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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.
| 1001. |
Assertion (A) : An ac emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero. Reason (R ) : In any circuit element, current is always in the phase with voltageA. 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 In inductor current lags the voltage by `(pi)/(2)` In capacitor current leads the voltage by `(pi)/(2)` |
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| 1002. |
A radio can be turned over a frequency range from `500 kHz to 1.5 MHz`. If its L-C circuit has an effective inductance of `400 (mu)H`, what is the range of its variable capacitor. |
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Answer» We know that resonant frequency of a tuning circuit is given by `f_(0)=(1)/(2 pi sqrt(LC))`. From here we can find C, which is given as `C=(1)/(4 pi^(2)f_(0)^(2)L)` We are given the range of frequency and we have to find the range of variable capacitor. By putting the extreme value of frequency, we can find the range of capacitor. By Putting `F_(0)=500KHz`, we get `(C_1)=(1)/(4 pi^(2)(500xx10^(3))^(2)xx400xx10^(-6))=253pF` `C_(2)=(1)/(4 pi^(2)(1.5xx10^(6))^(2)xx400xx10^(-6))=28 pF` so the range of variable capacitor comes as , 28 PF-253 pF`. |
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| 1003. |
A fully charged capacitor C with initial charge `underset(o)(q)` is connected to a coil of self-inductance L at t = 0 . The time at which the energy is stored equally between the electric and the magnetic fields isA. `pi``sqrt LC`B. `pi`/4`sqrt LC`C. `2pi``sqrt LC`D. `sqrt LC` |
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Answer» Correct Answer - B In LC oscillation energy is transferred C to I, I to C, maximum energy in L is = `1/2 LI_max^(2)` . Maximum energy in C is `q_max^(2)`/2C Equal energy will be when 1/2L`t^(2)` = 1/2.1/2L`t_max^(2)` I = 1/`sqrt2` `I_max` I = `I_max`sin`omega`t = `1/sqrt 2``I_max` `omega`t = `pi`/4 or 2 `pi`/T t = `pi`/4 or t = T/8 `implies` t = 1/8 `2pi``sqrt LC` = `pi`/4 `sqrt LC` |
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| 1004. |
In a circuit inductance `L` and capacitance `C` are connected as shown in figure and `A_(1)` and `A_(2)` are ammeters. When key `k` is pressed to complete the circuit, then just after closing key `k`, the readig of `A_(1)` and `A_(2)` will be: A. zero in both `A_(1) and A_(2)`B. maximum in both `A_(1)and A_(2)`C. zero in `A_(1) "and maximum in" A_(2)`D. maximum in both `A_(1)` and zero in `A_(2)` |
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Answer» Correct Answer - D |
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| 1005. |
During current growth in L-R circuit, match the following table, |
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Answer» Correct Answer - (A) Q, (B) R, (C) R, (D) R |
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| 1006. |
Match the following `{:(,,"Table-1",,"Table-2"),(,(A),L,(P),[M^(0)L^(0)T^(-2)]),(,(B),"Magnetic Flux",(Q),[ML^(2)T^(-2)A^(-1)]),(,(C),LC,(R),[ML^(2)T^(-2)A^(-2)]),(,(D),CR^(2),(S),"None"):}` |
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Answer» Correct Answer - (A) R, (B) Q, (C) S, (D) R |
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| 1007. |
Power delivered by an ac source of angular frequency `Omega_(0)` to an `LCR` series circuit is maximum when .A. `omegaL =omegaC`B. `omegaL = (1)/(omegaC)`C. `omegaL =R -(1)/(omegaC)`D. `omegaC = R - (1)/(omegaL)` |
| Answer» Correct Answer - B | |
| 1008. |
A conducting bar is slid at a constant velocity v along two conducting rods. The rods ar seprated by a distance l and connected across a resistor R. The entire apparatus is placed in an external magnetic field B directed into the page An increase in which of the following would NOT increase the current generated by the apparatus?A. vB. lC. RD. B |
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Answer» Correct Answer - C |
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| 1009. |
In the figure shown a uniform conducing rod of mass m and length l is suspended invertical plane by two conducing springs of spring constant k each. Upper end of springs are connected to each other by a capacitor of capacitance C. A uniform horizontal magnetic field `(B_(0))` perpendicular to plane of springs in space initially rod is in equillibrium. If the rod is pulled down and released, it performs SHM. (Assume resistance of springs and rod are negligible) Find the period of oscillation of rod.A. `2pisqrt((m)/(K))`B. `2pisqrt((B_(0)^(2)l^(2)C)/(K))`C. `2pisqrt((m+B_(0)^(2)l^(2)C)/(K))`D. None of these |
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Answer» Correct Answer - D |
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| 1010. |
In the circuit diagram shown in Figure E=18V , L=2H, `R_(1)=3Omega , R_(2)=6Omega`. Switch S is closed at t=0 Match the following: |
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Answer» Correct Answer - (A) R, (B) P, (C) Q, (D) Q |
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| 1011. |
Power delivered by an ac source of angular frequency `omega_(0)` to an `LCR` series circuit is maximum when .A. `omegaL=omegaC`B. `omegaL=1/(omega C)`C. `omegaL= -(1/(omegaC))^(2)`D. `omegaL=sqrt(omega C)` |
| Answer» Correct Answer - b | |
| 1012. |
The current and voltage functions in an `AC` circuit are `i=100 sin 100 tmA, V=100sin(100t+pi/2)V` The power disspitated in the circuit isA. `10 W`B. `2.5W`C. `5W`D. `5kW` |
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Answer» Correct Answer - B `P=V_(rms)I_(rms)cosphi` `=(100/sqrt2)((100xx10^-3)/sqrt2).cos(pi/3)` `=2.5W` |
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| 1013. |
In L-C-R series circuit suppose `omega_(r)` is resonance frequency, then match the following table, `{:(,,"Table-1",,"Table-2"),(,(A),"If " omegagtomega_(r),(P),"Current will lead the voltage"),(,(B),"If " omega=omega_(r),(Q),"Voltage will lead the current"),(,(C),"If " omega=2omega_(r),(R),X_(L)=2X_(C)),(,(D),"If " omegaltomega_(r),(S),"Current and voltage are in phase"),(,,,(T),"None"):}` |
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Answer» Correct Answer - (A) Q, (B) S, (C) Q, (D) P |
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| 1014. |
A `100 Omega` resistance is connected in series with a `4 H` inductor. The voltage across the resistor is, `V_(R ) = (2.0 V) sin (10^(3) t)`. Find amplitude of the voltage across the inductor.A. `40 V`B. `60 V`C. `80 V`D. `90 V` |
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Answer» Correct Answer - C Amplitude of voltage across inductor, `V_(0) = I_(00 X_(L)` `= (2.0 xx 10^(-2) A) (4.0 xx 10^(3)` ohm) = 80 volts |
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| 1015. |
Instantaneous voltage and instantaneoss current in an L-R circuit in AC is V=100sin (100)t and `i=10sin(100t-pi//4)`. Match the following table, `{:(,(A),R,(P),(1)/(10sqrt2)"SI units"),(,(B),X_(L),(Q),5sqrt2 "SI unit"),(,(C),L,(R),10sqrt2 "SI units"),(,(D),"Average power in one cycle",(S),"None"):}` |
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Answer» Correct Answer - (A) Q, (B) Q, (C) P, (D) S |
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| 1016. |
Magnetic flux in a circular coil of resistance `10Omega` changes with time as shown in figure. `otimes` direction indicates a direction perpendicular to paper inwards. Match the following table. |
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Answer» Correct Answer - (A) Q,S, (B) R, (C) P,S, (D) Q,S |
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| 1017. |
In an `L-R-C` circuit the current is given by `i = I cos omega t`. The voltage amplitudes for the resistor. Inductor and capacitor are `V_(R ), V_(L)` and `V_(C )` respectively. (a) The instantaneous power into the resistor is `P_(R ) = V_(R ) I cos^(2) omega t`. (b) The instantaneous into the inductor is `P_(L) = - V_(L) I sin omega t cos omega t` (c ) The instantaneous power into the capacitor is `P_(c ) = V_(c ) I sin omega t cos omega t`. (d) `p_(R ) + p_(L) + P_(c )` equals total power `p` supplied by the source at each instant of time.A. (a),(c ),(d) are correctB. (b),(c ) are correctC. (a) is correctD. (a),(b),(c ),(d) are correct |
| Answer» Correct Answer - D | |
| 1018. |
A conducting bar is slid at a constant velocity v along two conducting rods. The rods ar seprated by a distance l and connected across a resistor R. The entire apparatus is placed in an external magnetic field B directed into the page The induced current in the above circuit is:A. sinusoidalB. clockwiseC. counterclockwiseD. there is not enough information to determine the direction and nature of the current |
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Answer» Correct Answer - C |
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| 1019. |
In the figure `V_(ab)` versus time graph along an inductor is shown. Match the following |
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Answer» Correct Answer - (A) Q, (B) R, (C) Q |
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| 1020. |
A conducting bar is slid at a constant velocity v along two conducting rods. The rods ar seprated by a distance l and connected across a resistor R. The entire apparatus is placed in an external magnetic field B directed into the page Which of the following represents the current i generated by the apparantus?B. C. D. |
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Answer» Correct Answer - A |
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| 1021. |
The potentiak difference across a 2H inductor as a function of time is shown in the figure. At time t=0, current is zero. Current versus time graph across the inductor will beA. B. C. D. |
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Answer» Correct Answer - B |
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| 1022. |
The potential difference across a `2 H` inductor as a function of time is shown in figure. At time `t=0`, current is zero Current `t = 2` second is A. ` 1A`B. `3 A`C. `4 A`D. `5 A` |
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Answer» Correct Answer - 4 `|e| = L (di)/(dt) implies |e| dt = L (i_(1) - i_(2))` `|e| dt =` area of `Delta` le for `t = 0` to 2 sec. |
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| 1023. |
An `LCR` circuit has `L = 10 mH`. `R = 3` ohm and `C = 1 mu F` connected in series to a source is `15 cos omega t` volt. What is average power dissipated per cycle at a frequency that is `10%` lower than the resonant requency? |
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Answer» Here, `L = 10^(-2) H, R = 3 Omega, C = 10^(-6) F` Resonant frequency, `omega_(0) = (1)/(sqrt(LC)) = (1)/(sqrt(10^(-2) xx 10^(-6))) = 10^(4) "rad"//s` Actual frequency, `omega = (90%) omega_(0)` `= 9 xx 10^(3) "rad"//s` `X_(L) = omega L = 9 xx 10^(3) xx 10^(-2) = 90 Omega` `X_(C ) = (1)/(omega C) = (1)/(9 xx 10^(3) xx 10^(-6)) = (1000)/(9) Omega` `Z = sqrt(R^(2) + (X_(C ) - X_(L))^(2))` `= sqrt(3^(2) + ((100)/(9) - 90)^(2)) = 21.3 Omega` Power dissipated`//` cycle `=E_(v)I_(v) cos phi` `=E_(0) (E_(v)/Z) R/Z=(E_(v)/Z)^(2)xxR` `=(15/(sqrt(2)xx21.3))^(3)xx3=0.744 W` |
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| 1024. |
A circuit draws a power of 550 watt from a source of 220 volt, `50 Hz`. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, The capacitance should be connected in series with it isA. `75 mu F`B. `60 mu F`C. `50 mu F`D. `65 mu F` |
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Answer» Correct Answer - A As the current lags behind ghe potentail difference, the circuit contains resistance and inducntace. Power, `P = V_("rms") xx i_("rms") xx cos phi` , Here, `i_("rms") = (V_("rms"))/(Z)`, where `Z = sqrt([(R^(2) + (omega L)^(2))])` `P = (V_("rms")^(2) xx cos phi)/(Z)` or `Z = (V_("rms")^(2) xx cos phi)/(P)` So, `Z = ((220)^(2) xx 0.8)/(550) = 70.8` ohm Now, power factor `cos phi = (R )/(Z)` or `R = Z cos phi` `:. R = 70.4 xx 0.8 = 56.32` ohm Further, `Z^(2) = R^(2) + (omega L)^(2)` or `(omega L) = sqrt((Z^(2) - R^(2)))` or `omega L = sqrt((70.4)^(2) - (56.32)^(2)) = 42.2` ohm When the capacitor is connected in the circuit, `Z = sqrt([R^(2) + (omega L = (1)/(omega C))^(2)])` and `cos phi = (R )/(sqrt([R^(2) + (omega L - (1)/(omega C))^(2)]))` When `cos phi = 1, omega L = (1)/(omega C)` `:. C = (1)/(omega (omega L)) = (1)/(2 pi f (omega L))` `= (1)/((2 xx 3.14 xx 50) xx (42.2)) = 75 xx 10^(-6) F = 75 mu F` |
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| 1025. |
A series `R-L-C` circuit has `R = 100` ohm. `L = 0.2 mH` and `C = (1)/(2) mu F`. The applied voltage `V = 20 sin omega t`. Then When the currentsd lags the applied voltage by `45^(@)`, the value of `omega` is approximatelyA. `5xx 10^(5) "rad"//s`B. `3 xx 10^(5) "rad"//s`C. `4 xx 10^(5) "rad"//s`D. `4 xx 10^(10) "rad"//s` |
| Answer» Correct Answer - A | |
| 1026. |
A series `R-L-C` circuit has `R = 100 Ohm. L = 0.2 mH` and `C = (1)/(2) mu F`. The applied voltage `V = 20 sin omega t`. When the current lags the applied voltage by `45^(@)`, the equation of the current isA. `0.2 sin (omega t + tan^(-1) 0.3)`B. `0.2 sin (omega t - tan^(-1) 0.3)`C. `0.3 sin (omega t + tan^(-1) 0.3)`D. `0.3 sin (omega t - tan^(-1) 0.3)` |
| Answer» Correct Answer - B | |
| 1027. |
A series circuit has an impedance of `60.0 Omega` and a power factor of `0.720` at `50.0 Hz`. The source voltage lags the current.(a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor?(b) What size element will raise the power factor to unity? |
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Answer» Correct Answer - A::B::C::D (a) Voltage lags `:. X_CgtX_L` Power factor, `cosphi=R/Z` `=R/(sqrt(R^2(X_C-X_L)^2)` To increase the power factor denominator should decrease. Hence `X_L` should increase. Therefore, an inductor is required to be connected. (b) `cosphi=R/Z=0.72` `:. R=0.72Z=0.72xx60` `=43.2Omega` `(X_C-X_L)=sqrt((60)^2-(43.2)^(2))` `=41.64Omega` New inductor of inductane `41.64Omega` should be added in the circuit. `L=X_L/(2pif)` `=41.64/(2pi(50))=0.133H`. |
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| 1028. |
Assertion : In series `L-C-R` circuit, voltage will lead the current function for frequency greater than the resonance frequency. Reason : At resonance frequency, phase difference between current function and voltage function is zero.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B At resonance frequency `f_r`, `X_C=X_L` Now `X_L=2pifL` or `X_Lpropf` For `fgtf_r,X_LgtX_C` At resonance `X_L=X_Cimplies Z=R` `:. Cosphi=R/Z=1` or `phi=0^(@)` |
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| 1029. |
An `LCR` circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference ofA. zeroB. `pi` / 4C. `pi` / 2D. `pi` |
| Answer» Correct Answer - A | |
| 1030. |
At resonance, `V_(L)` and `V_(C )` both very much greater than the applied potential, `V` itself. The quantity factor for an `LCR` circuit in resonance is given by `Q = (X_(L))/(R )`. In pratice, `Q = 200` has been achieved. At resonance, the capacitor has been adjusted for (1). `200 xx 10^(-6) mu F` (2) `0.00013 mu F` (3). `0.0012 mu F` (4). `0.0013 F` At resonance, the potential difference across the inductance is (1) `1.3 V` (2) `13 V` (3). `0.3 V` (d) none of these The potential across the capacitor at resosnance is (1) `1.3 V` (2) `13 V` (3) `lt 13 V` (4) none of these The `Q` factor is (1) `(V_(L))/(V_(C ))` (2) `(V_(C ))/(V_(L))` (3) `(V_(C ))/(V)` (d) `(V_(L))/(V)` (e) choose the right statement. (1) `V_(L)+V_(C)` can be greater than `V_("applied")` (2) `V_(L)+V_(C)=V_("applied")` (3) `V_(L)+V_(C) lt V_("applied")` (4) none of these |
| Answer» Correct Answer - 2, 3, 1, (3,4) | |
| 1031. |
An `LCR` circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference ofA. `pi//4`B. zeroC. `pi`D. `pi//2` |
| Answer» Correct Answer - 2 | |
| 1032. |
An ac generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. If the ac source `G` is `100 V` rating at resonant frequency of the circuit, then average power supplied by the source isA. `0 J`B. `1 m J`C. `100 mJ`D. not possible to calcualte from the given information |
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Answer» Correct Answer - D As 1ms time duration is very less than time period `T` at resonance, thermal energy produced is not possible to calculate without information about start of the given time duration. |
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| 1033. |
An alternating voltage of 200 volt, at 400 cycyles`//`sec in applied in a circuit containing an inductance of 0.01 henry in series with a resistanceof 22.8 ohms. The voltage across the inductance isA. 148.2 voltB. 392.4 voltC. 74.1 voltD. 196.2 volt |
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Answer» Correct Answer - A `I = (V_("rms"))/(sqrt(R^(2) + (L omega)^(2))) = 5.9 amp` `V_(L) = I omega L = 148.2` volt. |
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| 1034. |
In an electrical circuit `R,L,C` and an `AC` voltage source are all connected in series. When `L` is removed from the circuit, the phase difference between the voltage and the current in the circuit is `pi//3`. If instead, `C` is removed from the circuit, difference the phase difference is again `pi//3`. The power factor of the circuit isA. `(1)/(2)`B. `(1)/(sqrt(2))`C. 1D. `(sqrt(3))/(2)` |
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Answer» Correct Answer - C When L is removed from the circuit, it becomes RC circuit. `tan phi = "tan" (pi)/(3)=(X_(C ))/(R ) therefore X_(C )=R "tan" (pi)/(3)=sqrt(3)R` When C is removed from the circuit, itbecomes RL circuit. `therefore tan phi = "tan"(pi)/(3)="tan"(pi)/(3)=(X_(L))/(R ), X_(L)=R "tan" (pi)/(3)=sqrt(3)R` Impedance of the circuit, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))=R` Power factor, `cos phi=(R )/(Z)=(R )/(R )=1`. |
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| 1035. |
An inductor 200 mH, capacitor `500mu F` and resistor `10 Omega` are connected in series with a 100 V variable frequency ac source. What is the frequency at which the power factor of the circuit is unity?A. `10.22 Hz`B. `12.4 Hz`C. `19.2 Hz`D. `15.9 Hz` |
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Answer» Correct Answer - D Here `L = 200 mH = 200xx10^(-3)H=0.2H` `C=500 mu F=500xx10^(-6)=5xx10^(-4)F` `R=10 Omega` and `V_("rms")=100 V` Power factor `=cos phi =(R )/(Z)=1` (Given) `therefore Z=R rArr sqrt(R^(2)+(X_(L)-X_(C ))^(2))=R` `R^(2)+(X_(L)-X_(C ))^(2)=R^(2) rArr X_(L)-X_(C )=0` or `X_(L)=X_(C )` (This is resonance condition) `2pi upsilon L=(1)/(2pi upsilon C) rArr 4pi^(2)upsilon^(2)LC=1` `therefore upsilon=(1)/(2pi sqrt(LC))=(1)/(2xx3.14sqrt(0.2xx5xx10^(-4))` `=(1)/(2xx3.14xx10^(-2))=(100)/(6.28)=15.9 Hz` |
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| 1036. |
Assertion: Capacitance serves as a block for `DC` and offerse an easy path to `AC`. Reason: Capacitance reactance is inversely proportional to frequency.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - A The capacitive reactance of capacitor is given by `X_(C)=1/(omega C)=1/(2pi f C)` So this is infinite for `DC (f=0)` and has a very small value of `AC`, therefore a capacitor blocks `DC`. |
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| 1037. |
The voltage of an AC source varies with time according to the equation `V=100sin 100pitcos100pit` where `t` is in seconds and `V` is in volts. ThenA. The peak voltage of the source is `100` voltsB. The peak voltage of the source is `50 `voltsC. The peak voltage of the source is `100sqrt(2)` voltsD. The frequency of the source is `50Hz` |
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Answer» Correct Answer - B `V=50xx2 sin 100pit cos 100pit=50sin 200pit` `implies V_(0)=50` volts and `v=10Hz` |
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| 1038. |
The capacitance of a pure capacitance is `1` farad. In DC circuits, its effective resistance will beA. ZeroB. InfiniteC. `1 ohm`D. `1//2 ohm` |
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Answer» Correct Answer - B `X_(C)=1/(2pi vC)=1/0=oo` |
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| 1039. |
Phasor diagram of a series `AC` circuit is shown in figureure. Then, A. The circuit must be containing resistor and capacitor onlyB. The circuit must be containing resistor and inductor onlyC. The circuit must be containing all three elements L, C and RD. The circuit cannot have only capacitor and inductor |
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Answer» Correct Answer - D In case of only capacitor and inductor phase difference between current and voltage should be `90^@` |
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| 1040. |
A condenser of capacity 20`mu`F is first charged and then discharged through a 10mH inductance. Neglecting the resistance of the coil, the frequency of the resulting vibrations will beA. 364 cycles/sB. 35.6 cycles/sC. `365 xx 10^(3)` cycles/sD. 3.56 cycles/s |
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Answer» Correct Answer - A `omega` = 1/`sqrt LC` `therefore` f = `1/2pi sqrt LC` = 1/`2pi``sqrt 20 xx 10^(-6) xx 10 xx 10^(-3)` = 356 Hz |
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| 1041. |
Asserition: In alternating current direction of motion free electron changes periodically. Reason: Alternating current changes its direction after certain time interval.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - B Motion of electron is random with drift velocity opposite to the direction of current. |
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| 1042. |
An inductor, a capacitor, and a resistor are connected in series. The combination is connected across an ac source. Statement 1: Peak current through each remains same. Statement 2: Average power dielivered by source is equal to average power developed across resistance.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - B Average power consumed by capacitor or inductor is zero. |
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| 1043. |
An inductor, a capacitor, and a resistor are connected in series. The combination is connected across an ac source. Statement 1: Peak current through each remains same. Statement 2: Average power dielivered by source is equal to average power developed across resistance.A. Statement 1 is true, statement 2 is true, Statement 2 is the correct explanation for statement 1.B. Statemet 1 is True, Statement 2 is true , Statement 2 is NOT the correct explanation for Statement 1C. Statement 1 is True, Statement 2 is False.D. Statement 1 is False, Statement 2 is true |
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Answer» Correct Answer - B Average power consumed by capacitor or inductor is zero. |
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| 1044. |
An ideal inductor of `(1//pi)` is connected in series with a `300gamma` resistor If a `20V.200H_(z)` ac source is connected across the combination the phase difference between the voltage and the current is .A. `tan^(-5)/(4)`B. `ta n-^(1)4/5`C. `ta n-^(1)3/4`D. `ta n-^(1)4/3` |
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Answer» Correct Answer - D `f = 200 Hz , omega = 2pi f = 400 pi rad//sec` `X_(L) = omegaL = 400 pi xx (1)/(pi) = 400 Omega` `tan phi = X_(l)/R = (400)/(300) implies phi =tan^(-1) ((4)/(3))` . |
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| 1045. |
An inductor, a capacitor, and a resistor are connected in series. The combination is connected across an ac source. Statement 1: Peak current through each remains same. Statement 2: Average power dielivered by source is equal to average power developed across resistance.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 In series current is same. Indcutor and capacitor does not consume power |
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| 1046. |
Compare between inductive reactance and capacitive reactance. |
| Answer» Inductive reactance ,`X_L=omegaL` increases with the increase of either frequency or inductance or both. Capacitive reactance, `X_C=1/(omegaC)` decreases with the increase of either frequency or capacitance or both. | |
| 1047. |
In an LCR series combination, R=400`Omega`, L=100 mH and C=`1muF`. This combination is connected to a 25sin2000t volt voltage source. Find the impedance of the circuit and the peak value of the circuit current. |
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Answer» Ist part: Frequency of the source , `omega=200rad//s` `therefore Impedance`, `Z=sqrt(R^2+(omegaL-1/(omegaC))^2)` `=sqrt(400^2+(2000times100times10^-3-1/(200times10^-6))^2)` `=sqrt(16times10^4+9times10^4)=500Omega` Peak value of current , `I_0=V_0/R=25/500=50mA` |
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| 1048. |
The impedance of a series circuit consists of `3 ohm` resistance and `4 ohm` reactance.The power factor of the circuit isA. `0.4`B. `0.6`C. `0.8`D. `1.0` |
| Answer» Correct Answer - B | |
| 1049. |
In a series LCR circuit containing an AC voltage source of frequency `omega`, current and voltage are measured. Then for the resistance, consider the statements- [a] current is maximum at `omega^(2)=1//LC` [b] current is minimum at `omega^(2)=1//LC` [c] voltage across R is maximum at `omega^(2) =1//LC` [d] voltage across R is minimum at `omega^(2)=1//LC` the voltage statement isA. a and c are both correctB. b and d are both correctC. a and d are both correctD. none of these |
| Answer» Correct Answer - A | |
| 1050. |
A resistor and a capacitor are connected to an ac supply of 200 V, 50 Hz, in series. The current in the circuit is 2A. If the power consumed in the circuit is 100 W then the resistance in the circuit isA. `100 Omega`B. `25 Omega`C. `sqrt(125xx75) Omega`D. `400 Omega` |
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Answer» Correct Answer - B `I^(2)R=100` `R=100/I^(2)=100/(2)^(2)=25` |
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