 3 : A resistance is connected to an ac source. If an inductor is included in the circuit in series. The average power dissipated in the resistance will decrease.A. TTTB. FFTC. TTFD. TFF |
| Answer» Correct Answer - A | |
| 885. |
Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance `1Omega//m`. Position of the conducting rod at `t=0` is shown. A time dependent magnetic field `B=2t` tesla is switched on at `t=0` The magnitude of the force required to move the conducting rod at constant speed 5cm/s at the same instant `t=2s`, is equal toA. `0.096N`B. `0.12N`C. `0.08N`D. `0.064N` |
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Answer» Correct Answer - C `i=e_("net")/R=0.08/((1)(30+30+20)xx10^-2)` `=0.1A` `F=ilB` `=(0.1)(0.2)(4)=0.08N` |
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| 886. |
Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance `1Omega//m`. Position of the conducting rod at `t=0` is shown. A time dependent magnetic field `B=2t` tesla is switched on at `t=0` At `t=0`, when the magnetic field is switched on, the conducting rod is moved to the left at constant speed `5cm//s` by some external means. At `t=2s`, net induced emf has magnitudeA. `0.12V`B. `0.12V`C. `0.04V`D. `0.02V` |
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Answer» Correct Answer - B At `t=2s,` rod will move `10cm`. Hence `40cm` side will become `30cm`. `|e|=e_1(say)=S((dB)/(dt))` `(0.2xx0.3)(2)=0.12V` At `t=2s, B=4T` `:. e_2=Bvl` `=(4)(5xx10^-2)(0.2)` `=0.04V` `:. e_("net")=e_1-e_2=0.08V` |
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| 887. |
In the above question find the force required to move the rod with constant velocity `v`, and also find the power delivered by the external agent. |
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Answer» The force needed to keep the velocity constant `F_(ext)=ilB=(B^(2)l^(2)v)/(R+r)` Power due to external force `=(B^(2)l^(2)v^(2))/(R+r)=epsilon^(2)/(R+r)=i^(2)(R+r)` Note that the power delivered by the external agent is converted into joule heating in the circuit.That means magnetic field helps in converting the mechanical energy into joule heating. |
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| 888. |
(a) Solve the above question for part (ii). (b) Solve the above question for part (iii). |
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Answer» (a) `B_(1)S, (b) `B_(2)S` |
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| 889. |
Electric power is transmitted over long distances through conducting wires at high voltage becauseA. it reduces the possibility of theft of wireB. this entails less power lossesC. AC generators produce electric power at high voltagesD. AC signal of high voltage travels faster |
| Answer» Correct Answer - B | |
| 890. |
A 20 H inductor is placed in series with 10 W resistor and an emf of 100 V is suddenly applied to the combination. At t = 1 s from t = 1 s from the start, find the rate at which energy is being stored in the magnetic field around the inductor (givne `e^(-0.5)=0.61`). |
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Answer» `i_(0)=(epsilon)/(R)=10A`. For growth of current, `i=i_(c)(1-e^((-Rt)/(L)))` `therefore" "e=10(1-e^(-0.5))=3.9A` Energy `U=(1)/(2)Li^(2)rArr (dU)/(dt)=Li(di)/(dt)=(R)/(L)i_(0)(1-e^((-Rt)/(L)))(i_(0)e^((-Rt)/(L)))=237.9W.` |
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| 891. |
In series with `20 ohm` resitor a `5` henry inductor is placed. To the combination an e.m.f. of `5` volt is applied. What will be the rate of increase of current at `t=0.25 sec`?A. `e`B. `e^(-2)`C. `e^(-1)`D. none of these |
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Answer» Correct Answer - C `i=i_(0)(1e^(-Rt//L))implies(di)/(dt)=i_(0)((R )/(L))e^(-Rt//L)=(E)/(L)e^(-Rt//L)` |
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| 892. |
The parallel combination of inductor and capacitor is called asA. rectifier circuitB. tank circuitC. acceptor circuitD. filter circuit |
| Answer» Correct Answer - B | |
| 893. |
The frequency of LC oscillation is given byA. `f=(1)/(2pisqrt(LC))`B. `f=2pisqrt(LC)`C. `f=(1)/(pisqrt(LC))`D. `f=(1)/(4piepsi_(0))sqrt(LC)` |
| Answer» Correct Answer - A | |
| 894. |
The instantaneous current from an a.c source is `I = 5 sin 100 pi t` What is the frequency of a.c. ? What is the rms value of current ? |
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Answer» Here, `I = 5 sin 100 pi t` Compare with `I = I_(0) sin omega t` ` I_(0) = 5 A, omega = 2 pi v = 100 pi`, `v = (100)/(2 pi) = 50 Hz` `I_(v) = (I_(0))/(sqrt 2) = (5)/(sqrt 2) = (5 sqrt 2)/(2) = 3.54 A` |
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| 895. |
The line the draws power supply to your house from street hasA. zero average currentB. 220 V average voltageC. voltage and current out of phasea by `90^(@)`D. voltage and current possibly differing in phase `phi` such that `|phi| lt (pi)/(2)` |
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Answer» Correct Answer - A::D The line that draws power supply to our house from street supplies alternating current, whose average `value//mean` value over a cycle is zero. As the line has some resistance `(R != 0)`, thereofore, voltage and current differ in phase `phi` such that `|phi| lt pi //2`. choice (a) and (d) are correct. |
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| 896. |
The current leads the voltage by an angle `phi` which is given byA. `tan^(-1)((1)/(omegaCR))`B. `tan^(-1)(omegaCR)`C. `tan^(-1)((omegaC)/(R))`D. `tan^(-1)((R)/(omegaC))` |
| Answer» Correct Answer - A | |
| 897. |
What is the root mean square value of current of effective current of an a.c. having a peak value of 5.0 amp? What will be the reading shown for this current by (i) an a.c. ammeter (ii) an ordinary moving coil ammeter ? |
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Answer» Here, `I_(v) = ? I_(0) = 5.0 A` As `I_(v) = (I_(0))/(sqrt 2)` `:. I_(v) = (5)/(sqrt2) xx (sqrt 2)/(sqrt 2) = (5 xx 1.414)/(2) = 3.54 A` (i) An a.c. ammeter will read 3.54 ampere (ii) An ordinary moving coil ammter will read zero becauseit records average value of current over a complete cycle, which is zero in case of alternating current. |
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| 898. |
A series L-C-R circuit is connected to an alternating voltage source of frequecy f. If the current leads the e.m.f. By `45^(@)`, then the value of C isA. `1/(2 pi f(2 pi fL+R))`B. `1/(2 pi fC(2 pi fL-R))`C. `1/(pi f(2 pi fL-R))`D. `1/(pi f(2 pi fL+R))` |
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Answer» Correct Answer - B In a series L-C-R circuit, `tan phi = (X_(C)-X_(L))/R` As the current leads the e.m.f. `X_(C) gt X_(L)`. `:. tan 45^(@) = (X_(C)-X_(L))/R =1` `:. X_(C) = X_(L)+R` `:. 1/(omega C) = omega L + R :. Omega C = 1/(omegaL+R)` `:. C = 1/(omega(omegaL+R)) = 1/(2pi f(2 pi f L+R))`. |
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| 899. |
A coil of 1000 turns each of area `1.2m^(2)` is rotating about an axis in its plane and perpendicular to uniform magnetic field of induction `2xx10^(-3)T`. If it performs 300 rpm. The peak value of induced emf isA. 65.4 VB. 25.4 VC. 70.4 VD. 75.4 V |
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Answer» Correct Answer - D `e_(0)=nAB2pif` `=(100xx1.2xx2xx10^(-3)xx2xx3.14xx300)/(60)` `=75.4V` |
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| 900. |
The values of current and voltage in an AC circuits are respectively I=4 sin `omegat` and `e=100cos[omegat+(pi//3)]`. The phase difference between voltage and current isA. `(7pi)/(6)`B. `(6pi)/(5)`C. `(5pi)/(6)`D. `(pi)/(3)` |
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Answer» Correct Answer - C Phase difference`=(pi)/(2)+(pi)/(3)=(5pi)/(6)` |
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