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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.
| 1051. |
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. (A) `100 Omega`B. (B) `25 Omega`C. (C) `sqrt(125xx75)Omega`D. (D) `400 Omega` |
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Answer» Correct Answer - B In an ac circuit capacitor does not consume any power. Therefore, power is cosumed by the resistor only. `:. P= I_(V)^(2) R or 100 =(2)^(2) R or R = 25 Omega` |
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| 1052. |
A resistor and an inductor are connected to an ac supply of 120 V and 50 Hz, the current in the circuit is 3 A. If the power consumed in the circuit is 108 W, then the resistance in the circuit isA. (A) `12 Omega`B. (B) `40 Omega`C. (C) `sqrt((52 xx 28))Omega`D. (D) `360 Omega` |
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Answer» Correct Answer - A In an ac circuit, a pure inductor does not consume any power. Therefore, power is consumed by the resistor only. `:. P=I_(V)^(2) R` ltbRgt or `108 =(3)^(2) R or R = 12 Omega`. |
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| 1053. |
A coil of inductance `5.0 mH` and negligible resistance is connected to an oscillator giving an output voltage `E=(10V) sin omegat`.Which of the following is correctA. for `omega=100 s^(-1)` peak current is `20 A`B. for `omega=500 s^(-1)` peak current is `4 A`C. for `omega=1000 s^(-1)` peak current is `2 A`D. for `omega=1000 s^(-1)` peak current is `4 A` |
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Answer» Correct Answer - A,B,C `I_(0)=V_(0)/(omegaL)=10/(omegaxx5xx10^(-3))` |
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| 1054. |
A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage `V=10 sin (100t)`. The peak current in the circuit will be:A. 2ampB. 1ampC. 10ampD. 20amp |
| Answer» Correct Answer - D | |
| 1055. |
For a series `RLC` circuit `R=X_(L)=2X_(C)`. The impedence of the current and phase different (between) `V` and `i` will beA. `(sqrt5R)/(2),tan^(-1)(2)`B. `(sqrt5R)/(2),tan^(-1) ((1)/(2))`C. `sqrt5X_(C),tan^(-1) ((1)/(2))`D. `sqrt5R,tan^(-1) ((1)/(2))` |
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Answer» Correct Answer - B |
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| 1056. |
Which of the following is true for an ideal capacitor connected to a sinusoidal voltage source ?A. Neither the average power nor the average current is zeroB. Average current is zero but not the average powerC. The average power is zero, but not the average currentD. Both the average power and average current are zero |
| Answer» Correct Answer - D | |
| 1057. |
The impedance of a circuit consister of `3Omega`resistance and `4Omega` reactance. The power factor of the circuit isA. 0.4B. 0.6C. 0.8D. 1 |
| Answer» Correct Answer - B | |
| 1058. |
In a series RLC circuit, if the frequency is increased to a very large value, what value does the phase angle between current and voltage approach ?A. `90^(@)`B. `0^(@)`C. `30^(@)`D. `45^(@)` |
| Answer» Correct Answer - A | |
| 1059. |
Consider an ac circuit where an incandescent light bulb is in series with an inductor. If the frequency of generator is increased, what will happen to the brightness of bulb ?A. The bulb will have same brightnessB. The bulb will shine brightenC. The bulb will become dimmerD. Depends on the value of L and R |
| Answer» Correct Answer - C | |
| 1060. |
A series LCR circuit, has equal resistance capacitive reactance. What is the phase angle between voltage across generator and resistor ?A. `0^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - A | |
| 1061. |
In an `AC` circuit the reactance of a coil is `sqrt(3)` times its resistance, the phase difference between the voltage across the current through the coil will beA. `pi//3`B. `pi//2`C. `pi//4`D. `pi//6` |
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Answer» Correct Answer - A `tan varphi=(X_(L))/R =(sqrt(3)R)/Rimplies sqrt(3) implies varphi=60^(@)=pi//3` |
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| 1062. |
The figure shows variation of `R,X_(L)` and `X_(C)` with frequency `f` in a series `L,C,R` circuit. Then for what frequency point, the circuit is inductive ? A. `A`B. `B`C. `C`D. All points |
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Answer» Correct Answer - C At `A:X_(C)gtX_(L)` At `B:X_(C)=X_(L)` At `C:X_(C) lt X_(L)` |
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| 1063. |
A capacitor of capacitnace C is charged to a potential difference V and then disconnected from the battery. Now it is connected to an inductor of inductance L at t=0. ThenA. Energy stored in capacitor and inductor will be equal at time `t=(pi)/(2)sqrtLC`B. Potential difference across inductor will be `(V)/(2)` at time `t=(pi)/(3)sqrtLC`C. The rate of increase of energy in magnetic field will be maximum at `(pi)/(4)sqrtLC`D. When the potentail difference across the capacitor `sqrt(((3C)/(L)))` |
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Answer» Correct Answer - B::C::D |
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| 1064. |
A resonant `AC` circuit contains a capacitor of capacitance `10^(-6)F` and an inductor of `10^(-4)H`. The frequency of electrical oscillation will beA. `105 Hz`B. `10 Hz`C. `(10^(5))/(2pi) Hz`D. `10/(2pi)Hz` |
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Answer» Correct Answer - C `v=1/(2pisqrt(LC))=1/(2pisqrt(10^(-6)xx10^(-4)))=(10^(5))/(2pi)Hz` |
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| 1065. |
The figure shows variation of `R,X_(L)` and `X_(C)` with frequency `f` in a series `L,C,R` circuit. Then for what frequency point, the circuit is inductive ? A. AB. BC. CD. All points |
| Answer» Correct Answer - c | |
| 1066. |
Which of the following plots may represent the reactance of a series `LC` combination? A. `a`B. `b`C. `c`D. `d` |
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Answer» Correct Answer - D Reactance `= X=X_(L)-X_(C)=2pifL-1/(2pifC)` |
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| 1067. |
The figure shows variation of `R,X_(L)` and `X_(C)` with frequency `f` in a series `L,C,R` circuit. Then for what frequency point, the circuit is inductive ? A. `A`B. `B`C. `C`D. All points |
| Answer» Correct Answer - C | |
| 1068. |
Which of the following plots may represent the reactance of a series `LC` combination? A. aB. bC. cD. d |
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Answer» Correct Answer - D `X = X_(L) ~ X_(C)` It may be `X_(L) - X_(C) = + ve, X is + ve` `X_(L) - X_(C) = -ve, X is - ve` . |
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| 1069. |
A series combination of `R,LC` is connected to an a.c source. If the resistance is `3 Omega` and the reactance is `4 Omega`, the power factor of the circuit isA. 0.4B. 0.6C. 0.8D. `1.0` |
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Answer» Correct Answer - B `x = 4 Omega, R = 3 Omega` `Z = sqrt(R^(2) + X^(2)) = sqrt(3^(2) + 4^(2)) = 5` Power factor `= cos phi = (R )/(Z) = (3)/(5) = 0.6` |
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| 1070. |
Which of the following plots may represent the reactance of a series `LC` combination? A. aB. bC. cD. d |
| Answer» Correct Answer - D | |
| 1071. |
A conducting light string is wound on the rim of a metal ring of radius `r` and mass `m`. The free end of the string is fixed to the ceiling. A vertical infinite smooth conducting plane is always tangent to the ring as shown in the figure. A uniform magnetic field Bis applied perpendicular to the plane of the ring. The ring is always inside the magnetic field. The plane and the strip are connected by a resistance `R`. When the ring is released, find a. the curent in the resistance `R` as as function of time. b. the terminal velocity of the ring. |
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Answer» Correct Answer - 4 |
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| 1072. |
The circuit shows a resistance, `R=0.01Omega` and inductance L=3mH connected to a conducting rod PQ of length l=wm which can slide on a perfectly conducting circular arc of radius l with its center at P. Assume that friction and gravity are absent and a constant uniform magnetic field b=0.1T exists as shown in the figure. At t=0, the circuit is seitched on a simultaneously an external torque is applied on the rod so that it rotates about P with a constant angular velocity `omega2 rad//sec`. Find the magnitude of this torque (inN-m) at t=(0.3In 2) second. |
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Answer» Correct Answer - 4 |
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| 1073. |
If an alternating voltage is represented as `E=141 sin (628 t)`, then the rms value of the voltage and the frequency are respectivelyA. 141 Hz, 628 HzB. 100 V, 50 HzC. 100 V, 100 HzD. 141 V, 100 Hz |
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Answer» Correct Answer - C E = 141sin(628t) `underset(rms)(E)` = `underset(o)(E)`/`sqrt 2` = 141/141 = 100 V and `2pi`f = 628 `adr` f = 100 Hz |
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| 1074. |
In an circuit, V and I are given by `V=150sin(150t) V` and `I =150sin(150t+(pi)/(3))A`. The power dissipated in the circuit isA. 106 WB. 150 WC. 5625 WD. zero |
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Answer» Correct Answer - C Compare C = 150 sin (150 t) with `V = V_(0)sin omega t`, we get, `V_(0)=150 V` Compare `I = 150 sin (150 t+(pi)/(3))` with `I = I_(0)sin(omega t+ phi)`, we get `I_(0)=150A, phi = (pi)/(3)=60^(@)` The power dissipated in ac circuit is `P=(1)/(2)V_(0)I_(0)cos phi=(1)/(2)xx150xx150xx cos 60^(@)` `=(1)/(2)xx150xx150xx(1)/(2)=5625 W` |
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| 1075. |
In the case of an inductorA. Voltage lags the current by `(pi)/(2)`B. Voltage leads the current by `(pi)/(2)`C. Voltage leads the current by `(pi)/(3)`D. Voltage leads the current by `(pi)/(4)` |
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Answer» Correct Answer - B In an inductor voltage leads the current by `(pi)/(2)` or current lags the voltage by `(pi)/(2)`. |
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| 1076. |
A `20 V` 5 watt lamp is used in ac main `220 V` and frequency 50 c.p.s. Inductance of inductor,to be put in series to run tha lamp.A. `2.53 H`B. `5 H`C. `7.5 H`D. `9 H` |
| Answer» Correct Answer - A | |
| 1077. |
An ideal inductor is in turn put across 220 V, 50 Hz and 220 V, 100 Hz supplies. The current flowing through it in the two cases will beA. equalB. differentC. zeroD. infinite |
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Answer» Correct Answer - B The current in the inductor coil is given by `I = (V)/(X_(L))=(V)/(2pi upsilon L)` Since frequency `upsilon`in the two cases is different, hence the current in two cases will be different. |
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| 1078. |
A conductor of capacity 1pF is connected to an `A.C` source of `220 V` and `50 Hz` frequency. The current flowing in the circuit will beA. `6.9 xx 10^(-8) A`B. `6.9 A`C. `6.9 xx 10^(-6) A`D. zero |
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Answer» Correct Answer - 1 `i_("rms") = (E_("rms"))/(X_(c))` |
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| 1079. |
In a circuit, the frequency is `f = (1000)/(2 pi) Hz` and the inductance is 2 henry, then the reactance will beA. `200 Omega`B. `200 mu Omega`C. `2000 Omega`D. `2000 mu Omega` |
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Answer» Correct Answer - 3 `X_(L) = omega L = 2 pi fl = 2 pi xx (1000)/(2 pi) xx 2 = 2000 Omega` |
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| 1080. |
An `LC` source rated `100 V` (rms) supplies a current of `10 A` (rms) to a circuit. The average power delivered by the sourceA. must be `1000 W`B. may be `1000 W`C. may be greater than `1000 W`D. may be less than `1000 W` |
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Answer» Correct Answer - B,D `P_(av) = V_("rms") I_("rms") cos phi` If `cos phi = 1` then maximum average power `= V_("rms") I_("rms") = 1000` watt. If `cos phi lt 1` then average power will be less than 1000 watt. So correct option is `b,d` |
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| 1081. |
An `AC` source rated `100 V (rms)` supplies a current of `10 A(rms)` to a circuit. The average power delivered by the sourceA. must be `1000W`B. may be `1000W`C. may be greater than `1000W`D. may be less than `100 W` |
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Answer» Correct Answer - C `P = V_(rms) i_(rms)cos phi` If `phi = 100 xx 10 xx 1 =1000 W` `cos phi le 1` `P le 1000W`. |
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| 1082. |
Quantity that remains unchanged in a transformer isA. voltageB. currentC. frequencyD. None of these |
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Answer» Correct Answer - C A transformer does not change the frequency of ac. |
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| 1083. |
The core of a transformer is laminated to reduceA. flux leakageB. hysteresisC. copper lossD. eddy current |
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Answer» Correct Answer - D The core of a transformer is laminated to reduce eddy current. |
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| 1084. |
The resistance of a coil for DC is in ohms. In AC, the resistanceA. Will remain sameB. Will increasesC. Will decreaseD. Will be zero |
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Answer» Correct Answer - B The coil having inductance `L` besides the resistance `R`. Hence for AC its effective resistance `sqrt(R^(2)+X_(L)^(2))` will be larger then its resistance `R ` for `DC`. |
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| 1085. |
When a coil is connected to a 100 V D.C. supply, the current is 2 A. When the same coil is connected to A.C. source `E=100sqrt(2) sin omega t`, the current is 1 A. Find the inductive reactance used in the circuit. |
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Answer» In D.C. circuit `I =(V)/(R) rArr "So, " R=(V)/(I) = 50 Omega` In A.C. `I_("rms")=(V_("rms"))/(Z)rArr 1 = (100)/(z) rArr Z=100 Omega` Now `Z=sqrt(R^(2)+X_(L)^(2))rArr X_(L)=50sqrt(3)Omega` |
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| 1086. |
In an LCR series a.c. Circuit the voltage across each of hte components L,C and R is 50V. The voltage across the LC combination will beA. `50 V`B. `50sqrt(2) V`C. `100 V`D. `0V` (Zero) |
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Answer» Correct Answer - D Net voltage across `LC` combination `=V_(L)-V_(C)=0V`. From all the given options only options (a) is correct. |
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| 1087. |
in a LCR circuit capacitance is chagned from C to 2C. For the resomat frequency to remain unchaged, the inductance should be chagned from L toA. 4LB. 2 LC. `L//2`D. `L//4` |
| Answer» Correct Answer - c | |
| 1088. |
The resistance of a coil for DC is in ohms. In AC, the resistanceA. Will remain sameB. Will increaseC. Will decreaseD. Will be zero |
| Answer» Correct Answer - b | |
| 1089. |
A resistor of `500 Omega` and an inductance of 0.5 H are in series with an ac source which is given by `V=100sqrt(2)sin(1000t)`. The power factor of the combination isA. `(1)/(sqrt(2))`B. `(1)/(sqrt(3))`C. `0.5`D. `0.6` |
| Answer» Correct Answer - A | |
| 1090. |
in a LCR circuit capacitance is chagned from C to 2C. For the resomat frequency to remain unchaged, the inductance should be chagned from L toA. `4L`B. `2L`C. `L//2`D. `L//4` |
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Answer» Correct Answer - C `V_(0)=1/(2pisqrt(LC))` If `C` changes to `2C` then for keeping `V_(0)` constant `L`must change to `L//2` |
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| 1091. |
Alternating current can not be measured by D.C. Ammeter becauseA. ac cannot pass through dc ammeterB. Average value of complete cycle is zeroC. ac is virtualD. ac changes its direction |
| Answer» Correct Answer - b | |
| 1092. |
There is a `5 Omega` resistance in an `AC`, circuit. Inductance 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` |
| Answer» Correct Answer - c | |
| 1093. |
In a ac circuit of capacitance the current from potential isA. ForwardB. BackwardC. Both are in the same phaseD. None of these |
| Answer» Correct Answer - a | |
| 1094. |
The average power is dissipated in a pure inductor isA. `(VI^(2))/4`B. `1/2VI`C. zeroD. `VI^(2)` |
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Answer» Correct Answer - C In the inductor circuit, phase difference `(phi)= 90^(@)` or `pi/2` `therefore P_(av) = 0` |
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| 1095. |
Alternating current can not be measured by D.C. Ammeter becauseA. `AC` voltage pass through `DC` ammeterB. Average value of complete cycle is zeroC. `AC` is virtualD. `AC` change its direction |
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Answer» Correct Answer - B In `DC` ammeter, a coil is free to rotate in the magnetic field of a fixed magnet. If an alternating current is passed through such a coil, the torque will reverse its direction each time the current changes direction and the average the of the torque will be zero. |
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| 1096. |
An inductor of inductance `L` and ressistor of resistance `R` are joined in series and connected by a source of frequency `omega`. Power dissipated in the circuit isA. `((R^(2)+omega^(2)L^(2)))/(V)`B. `(V^(2)R)/((R^(2)+omega^(2)L^(2)))`C. `(V)/((R^(2)+omega^(2)L^(2)))`D. `sqrt(R^(2)+omega^(2)L^(2))/V^(2)` |
| Answer» Correct Answer - b | |
| 1097. |
The average power dissipated in a pure inductor `L carrying an alternating current of rms value I is .A. `1/2LI^(2)`B. `I^(2)X_(1)`C. `LI^(2)//4`D. zero |
| Answer» Correct Answer - D | |
| 1098. |
The self-inductance of a choke coil is `10mH`. When it is connected with a `10 V DC` source, then the loss of power is `20 "watt"`. When it is connected with `10 "volt" AC` source loss of power is `10"watt"`. The frequency of `AC` source will beA. 80 HzB. 100 HzC. 120 HzD. 220 Hz |
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Answer» Correct Answer - A With DC, power, `P = V^(2)/R` `R=(10)^(2)/20 = 5 Omega` With AC power, `P= (V_(rms)^(2)R)/Z^(2) rArr Z^(2) = ((10^(2) xx 5)/10 = 50 Omega` The impedance, `Z^(2) = R^(2)+4pi^(2)v^(2)L^(2)` `50 = (5)^(2) + 4(3.14)^(2) xx v^(2) xx (10 xx 10^(-3))^(2)` v=80 Hz |
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| 1099. |
An inductor of inductance `L` and ressistor of resistance `R` are joined in series and connected by a source of frequency `omega`. Power dissipated in the circuit isA. `((R^(2)+omega^(2)L^(2)))/V`B. `(V^(2)R)/((R^(2)+omega^(2)L^(2)))`C. `V/((R^(2)+omega^(2)L^(2)))`D. `(sqrt(R^(2)+omega^(2)L^(2)))/(V^(2))` |
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Answer» Correct Answer - B `P= Vi cos varphi=V(V/Z)(R/Z)=(V^(2)R)/(Z^(2))=(V^(2)R)/((R^(2)+omega^(2)L^(2)))` |
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| 1100. |
Assertion: The quantity `L//R` possesses dimensions of time. Reason: To reduce the rate of increases of current through a solenoide should increase the time constant `(L//R)`.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 The relation of induced emf is `e=(Ldi)/(dt)` and current `i` is given by `i=e/R=1/R.(Ldi)/(dt)` `implies (di)/(dt) =i R/L=i/(L//R)` In order to decreases the rate of increases of current through solenoid. We have to increases the time constant `L//R`. |
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