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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.
| 801. |
Assertion (A) : The r.m.s value of alternating current is defined as the square root of the average of `I^(2)` during a complete cycle. Reason (R ) : For sinusoidal a.c. `(I = I_(0) sin wt) I_("rms") = (I_(0))/(sqrt(2))`A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. Assertion is true but Reason is falseD. Assertion is false but Reason is true |
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Answer» Correct Answer - 2 Definition of rms current, `I_("rms") = sqrt((int_(0)^(T) l^(2) dt)/(int_(0)^(T) dt))` If `l = l_(0) sin omega t` then `I_("rms") = sqrt((1)/(T) int_(0)^(t) l^(2) dt)`, `l_("rms") = sqrt((1)/(T) int_(0)^(T) sin omega t dt) = (l_(0))/(sqrt(2))` So both statements are ture but statement-2 is not correct explanation of statement-1 |
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| 802. |
Assertion: When frequency is greater than resonace frequency is a series `LCR` circuit, it will the an inductive circuit. Reason : Resultant voltage will lead the current.A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. Assertion is true but Reason is falseD. Assertion is false but Reason is true |
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Answer» Correct Answer - 1 At resonance `X_(L) = X_(C )` and frequency `f_(0) = (1)/(2 pi) sqrt((1)/(LC))` If `f gt f_(0)` then `X_(L) gt X_(C )`, so it will be an inductive circuit. `AC` current must lag `AC` voltage. |
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| 803. |
Assertion (A) : Maximum power is dessipated in a circuit (through `R`) in resonace Reason (R ) : At resonance in a series `LCR` circuit, the voltage across inductor and capacitor are out phase.A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. Assertion is true but Reason is falseD. Assertion is false but Reason is true |
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Answer» Correct Answer - 1 At resonance `P = I_("max")^(2)` and `V_(L)` and `V_(C )` are out of phase. `I_("max")` is due to `Z_("min") = R` which is due to out of phase of `V_(L)` and `V_(C )` |
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| 804. |
If the voltage in an ac circuit is reparesented by the equation `V=220 sqrt(2) sin (314 t - phi)`, calculate frequency of acA. `50 Hz`B. `50 sqrt(2) Hz`C. `50//sqrt(2) Hz`D. `75 Hz` |
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Answer» Correct Answer - A As `omega = 2 pi F, 2 pi f= 314, i.e. , f=(314)/(2 xx pi) = 50 Hz`. |
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| 805. |
A light bulb is rated at 100 W for a 220V ac supply. Calculate the resistance of the bulb. |
| Answer» `P=V^2/R`or,`R=V^2/P=(220)^2/100=484Omega` | |
| 806. |
The electric field part of an electromagnetic wave in a medium is represented by `E_(x)=0,` `E_(y)=2.5(N)/(C) cos[(2pixx10^(6)(rad)/(m))t-(pixx10^(-2)(rad)/(s))x]` `E_(z)=0`. The wave isA. moving along y-direction with frequency `2pixx10^(6)` Hz and wavelenght `200m`B. `"moving along" `x-direction` "with frequency" `10^(6)` Hz and "wavelenght" `100m`C. moving along x-direction with frequency `10^(6)` Hz and wavelength `200m`D. moving along x-direction with frequency `10^(6)` Hz and wavelenght `200m` |
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Answer» Correct Answer - C |
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| 807. |
In any `AC` circuit the emf `(e)` and the current `(i)` at any instant are given respectively by `e= E_(0)sin omega t` `i=I_(0) sin (omegat-phi)` The average power in the circuit over one cycle of `AC` isA. `(V_(0)i_(0))/(2)`B. `(V_(0)i_(0))/(2)sin phi`C. `(V_(0)i_(0))/(2)cos phi`D. `V_(0)i_(0)` |
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Answer» Correct Answer - C |
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| 808. |
In any `AC` circuit the emf `(e)` and the current `(i)` at any instant are given respectively by `e= E_(0)sin omega t` `i=I_(0) sin (omegat-phi)` The average power in the circuit over one cycle of `AC` isA. `(E_(0)I_(0))/2`B. `(E_(0)I_(0))/2 sin phi`C. `(E_(0)I_(0))/2 cos phi`D. `E_(0)I_(0)` |
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Answer» Correct Answer - C The power is defined as the rate at which work is being done in the circuit. Power =rate of work done in one complete cycle. or `P_(av)=W/T` or `P_(av)=((E_(0)I_(0)cos phi)T//2)/T` or `P_(av)=(E_(0)I_(0)cos phi)/2` where `cos phi` is called the power factor of an `AC` circuit. |
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| 809. |
The figure represents the voltage applied across a pure inductor. The diagram which correctly represents the variation of curent `i` with time `t` is given by .B. C. D. |
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Answer» Correct Answer - C `V_L` function is cos function, which is `90^@` ahead of the current function. Hence, current function should be sin function. |
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| 810. |
An `L-C-R` series circuit with `100Omega` resisance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the current and the power dissipated in the `L-C-R` circuit |
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Answer» When capacitance is removed, then `tanphi=X_L/R` or `tan60^@=X_L/R` `:. X_L=sqrt3R` When inductance is removed, then `tanphi=X_C/R` or `tan60^@C=X_C/R` `:. X_C=sqrt3R` From Eq.(i) and (ii), we see that `X_C=X_L` So, the `L-C-R` circuit is in resonance Hence, `Z=R` `:. i_("rms")=V_("rms")/Z=200/100=2A` `(:P:)=V_("rms")i_("rms")cosphi` At resonance current and voltage are in phase, or `phi=0^@` `:. (:P:)=(200)(2)(1)=400W` |
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| 811. |
In a series circuit `R=300Omega, L=0.9H, C=2.0 muF` and `omega=1000rad//sec` . The impedence of the circuit isA. `1300 Omega`B. `900 Omega`C. `500 Omega`D. `400 Omega` |
| Answer» Correct Answer - c | |
| 812. |
A constant voltage at a frequency of `1 MHz` is applied to an inductor in series with variable capacitor, when capacitor is `500 pF`, the current has its maximum value, while it is reduced to half when capacitance is `600 pF`. Find `Q` factor of the circuit isA. 10.4B. 20.8C. 5.2D. 9.4 |
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Answer» Correct Answer - A `Q = (omega_(0) L)/(R ) = (314)/(30.01) = 10.4` |
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| 813. |
When an AC voltage, of variable frequency is applied to series L-C-R circuit , the current in the circuit is the same at 4 kHz and 9 kHz. The current in the circuit is maximum atA. 5 kHzB. 6.5 kHzC. 4.2 kHzD. 6 kHz |
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Answer» Correct Answer - D (d) `I_(1) = I_(2)` `therefore omega_(1)L - 1/(omega_(1)C) = 1/(omega_(2)C) - omega_(2)L` Solving this, we get `1/sqrt (LC)`= responance frequency = `sqrt (omega_(1)omega_(2))` At resonance frequency , current will be maximum . |
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| 814. |
In the circuit shown in figure , the supply has a constant rms value V but variable frequency f. The frequency at which the voltage drop across R is maximum is A. 100 HzB. 500 HzC. 300 HzD. None of these |
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Answer» Correct Answer - B (b) Voltage drop across R is maximum at resonance, i.e., at `omega = 1/sqrt (LC) or f = omega/(2pi) = 1/(2pi sqrt (LC))` `= 1/(2pi sqrt ((1/pi)(1/pixx10^(-6))) = 500 Hz` |
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| 815. |
In the a.c circuit shown in figure, the supply voltage has a constant r.m.s value but variable frquency f. Resonance frequency is A. `10 Hz`B. `100 Hz`C. `1000 Hz`D. `200 Hz` |
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Answer» Correct Answer - C `f_(0) = (1)/(2pi sqrt(LC)) = (1)/(2 pi sqrt((1)/(2 pi) . (1)/(2pi) xx 10^(-6))) = 1000 Hz` |
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| 816. |
In an ac circuit shown in fig , the supply voltage has a constant rms value v but variable frequency f. At resonance, the circuit A. has a current `I` given by `I=V/R`B. has a resonance frequency `500 Hz`C. has a voltage across the capacitor which is `180^(@)` out of phase with that across the inductorD. has a current given by =`V/sqrt(R^(2)+(1/pi+1/pi)^(2))` |
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Answer» Correct Answer - A,B,C Resonance frequency `f=1/(2pi)1/sqrt(LC)=500 Hz` At resonance `Z=R` &`I=V/z=V/R` `L` & `C` are in out of phase. |
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| 817. |
A circuit is set up by connecting L=100mH, C`=5muF` and R`=100Omega` in series. An alternating emf of 150` sqrt(2)V,(500)/(pi) Hz is applied across this series combination. Which of the following is correct:A. the impedance of the circuit is `141.4 Omega`B. the average power dissipated across resistance `225 W`C. the average power dissipated across inductor is zero.D. the average power dissipated across capacitor is zero |
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Answer» Correct Answer - A,B,C,D `Z=sqrt(R^(2)+(omegaL-1/(omegaC))^(2))=sqrt((100)^(2)+(100-200)^(2))=100sqrt(2)` `I_(rms)=V_(rms)/Z` `P_(R)=I_(rms)^(2)R rArr P_(L)=0 rArr P_(C)=0` |
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| 818. |
In a series `RC` circuit with an `AC` source (peak voltage `E_(0)=50 V` and `f=50//pi Hz),R=300 Omega`. C=25 muF`.Then:A. the peak current is 0.1 AB. the peak current is 0.7 AC. the average power dissipated is 1.5 WD. the average power dissipated is 3 W |
| Answer» Correct Answer - A::C | |
| 819. |
In a series `RC` circuit with an AC source, `R = 300 Omega, C = 25 muF, epsilon_0= 50 V` and `v = 50/(pi) Hz`. Find the peak current and the average power dissipated in the circuit.A. the `rms` current in the circuit is `0.1 A`B. the `rms` potential difference across the capacitor is `50 V`C. the `rms` potential difference across the capacitor is `14.1 V`D. the `rms` current in the circuit is `0.14 A` |
| Answer» Correct Answer - A,B | |
| 820. |
In a series `RC` circuit with an `AC` source (peak voltage `E_(0)=50 V` and `f=50//pi Hz),R=300 Omega`.`C=25 muF`.Then:A. the peak current is `0.1 A`B. the peak current is `0.7 A`C. the average power dissipated is `1.5 W`D. the average power dissipated is `3 W` |
| Answer» Correct Answer - A,C | |
| 821. |
If a capacitor of `8 mu F` is connected to a 220 V, 100 Hz ac source and the current passing through it is 65 mA, then the rms voltage across it isA. `129.4 V`B. `12.94 V`C. `1.294 V`D. 15 V |
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Answer» Correct Answer - B Here, `V_("rms")=220 V,I_("rms")=65 mA = 0.065 A` `C = 8 mu F = 8xx10^(-6)F, upsilon = 100 Hz` Capacitive reactance, `X_(C )=(1)/(2pi upsilon C)=(1)/(2xx3.14xx100xx8xx10^(-6))=199 Omega` Then rms voltage across the capacitor is `V_("Crms")=I_("rms")X_(C )=0.065xx199=12.94 V` |
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| 822. |
An ac generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. Average energy stored by the inductor `L_(2)` (source is at resonance frequency) is equal toA. zeroB. 1.2mJC. 2.4mJD. 4mJ |
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Answer» Correct Answer - B |
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| 823. |
A uniform conducting ring of mass `pi` kg and radius 1 m is kept on smooth horizontal table. A uniform but time varying magnetic field `B = (hat (i) + t^(2) hat (j))T` is present in the region, where t is time in seconds. Resistance of ring is `2 (Omega)`. Then Time (in second) at which ring start toppling isA. `(10)/(pi)s`B. `(20)/(pi)s`C. `(5)/(pi)s`D. `(25)/(pi)s` |
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Answer» Correct Answer - A |
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| 824. |
A choke coil and capacitor are connected in series and the current through the combination is maximum for AC of frequency n. If they are connected in parallel, at what frequency is the current through the combination minimum?A. nB. n/2C. 2nD. None of these |
| Answer» Correct Answer - A | |
| 825. |
If an emf=E=`E_0cosomegat` is applied to a circuit, the current becomes `I=I_0cosomegat`. What is the power factor of the circuit?A. zeroB. `1/sqrt2`C. 1D. `infty` |
| Answer» Correct Answer - C | |
| 826. |
The inductance and capacitance in a closed circuit are 20 mH and `2muF` respectively. The natural frequency will beA. 796 HzB. 5000 HzC. 40 HzD. 31400 Hz |
| Answer» Correct Answer - A | |
| 827. |
Suppose the frequency of the source in the above example can be varied (a) What is the frequency of the source at which resonance occurs ? (b) Calculate the impedeance, the current and power dissipated at the resonant condition. |
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Answer» (a) The frequency at which the resonance occurs is `omega_(0)=(1)/sqrt(LC)=(1)/(sqrt(25.48xx10^(-3)xx796xx10^(-6)))` =222.1rad/s `v_(r)=(omega_(0))/(2pi)=(221.1)/(2xx3.14)Hz=35.4Hz` (b) The impedance Z at resonant condition is equal to the resistance: Z=R=`3Omega` The rms current at resonance is `=(V)/(Z)=(V)/(R)=((283)/(sqrt(2)))(1)/(3)=66.7A` The power dissipated at resonance is `P=I^(2)xxR=(66.7)^(2)xx3=13.35kW` You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 7.8. |
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| 828. |
At an airport, a person is made to walk through the door way of a metal detector, for security reasons. If `she//he` is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work ? |
| Answer» The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm. | |
| 829. |
The r.m.s. voltage of the wave form shown is A. 10VB. 7VC. 6.37VD. none of these |
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Answer» Correct Answer - A |
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| 830. |
A rectangular frame ABCD, made of a uniform metal wire, has a straight connection between E and F made of the samae wire, as shown in fig. AEFD is a square of side 1m, and EB=FC=0.5m. The entire circuit is placed in steadily increasing, uniform magnetic field directed into the plane of the paper and normal to it. The rate of change of the magnetic field is `1T//s`. The resistance per unit length of the wire is `1omega//m`. Find the magnitude and directions of the currents in the segments AE, BE and EF. A. 2B. 4C. 43468D. 43471 |
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Answer» Correct Answer - D |
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| 831. |
In the circuit shown. Initially the capacitor is uncharged. The switch S is closed at time t=0. Then A. `(V_(a)-V_(b))` is increasing with time0B. `(V_(a)-V_(b))` is deceasing with timeC.D. |
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Answer» Correct Answer - D |
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| 832. |
`2.5/(pi) muF` capacitor and `3000-`ohm resistance are joined in series to an `AC` source of `200 "volts"` and `50sec^(-1)` frequency. The power factor of the circuit and the power dissipated in it will respectivelyA. `0.6, 0.06 W`B. `0.06, 0.6 W`C. `0.6, 4.8 W`D. `4.8, 0.6 W` |
| Answer» Correct Answer - c | |
| 833. |
In the circuit below, the `AC` source the voltage `V=20cos (omegat)` volts with `omega=2000rad//sec`. The amplitude of the current will be nearest to A. 2AB. 3.3AC. `2//sqrt(5)A`D. `sqrt(5)A` |
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Answer» Correct Answer - A `X_(L)=L omega=10 ` `X_(C)=1/(C omega)=10 ` `X_(L)-X_(C)=0` So, Z=R=6+4 =10 `I_(0)=V_(0)//Z =2A` |
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| 834. |
In an `AC` circuit, `V` and `I` are given by `V=100sin(100t)volts, I=100sin(100t+(pi)/3)mA`. The power dissipated in circuit isA. `10^(4)` wattB. 10 wattC. 2.5 wattD. 5 watt |
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Answer» Correct Answer - C `P_(av)=V_(rms) I_(rms) cos phi` `=100/(sqrt(2))xx100/(sqrt(2)) xx10^(-3) cos ((pi)/3) =2.5` watt. |
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| 835. |
An `LC` circuit contains a `20 mH` inductor asn a `50 mu F` capacitor with initial change of `10 mC`. The resistance of the circuit is negligible. Let the instant the circuit is closed be `t = 0`. A. Energy stored in the circuit in completely electrical at `t = (n pi)/(2000)`B. Energy stored in the circuitin completely magnetic at `t = ((2n + 1)pi)/(2000)`C. Energy stored in the circuit in shared equally between the inductor and capacitor at `t = ((2n + 1)pi)/(4000)`D. Energy stored in the circuit is shared euqally between the inductor and capacitor at `t = (n pi)/(2000)` |
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Answer» Correct Answer - 1,2,3 Instantaneous electrical energy `U_(E) = (q_(0)^(2) cos^(2) omega t)/(2C)` At `omega t = 0, pi, 2 pi, 3 pi`….. The energy is completely electrical. `t = (2 pi)/(2 pi f) = (n)/(2 f) = (n pi)/(1000)` sec , `n = 1, 2, 3, 4` or `t = 0, T//2, 3T//2`.... Instantaneous magnetic energy `U_(B) = (1)/(2) Lq_(0)^(2) omega^(2) sin^(2) omega t` or `U_(B) = (q_(0)^(2))/(2C) sin^(2) omega t` So, at `omega t = pi//2, 3 pi//2, 5pi //2` The energy is completely magnetic `t = ((2n + 1) pi)/(2(2pi f)) = ((2n + 1))/(4f) = ((2n + 1))/(2000) sec` Where `n = 0,1,2,3,4`...... or `t = T//4, 3T//4, 5T//4`....... Timings for energy shared equally between inductor and capacitor. `U_(B) = U_(E)` `(q_(0)^(2))/(2C ) sin^(2) omega t = (q_(0)^(2))/(2C) cos^(2) omega t` `tan^(2) omega t = 1` or `tan omega t = tan pi//4` `t = (pi)/(4 omega), (3 pi)/(4 omega),(5 pi)/(4 omega)`,............or `t = (T)/(8), (3T)/(8), (5T)/(8)`............ |
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| 836. |
A capacitor, a resistor and a 40 mH inductor are connected in series to an a.c. source of frequency 60 Hz. Calculate the capacitance of the capacitor,if current is in phase with the voltage. |
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Answer» If the current is in phase with the voltage, `V_(L)=V_(C)` `rArr X_(L)=X_(C)` `rArr f_(0) = (1)/(2pi sqrt(LC))=f` ` 4 pi^(2) f^(2)LC = 1` `rArr C = (1)/( 4 pi^(2) f^(2) xx L)=(1)/(4xx(3.14)^(2)xx60xx60xx40x10^(-3))=175.7 mu F` |
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| 837. |
An LC circuit contains a 20 mH inductor and a `50 mu F` capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant at which the circuit which is closed be t=0. At what time the energy stored is completely magnetic ?A. 3AB. t=0C. `t=1.54 ms`D. `t=3.14ms` |
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Answer» For L-C , oscillator , T = `2pi sqrt LC` From one extreme to another when electric field is maximum magnetic field is zero. When electric field is zero, magnetic field is maximum. At time` t = T/4`, energy stored is completely magnetic `therefore` Time, `t = T / 4 = 2pi sqrt LC/4` `implies t = pi sqrt( 20 xx 10^(-3) xx 50 xx 10^(-6)) / 2 = pisqrt (1 xx 10^(-6))/ 2` `implies` t = 1.57 ms |
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| 838. |
An inductor `(L = 200 mH)` is connected to an `AC` source of peak emf `210 V` and frequency `50 Hz`. Calculate the peak current. What is the instantaneous voltage of the source when the current is at its peak value? |
| Answer» Because phase difference between voltage and current is `pi//2` for pure inductor. So, | |
| 839. |
An alternating voltage (in volts) given by `V=200sqrt(2)sin(100t)` is connected to `1 mu F` capacitor through an ideal ac ammeter in series. The reading of the ammeter and the average power consumed in the circuit shall beA. 20 mA, 0B. 20 mA, 4WC. `20sqrt(2)mA, 8W`D. `20sqrt(2)mA, 4sqrt(2)W` |
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Answer» Correct Answer - A On comparing `V = 200sqrt(2)sin(100 t)` with `V = V_(0)sin omega t`, we get `V_(0)=200sqrt(2) V, omega = 100" rad " s^(-1)` `therefore V_("rms")=(V_(0))/(sqrt(2))=(200sqrt(2)V)/(sqrt(2))=200 V` The capacitive reactance is `X_(C )=(1)/(omega C)=(1)/(100xx1xx10^(-6))=10^(4)Omega` ac ammeter reads teh rms value of current. Therefore, the reading of the ammeter is `I_("rms")=(V_("rms"))/(X_(C ))=(200 V)/(10^(4)Omega)=20xx10^(-3)A=20 mA` The average power consumedin the circuit, `P = I_("rms")V_("rms")cos phi` In an pure capacitive circuit, the phase difference between current and voltage is `(pi)/(2)`. `therefore cos phi = 0 therefore P = 0` |
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| 840. |
An alternating voltage given by `V=300sqrt2sin (50t)` (in volts) is connected across a `1muF` capacitor through an AC ammeter. The reading of the computer will beA. 10mAB. 40mAC. 100mAD. 15mA |
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Answer» Correct Answer - D |
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| 841. |
In figure below if `Z_(1) = Z_(C )` and reading of ammeter is `1A`. Find value of source voltage `V`. A. 80 voltB. 60 voltC. 100 voltD. None |
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Answer» Correct Answer - C `iZ_(L) - iZ_(C ) = 0` So, `V_(z_(L)) = V_(z_(c))` `:. V_(L) = 1 xx 2 pi xx 30 xx (1)/(pi) = 60` volt `:. V_(R ) = 80 xx 1 = 80` volt `V = sqrt(V_(L)^(2) + V_(R )^(2)) = sqrt((80)^(2) + (60)^(2)) = 100` volt |
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| 842. |
State wheather the following two statement are true or falseA. TFB. FFC. TRD. FT |
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Answer» Correct Answer - A |
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| 843. |
In a series `C-R` circuit shown in figureure, the applied voltage is `10 V` and the voltage across capacitor is found to`8 V`. The voltage across `R`, and the phase difference between current and the applied voltage will respectively be .A. `6V, tan^-1(4/3)`B. `3V,tan^-1(3/4)`C. `6V,tan^-1(3/4)`D. none of these |
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Answer» Correct Answer - A `V_R=sqrt(V^2-V_C^2)=sqrt((10)^2-(8)^2)=6V` `tanphi=X_C/X_R=V_C/V_R=8/6=4/3` |
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| 844. |
In the circuit shown in Figure, the source has a rating of 15 V, 100 Hz. The resistance R is 3 `Omega` and the reactance of the capacitor is 4 `Omega`. It is known that the box certainly contains one or more element (resistance, capacitance or inductance). Which element/s are present inside the box? |
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Answer» Correct Answer - The box has L and C in series |
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| 845. |
Find the rms value of current `i=I_(m)sin omegat` from `(i) t=0 "to" t=(pi)/(omega)` (ii) `t=(pi)/(2omega) "to" t=(3pi)/(2omega)` |
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Answer» `i_(rms) =sqrt((underset((pi)/(2omega))overset((pi)/(omega))int I_(m)^(2) sin omegatdt)/((2pi)/(omega)))= sqrt((i_(m)^(2))/(2))=(I_(m))/(sqrt2)` `(ii) i=sqrt((underset((pi)/(2omega))overset((3pi)/(2omega))int I_(m)^(2) sin omegatdt)/((pi)/(omega)))= sqrt((I_(m)^(2))/(2))=(I_(m))/(sqrt2)` |
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| 846. |
A FM radio receiver has a series LCR circuit with L = 1 `muH`, and R = 100 `Omega`. The antenna receives radio waves and induces a sinusoidally alternating emf of ampli- tude 10 `muV`. The induced voltage is fed to the series LCR circuit. The capacitance in the circuit is adjusted to a value of C = 2 pF. (a) Find the frequency of radio wave to which the radio will tune to. (b) Find the rms current in the circuit. (c) Find quality factor of the resonance. |
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Answer» Correct Answer - (a) 112.9 MHz (b)70nA (c )7.07 |
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| 847. |
The coefficient of induction of a choke coil is `0.1 H` and resistance is `12 Omega`. If it is connected to an alternating current source of frequency `60 Hz`, then power factor will beA. `0.32`B. `0.30`C. `0.28`D. `0.24` |
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Answer» Correct Answer - B `cos varphi=R/Z=R/(sqrt(R^(2)+omega^(2)L^(2)))` `=12/(sqrt((12)^(2)+4xxpi^(2)xx(60)^(2)xx(0.1)^(2)))implies cos varphi=0.30` |
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| 848. |
Assertion (A) : If current varies sinusoidally the average power consumed in a cycle is zero. Reason (R ) : If current sinusoidally the average power consumed is zeroA. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. Assertion is true but Reason is falseD. Assertion is false but Reason is true |
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Answer» Correct Answer - 4 `lt P gt = (1)/(2) i_(m)^(2) R` |
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| 849. |
There is a `5 Omega` resistance in an `AC`, circuit. Impedence of `0.1 H` is connected with it in series. If equation of `AC` emf is `5 sin 50 t` then the phase difference between current and e.m.f. isA. `(pi)/2`B. `(pi)/6`C. `(pi)/4`D. `0` |
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Answer» Correct Answer - C `cos varphi=R/Z=R/(sqrt(R^(2)+omega^(2)L^(2)))=5/(sqrt(25+(50)^(2)xx(0.1)^(2)))` `=5/(sqrt(25+25))=1/(sqrt(2)) implies varphi=pi//4` |
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Figure shows an iron-cored transformer assumed to be 100% efficient. The ratio of the secondaty turns to the primary turns is 1:20. A 240 V ac supply is connected to the primary coil and a `6 Omega` resistor is connected to the secondary coil. What is the current in the primary coil?A. 0.10AB. 0.14AC. 2AD. 40A |
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Answer» Correct Answer - A the equivlent primary load is `(R_1)=((N_1)/(N_2))^(2) (R_2)=((20)/(1)^(2)) (6.0)=2400 Omega` Current in the primary coil `=(240)/(R_1) =(240)/(2400)=0.1 A` |
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