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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.
| 51. |
A coil has an inductance of 0.7 H and is joined in series with a resistance of 220 ohm. Find the wattless component of current in the circuit, when an alternating e.m.f. of 220 V at a frequency of 50 Hz is supplied to it. |
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Answer» `tan phi = (X_(L))/(R ) = (omega L)/(R ) = (2 pi xx 50 xx 0.7)/(220) = 1` `:. phi = 45^(@), Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(220 + 220^(2))` `= 220 sqrt(2) Omega` Wattless component of current `= I_(v) sin phi` `= (E_(v))/(Z) sin 45^(@) = (220)/(220sqrt(2)) xx (1)/(sqrt(2)) = 0.5 A` |
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| 52. |
What is meant by wattless component of current ? |
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Answer» Average power `(P)=V_("rms")(I_("rms")sin phi)cos.(pi)/(2)` The average power consumed in the circuit due to `(I_("rms")sin phi)` component of current is zero. This component of current is known as wattless current. `(I_("rms")sin phi)` is the wattless component of current. |
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| 53. |
An alternating voltage given by V=140sin314t is connected across a pure resistor of `50Omega`. Find the rms current through the resistor. |
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Answer» `I=V/R=140/50sin314t=2.8sin314t=I_0sinomegat` `thereforeI_(rms)=I_0/sqrt2approx2.8/1.4=2A` |
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| 54. |
A resistor `R` and the capacitor `C` are connected in series across an ac source of rms voltage `5V` if the rms voltage across `C` is `3V` then that across `R` is .A. `1V`B. `2V`C. `3V`D. `4V` |
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Answer» Correct Answer - D `V^(2) = V_(R)^(2) + V_(C)^(2) implies (5)^(2) = V_(R)^(2) + (3)^(2)` `V_(R) = 4V` . |
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| 55. |
A resistor `R` and the capacitor `C` are connected in series across an ac source of rms voltage `5V` if the rms voltage across `C` is `3V` then that across `R` is .A. `1 V1`B. `2 V`C. `3 V`D. `4 V` |
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Answer» Correct Answer - 4 `E = sqrt(E_(R )^(2) + E_(C )^(2))` |
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| 56. |
A `500 Omega` resistor and a capacitor `C` are connected in series across `50 Hz AC` supply mains. The r.m.s potential difference recorded on voltmeter `V_(1)` and `V_(2)` are `V_(1)=120 V` and `V_(2)=160 V`. The power taken for the mains is: A. `480 W`B. `240 W`C. `28.8 W`D. `14.4 W` |
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Answer» Correct Answer - C `V_(r.m.s)sqrt(V_(1)^(2)+V_(2)^(2))=sqrt(120^(2)+160^(2))` `I_(V)xx500=120, I_(V)=24A` `P_(av)=I_(V)^(2)R` |
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| 57. |
A coil has an inductance of 0.7 H and is joined in series with a resistance of 220 ohm. Find the wattless component of current in the circuit, when an alternating e.m.f. of 220 V at a frequency of 50 Hz is supplied to it.A. `0.2 A`B. `0.4A`C. `0.5A`D. `0.7A` |
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Answer» Correct Answer - C `X_(L)=omega_(L)=2pifL=220 Omega` `phi=tan^(-1)((X_(L))/R)= tan^(-1).(220)/(200)=45^(@)` `I=E/(sqrt(R^(2)+X_(L)^(2)))=220/((220)^(2)+(220)^(2))` Wattless component of current `=I sin phi` `=1/(sqrt(2)). Sin 45^(@)=0.5A` |
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| 58. |
An alternating current of rms value 10 A is passed through a 12`Omega` resistor. The maximum potential difference across the resistor isA. 20 VB. 90 VC. 169.68 VD. None of these |
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Answer» Correct Answer - C Maximum potential difference , `underset(o)(V)` = `underset(o)(I)`R = (`underset(rms)(I)`sqrt 2`)R = (`10sqrt 2`) (12) = 169.68 V |
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| 59. |
Two resistors are connected in series across `5V` rms source of alternating potentail. The potential difference across `6 Omega` resistor is `3 V_(m)`. If `R` is replaced by a pure inductor `L` of such magnitude that current remains same, then the potential difference across `L` is |
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Answer» Correct Answer - 4 Initially `i=(2)/(R )=(3)/(6) therefore R=4Omega , i=(1)/(2)A` Whebn `R` is replaced by `L,i` does not change `therefore (1)/(2)=(5)/(sqrt(4^(2)+X_(L)^(2)))," " therefore X_(L)=8Omega` `therefore V_(L)=iX_(L)=(1)/(2)xx8=4V` |
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| 60. |
Two resistor are connected in series across a `5 V` rms source of alternating potential. The potential difference across `6 Omega` resistor is `3V`. If `R` is replaced by a pure inductor `L` of such magnitude that current remains same. Then the potential difference across `L` is A. `1 V`B. `2V`C. `3V`D. `4V` |
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Answer» Correct Answer - D `V_(rms)(6 Omega)=3V=6I_(V) :. I_(V)=0.5A` `I_(V)=1/2=5/(sqrt(6^(2)+X_(L)^(2))), X_(L)=8 Omega` Now, `V_(L)=I_(V), X_(L)=1/2xx8=4V` |
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| 61. |
Two resistor are connected in series across a 5 V rms source of alternating potential. The potential difference across `6 Omega` resistor is 3V. If R is replaced by a pure inductor L of such magnitude that current reamins same. Then the pontential difference across L is A. (A) 1 VB. (B) 2 VC. (C) 3 VD. (D) 4 V |
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Answer» Correct Answer - D `V_(6 Omega)=3 =6(I_V) :. (I_V) = 0.5A` `(I_V)=1/2 = (5)(sqrt(6^(2)+X_(L)^(2))), (X_L)=8 Omega` ltbRgt Now, `(V_L) = (I_V)*(X_L) =1/2 xx 8 = 4 V`. |
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| 62. |
A resistance of `20 Omega` is connected to a source of alternating current rated `110 V, 50 Hz`. Find (a) the `rms` current, (b) the maxium instantaneous current in the resistor and the time taken by the current to change from its maximum value to the rms value.A. `2.5 xx 10^(-3)sec`B. `2.5 xx 10^(-2) sec`C. `5 xx 10^(-3) sec`D. `25 xx 10^(-3) sec` |
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Answer» Correct Answer - 1 `E = E_(0) cos omega t, i = i_(0) cos (2 pi ft)` but `i = (i_(0))/(sqrt(2))` |
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| 63. |
The current through an inductor of `1H` is given by `i =31 sin t`. Find the voltage across the inductor.A. `3sin t+3cos t `B. `3cos t+3sin t `C. `3sin t+3tcos t`D. `3tcos t+sin t` |
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Answer» Correct Answer - C |
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| 64. |
An rms voltage of 110 V is applied across a series circuit having a resistance `11 Omega` and an impedance `22 Omega`. The power consumed isA. `275 W`B. `366 W`C. `550 W`D. `1100 W` |
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Answer» Correct Answer - A `P=E_(v)I_(V) cos phi , P=E_(v) (E_(v))/R R/Z` `P=(E_(v)^(2)R)/(Z^(2)) =(110xx110xx11)/(22xx22)W =275W`. |
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| 65. |
110`underset(rms)(V)` is applied across a series circuit having resistance `11 Omega` and independence `22 Omega` . The power consumed isA. 275 WB. 366 WC. 550 WD. 1100 W |
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Answer» Correct Answer - A P = `underset(rms)(V)``underset(rms)(V)``underset(rms)(I)`cos`phi` = `underset(rms)(V)`(`underset(rms)(V)`/Z)(R/Z) = `(110)^(2)`(11)/`(22)^(2)` = 275 W |
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| 66. |
An alternating supply of 220 V is applied across a circuit with resistance `22 Omega` and impedance `44 Omega`. The power dissipated in the circuit isA. 1100 WB. 550 WC. 2200 WD. (2200/3) W |
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Answer» Correct Answer - B Here, V = 220 V Resistance, `R = 22 Omega` Impedance, `Z = 44 Omega` Current in circuit, `I=(V)/(Z)=(220V)/(44 Omega)=5 A` Power dissipated in the circuit, `P = I^(2)R=(5)^(2)xx22=550 W` |
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| 67. |
An rms voltage of 110 V is applied across a series circuit having a resistance `11 Omega` and an impedance `22 Omega`. The power consumed isA. 275WB. 366WC. 550WD. 1100W |
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Answer» Correct Answer - A |
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| 68. |
An inductor `(L = (1)/(100 pi) H)`, a capacitor `(C = (1)/(500 pi) F)` and a resistance `(3 Omega)` is connected in series with an AC voltage source as shown in the figure. The voltage of the AC source is given as `V = 10 cos(100 pi t)` volt. What will be the potential difference between A and B ? A. `8cos(100pit-127^(@))"volt"`B. `8cos(100pit-53^(@))"volt"`C. `8cos(100pit-37^(@))"volt"`D. `8cos(100pit+37^(@))"volt"` |
| Answer» Correct Answer - C | |
| 69. |
In an `L-C-R` series circuit, `R = sqrt(5) Omega, = 9 Omega, X_(C ) = 7 Omega`. If applied voltage in the circuit is `50 V` then impedance of the circuit in ohm will beA. 2B. 3C. `2 sqrt(5)`D. `3 sqrt(5)` |
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Answer» Correct Answer - 2 Impedance, `Z = R + X_(c ) + X_(L)` `= (5 sqrt(5 i) = 7 j + 9 j) = sqrt(5 i) + 2 j` `|Z| = sqrt(5 + 4) = sqrt(9) = 3` |
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| 70. |
An inductor `X_(L)` = 2`Omega` ) , a capacitor (`X_(C)` = 8`Omega`) and a resistance (R = 8 `Omega` ) are connected in series with an AC source . The voltage output of AC source is given by V = 10 cos(100`pi`t) The instantaneous potential difference between points A and B, when the applied voltage is 3/5th of the maximum value of applied voltage is |
| Answer» Correct Answer - 48 | |
| 71. |
Match the following `{:(,"Currents","r.m.s. values"),((1),x_(0) sin omega t,(i)" "x),((2),x_(0) sin omega t cos omega t,(ii)" "x_(0)/sqrt(2)),((3),x_(0) sin omega t+x_(0) cos omega t,(iii)" "x_(0)/((2sqrt(2)))):}`A. 1.(i),2.(ii),3.(iii)B. 1.(ii),2.(iii),3.(i)C. 1.(i),2.(iii),3.(ii)D. none of these |
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Answer» Correct Answer - B |
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| 72. |
In a certain circuit current changes with time accroding to `i=2sqrt(t)`. r.m.s. value of current between `t=2` to `t=4s` will beA. 3AB. `3sqrt3A`C. `2sqrt3A`D. `(2-sqrt2)A` |
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Answer» Correct Answer - C |
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| 73. |
In an `AC` circuit, the power factorA. is zero when the circuit contain an ideal resistance onlyB. is unity when the circuit contains an ideal resistance onlyC. is zero when the circuit contains an ideal inductance onlyD. is unity when the circuit contains an ideal inductance only |
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Answer» Correct Answer - B::C Power factor of an AC circuit is given by cos`phi` = R/Z For circuit containing an ideal resistance only, Z = R `therefore` cos`phi` = 1 For circuit containing an ideal inductance , R = 0 `therefore` cos`phi` = 0/`underset(L)(X)` = 0 |
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| 74. |
The average power dissipation in a pure capacitance in `AC` circuit isA. CVB. zeroC. 1/`CV^(2)`D. 1/4 `CV^(2)` |
| Answer» Correct Answer - B | |
| 75. |
The average power dissipation in a pure capacitance in `AC` circuit isA. `1/2CV^(2)`B. `CV^(2)`C. `1/4 CV^(2)`D. Zero |
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Answer» Correct Answer - D Average power in `AC` circuits is given by `P=V_(rms)i_(rms) cos varphi` For pure capacitive circuit `varphi=90^(@)` so `P=0` |
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| 76. |
In the series LCR circuit shown the impedance is A. `200 Omega`B. `100 Omega`C. `300 Omega`D. `500 Omega` |
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Answer» Correct Answer - D `X_(L) = 2pifL=2pi(50/pi) xx 1=100Omega` `X_(C) = 1/(2pifC) = 1/(2pi(50/pi)20xx10^(-6)) = 500 Omega` Impedance , `Z= sqrt(R^(2)+(X_(C)-X_(L))^(2)` `rArr = Z = sqrt((300)^(2)+(400)^(2))` = `500Omega` |
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| 77. |
A series LCR circuit is connected to a variable frequency 230 V source , L=5.0 H, `C=80muF,R=40Omega` Obtain impedance of the circuit and aplitude of current at resonance. |
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Answer» At resonance impedance, `Z=R=40Omega` `thereforeI_(rms)=V_(rms)/R=(sqrt2times230)/40=8.1 A` |
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| 78. |
When the frequency of the ac voltage applied to a series LCR circuit is gradually increased from a low value , the impedance of the circuit.A. monotonically increasesB. first increases and then decreasesC. first decreases and then increasesD. monotonically decreases |
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Answer» Correct Answer - C We Know`Z=sqrt((omegaL-1/(omegaC))^2+R^2)` IF , `(dZ)/(domega)=0`then `omega=1/sqrt(LC)` when `omegalt1/sqrt(LC)` then `(dZ)/(domega)le0` i.e., Z is a decreasing function. Again, when `omegagt1/sqrt(LC)`then`(dZ)/(domega)gt0` i.e., Z is a increasing function. |
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| 79. |
In which of following circuits the maximum power dissipation is observed?A. Pure capacitive circuitB. Pure inductive circuitC. Pure resistive circuitD. None of these |
| Answer» Correct Answer - C | |
| 80. |
What do you mean by the impedance of `LCR`-circuit |
| Answer» Impedance: It is the effective resistance offered by the element series `LCR` circuit `Z=sqrt(R^(2)(omegaL-1/(omegac))^(2))` | |
| 81. |
An alternating voltage is given by: `e = e_(1) sin omega t + e_(2) cos omega t`. Then the root mean square value of voltage is given by:A. `sqrt(e_(1)^(2)+e_(2)^(2))`B. `sqrt(e_(1)+e_(2))`C. `sqrt(e_(1)+e_(2)/2)`D. `sqrt(e_(1)^(2)+e_(2)^(2)/2)` |
| Answer» Correct Answer - D | |
| 82. |
An alternating voltage is given by: `e = e_(1) sin omega t + e_(2) cos omega t`. Then the root mean square value of voltage is given by:A. `sqrt(e_(1)^(2) + e_(2)^(2))`B. `sqrt(e_(1)e_(2))`C. `sqrt((e_(1)e_(2))/(2))`D. `sqrt((e_(1)^(2) e_(2)^(2))/(2))` |
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Answer» Correct Answer - D `e = e_(1) sin omega t + e_(2) cos omega t` `= sqrt(e_(1)^(2) + e_(2)^(2)) [sin omega t (e_(1))/(sqrt(e_(1)^(2) + e_(2)^(2))) + cos omega t (e_(2))/(sqrt(e_(1)^(2) + e_(2)^(2)))]` `implies sqrt(e_(1)^(2) + e_(2)^(2)) [sin omega t cos phi + cos omega t sin phi]` `implies e = sqrt(e_(1)^(2) + e_(2)^(2)) sin (omega t + phi)` where `phi = "tan"^(-1) (e_(2))/(e_(1))` `e_("rms") = (e_("max"))/(sqrt(2)) = sqrt(e_(1)^(2) + e_(2)^(2))/(sqrt(2)) = sqrt((e_(1)^(2) + e_(2)^(2))/(2))` |
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| 83. |
The r.m.s value of Potential due to superposition of given two alternating potentials `E_(1) = E_(0) sin omega t` and `E_(2) = E_(0) cos omega t` will beA. `E_(0)`B. `2 E_(0)`C. `E_(0)sqrt(2)`D. Zero |
| Answer» Correct Answer - 1 | |
| 84. |
An alternating voltage is given by: `e = e_(1) sin omega t + e_(2) cos omega t`. Then the root mean square value of voltage is given by:A. `sqrt(e_(1)^(2)+e_(2)^(2))`B. `sqrt(e_(2)e_(2))`C. `sqrt(e_(1)e_(2))/(2)`D. `sqrt(e_(1)^(2)+e_(2)^(2))/(2)` |
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Answer» Correct Answer - D d) `e=sqrt(e_(1)^(2)+e_(2)^(2))sin(omegat+phi)` Where, `cosphi = (e_(1))/(sqrt(e_(1)^(2)+e_(2)^(2)))` and `sinphi = (e_(2))/(sqrt(e_(1)^(2)+e_(2)^(2)))` i.e., `e_(0) = sqrt(e_(1)^(2)+e_(2)^(2))` or `e_(rms) = sqrt(e_(1)^(2)+e_(2)^(2))/(sqrt(2))` |
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| 85. |
Voltage and current in an ac circuit are given by `V=5 sin (100 pi t-pi/6)` and `I=4 sin (100 pi t+pi/6)`A. voltage leads the current by `30^(@)`B. current leads the voltage by `30^(@)`C. current leads the voltage by `60^(@)`D. voltage leads the current by `60^(@)` |
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Answer» Correct Answer - C c) Phase difference between current and voltage. `Deltaphi-phi_(2)-phi_(1)=pi/6-(-pi/6)=pi/3` |
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| 86. |
The phase diffenernce between the alternating current and emf is `(pi)/(2)`. Which of the following cannot be the constiuent of the circuit?A. `C` aloneB. `R,L`C. `L,C`D. `L` alone |
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Answer» Correct Answer - 2 `tan phi=X/R=prop =1/0 rArr R=0` |
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| 87. |
An alternating voltage is represented as `E=20 sin 300 t`. The average value of voltage over one cycle will beA. ZeroB. 10 voltsC. `20sqrt(2)` voltD. `20/sqrt(2)` volt |
| Answer» Correct Answer - a | |
| 88. |
In an `AC` circuit, the mass value of the current `I_("rms")` is related to the peak current `I_(0)` asA. `I_(rms)=1/pi I_(0)`B. `I_(rms)=1/sqrt(2) I_(0)`C. `I_(rms)=sqrt(2) I_(0)`D. `I_(rms)=pi I_(0)` |
| Answer» Correct Answer - b | |
| 89. |
In an `AC` circuit, the mass value of the current `I_("rms")` is related to the peak current `I_(0)` asA. `i_(rms)=sqrt2i_(0)`B. `i_(rms)=pii_(0)`C. `i_(rms)=(i_(0))/(pi)`D. `i_(rms)=(1)/(sqrt2)i_(0)` |
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Answer» Correct Answer - D |
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| 90. |
In an `AC` circuit, the mass value of the current `I_("rms")` is related to the peak current `I_(0)` asA. `I_("rms") = (1)/(pi) I_(0)`B. `I_("rms") = (1)/(sqrt(2)) I_(0)`C. `I_("rms") = sqrt(2) I_(0)`D. `I_("rms") = pi I_(0)` |
| Answer» Correct Answer - 2 | |
| 91. |
It is known to all of you that the impedance of a circuit is dependent on the frequency of source. In order to study the effect of frequency on the impedance, a student in a lab took 2 impedance boxes `P` and `Q` and connected them in series with an `AC` source of variable frequency. The emf of the source is constant at `10 V` Box `P` contains a capacitance of `1muF` in series with a resistance of `32 Omega`. And the box `Q` has a coil of self-inductance `4.9 mH` and a resistance of `68 Omega`in series. He adjusted the frequency so that the maximum current flows in `P` and `Q`. Based on his experimental set up and the reading by him at various moment, answer the following questions. Impedance of box P at the above frequency isA. `70Omega`B. `77 Omega`C. `90 Omega`D. `100 Omega` |
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Answer» Correct Answer - B `X_C=1/(omegaC)=1/((10^5/7)(10^-6))=70Omega` `Z_P=sqrt(R_P^2+X_C^(2))` `=sqrt((32)^+(70)^(2))` `~~ 77Omega` |
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| 92. |
A filament bulb `(500 W, 100 V)` is to be used in a `230 V` main supply. When a resistance `R` is connected in series, it works perfectly and the bulb consumers `500 W`. The value of `R` isA. `230 Omega`B. `46 Omega`C. `25 Omega`D. `13 Omega` |
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Answer» Correct Answer - C If a rated voltage and power are given, then `P_("rated") = (V_("rated")^(2))/R` `therefore` Current in the bulb, `I=P/V` `[therefore`P=VI] `I= 500/100= 5A` `therefore` Resistance of bulb, `R_(b) = (100 xx 100)/(500) = 20 Omega` `therefore` Resistance R is connected in series. `therefore` Current, `I= E/R_("net") = 230/(R+R_(b))` `therefore rArr R+20 = 230/5 =46` `R= 26Omega` |
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| 93. |
An L-C circuit contains 20 mH inductor and a `50 muF` capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. what is the total energy stored initially ? At what times is the total energy shared equally between the inductor and the capacitor ? |
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Answer» Given Inductance `L = 20 mH = 20xx10^(-3)H` Capacitance of capacitor `C=50 mu f=50xx10^(-6)F` Initial charge on the capacitor, `Q_(i)=10mc =10xx10^(3)C` Equal sharing of energy between inductor and capacitor means the energy stored in capacitor `= (1)/(2)xx` Maximum energy `(Q^(2))/(2C)=(1)/(2)xx(Q_(0)^(2))/(2C) " " Q=(Q_(0))/(sqrt(2))` .....(ii) From `Q=Q_(0)cos omega t=Q_(0)cos.(2pi)/(T)t` `(Q_(0))/(sqrt(2))=Q_(0).cos.(2pi)/(T)t` `(1)/(sqrt(2))=cos.(2pi t)/(T)` or `cos(2n+1)pi//4 = cos.(2pi t)/(T)` `((2n+1)pi)/(4)=(2pi t)/(T)` `t=T//8(2n+1) (n=0,1,2,3,....)` Hence the energy will be shared half on capacitor and half on inductor. `t=(T)/(8), (3T)/(8), (5T)/(8)`, ....... |
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| 94. |
A dc ammeter and a hot wire ammeter are connected to a circuit in series. When a direct current is passed through circuit, the dc ammeter shows 6 A. When ac current flows through circuit, the ac ammeter shows 8A. What will be reading of each ammeter if dc and ac current flow simulataneously through the circuit?A. the DC ammeter will shown zero currentB. the DC ammeter will shown 6A currentC. the AC ammeter will shown 14A currentD. the AC ammeter will shown zero current |
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Answer» Correct Answer - B |
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| 95. |
A dc ammeter and a hot wire ammeter are connected to a circuit in series. When a direct current is passed through circuit, the dc ammeter shows 6 A. When ac current flows through circuit, the ac ammeter shows 8A. What will be reading of each ammeter if dc and ac current flow simulataneously through the circuit?A. dc=6 A, ac=10AB. dc=3 A, ac=5AC. dc=5 A, ac=8AD. dc=2 A, ac=3A |
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Answer» Correct Answer - A Resultant current is superposition of tow currents, i.e. I(instantaneous total current ) `=6+(I_0) sin omega t` dc ammeter will read average value `=bar(6+(I_0)sin omega t) = 6` `(: Bar((I_0) sin omega t =0))` ac ammeter will read `=sqrt((bar(6+(I_0)sin omega t^(2)))` `=sqrt(bar(36+12 I_(0) sin omega t +I_(0)^(2) sin ^(2)omega t))` (`because bar(I_(0)si omega t =0))` since `bar(sin^(2) ometa t)=1/2 and (I_(rms))=8=(I_0)/(sqrt(2) implies (I_0)= 8 sqrt(2)A` Therefore ac reading `=sqrt(36 + (I_(0)^(2))/(2)) = sqrt(36 + 64) = 10 A`. |
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| 96. |
In a series LCR circuit with an AC so `(E_(rms)=50V)` and `f=50//pi` Hz), R=30 C=0.02mF, L=1.0H, which of the follwing is correctA. the rms current in the circuit is 0.1 AB. the rms potential difference acrosss the capacitor isC. the rms potential difference acrosss the capacitor isD. the rms current in the circuit is 0.14A |
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Answer» Correct Answer - A::B |
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| 97. |
A sinusoidal ac current flows through a resistor of resistance R . If the peak current is `I_(p)` , then the power dissipated isA. `I_(p)^(2)R cos theta`B. `1/2 I_(p)^(2) R`C. `4/pi I_(p)^(2) R`D. `1/pi I_(p)^(2) R` |
| Answer» Correct Answer - b | |
| 98. |
A sinusoidal ac current flows through a resistor of resistance R . If the peak current is `I_(p)` , then the power dissipated isA. `I_(1)^(p)R cos theta`B. `1/2I_(p)^(2)R`C. `4/pi I_(p)^(2)R`D. `1/pi^(2)I_(p)^(2)R` |
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Answer» Correct Answer - B `ltP gt = I_(rms)^(2) R=(I_(p)/sqrt2)^(2) R=(I_(p)^(2)R)/2` |
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| 99. |
An ideal inductor of inductance `50muH` is connected to an AC source of 220V, 50 Hz. Find the inductive reactance. |
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Answer» `X_(L)=omega L = 2 pi f L = 2 xx pi xx 50xx 50 xx 10^(-6)= 2 pi xx 25 xx 25 xx 10^(-4) Omega` `= 5 pi xx 10^(-3) Omega = 5 pi m Omega` |
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| 100. |
IF the value of inductor L is 1mH and the applied ac source frequency is 50 Hz, find the inductive reactance in the above case. |
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Answer» Here,`L=1mH=10^-3H` and `omega=2pif=2times3.14times50=314Hz` `thereforeX_L=omegaL=314times10^-3=0.314Omega` |
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