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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.
| 551. |
With increase in frequency of an `AC` supply, the inductive reactance:A. decreasesB. increases directly proportional to frequencyC. increases as square of frequencyD. decreases inversely with frequeny |
| Answer» Correct Answer - B | |
| 552. |
In previous question the resistance of coil `A` is .A. `0.01H`B. `0.02H`C. `0.03H`D. `0.04H` |
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Answer» Correct Answer - A `Z^(2) = R^(2) + X_(L)^(2) = (R_(A) + R_(B))^(2) + omega^(2) (L_(A) + L_(B))^(2)` `(14.4)^(2) = (5 + 5.8)^(2) + (2pi xx 50)^(2) (L_(A) + 0.02)^(2)` `L_(A) = 0.01 H` . |
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| 553. |
Choke coil is used to controlA. `AC`B. `DC`C. Both `AC` and `DC`D. Neither `AC` nor `DC` |
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Answer» Correct Answer - A The choke coil can be used only in AC circuits, not in DC circuits, because for DC `(omega=0)` the inductive reactance `X_(L)=omegaL`. Of coil is zero, only the resistance the coil remains effective which too is almost zero. |
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| 554. |
The vector diagram of current and voltage for a circuit as shown. The components of the circuit will be A. `LCR`B. `LR`C. `LCR` or `LR`D. None of these |
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Answer» Correct Answer - C From phasor diagram it is clear that current is lagging with respect to `E_(rms)`. This may be happen in `LCR` or `LR ` circuit. |
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| 555. |
With increase in frequency of an `AC` supply, the capacitive reactance:A. varies inversely with frequencyB. varies directly with frequencyC. varies directly as square of frequencyD. remains constant |
| Answer» Correct Answer - A | |
| 556. |
With increase in frequency of an `AC` supply, the inductive reactance:A. decreasesB. increases directly proportional to frequencyC. increases as square of frequencyD. decreases inversely with frequency |
| Answer» Correct Answer - B | |
| 557. |
The frequency for which a 5`mu`F capacitor has a reactance of `1/1000 Omega`is given byA. `100/pi` MHzB. `1000/pi` MHzC. `1/1000 Hz`D. 1000 Hz |
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Answer» Correct Answer - A Capacitive reactance , `underset(C )(X)` = 1/2`pi``nu`c `adr` 1/1000 = 1/`2pi xx nu xx 5 xx 10^(-6)` `therefore` `nu` = 100/`pi` MHz |
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| 558. |
A circuits contains a capacitor and inductance each with negligible resistance. The capacitor is initially charged and the charging battery is disconnected. At subsequent time , the charge on the capacitor willA. increase exponentiallyB. decrease exponentiallyC. decrease linearlyD. remain constant |
| Answer» Correct Answer - C | |
| 559. |
A capacitor becomes a perfect insulator forA. constant direct currentB. alternating currentC. direct as well as alternating currentD. variable direct current |
| Answer» Correct Answer - A | |
| 560. |
For series LCR circuit, correct statements areA. Applied e.m.f. and potential difference across resistance are in same phaseB. Applied e.m.f. and potential difference at inductor coil have phase difference of `pi//2`C. Potential difference at capacitor and inductor have phase difference of `pi//2`D. Potential difference across resistance and capacitor have phase difference of `pi//2` |
| Answer» Correct Answer - c | |
| 561. |
A choke coil has.A. high inductance and high resistanceB. low inductance and low resistanceC. high inductance and low resistanceD. low inductance and high resistance |
| Answer» Correct Answer - C | |
| 562. |
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. 0.32B. `0.30`C. 0.28D. 0.24 |
| Answer» Correct Answer - b | |
| 563. |
A `0.7` henry inductor is connected across a `120 V-60 Hz AC` source. The current in the inductor will be very nearlyA. 4.55 ampB. 0.355 ampC. 0.455 ampD. 3.55 amp |
| Answer» Correct Answer - c | |
| 564. |
A circuit has a resistance of `11 Omega`, an inductive reactance of `25 Omega`, and a capacitive resistance of `18 Omega`. It is connected to an `AC` source of `260 V` and `50 Hz`. The current through the circuit (in amperes) isA. `11`B. `15`C. `18`D. `20` |
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Answer» Correct Answer - D `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt((11)^(2)+(25-18)^(2))=13 Omega` Current `i=260/13=20A` |
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| 565. |
A circuit has a resistance of `11 Omega`, an inductive reactance of `25 Omega`, and a capacitive resistance of `18 Omega`. It is connected to an `AC` source of `260 V` and `50 Hz`. The current through the circuit (in amperes) isA. 11B. 15C. 18D. 20 |
| Answer» Correct Answer - d | |
| 566. |
A series circuit connected across a `200 V, 60 Hz` line consists of a capacitive reactance `30 Omega` non inductive resistor of `44 Omega` and a coil of inductive reactance `90 Omega` and resistance `36 Omega` as shown in the diagram The power dissipated in the inductance coil isA. zeroB. `320 W`C. `144 W`D. `160 W` |
| Answer» Correct Answer - C | |
| 567. |
In an `AC` circuit, the power factorA. Is zero when the circuit contains 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 |
| Answer» Correct Answer - bc | |
| 568. |
A series circuit connected across a `200 V, 60 Hz` line consists of a capacitive reactance `30 Omega` non inductive resistor of `44 Omega` and a coil of inductive reactance `90 Omega` and resistance `36 Omega` as shown in the diagram The potentail difference across the coil isA. `100 v`B. `194 V`C. `97 V`D. zero |
| Answer» Correct Answer - B | |
| 569. |
A series circuit connected across a `200 V, 60 Hz` line consists of a capacitive reactance `30 Omega` non inductive resistor of `44 Omega` and a coil of inductive reactance `90 Omega` and resistance `36 Omega` as shown in the diagram The power used in the circuit isA. `320 W`B. `144 W`C. `160 W`D. `96 W` |
| Answer» Correct Answer - A | |
| 570. |
A resistance of `40` ohms and an inductance of `95.5` millihenry are connected in series in a `50` cycle/sec AC circuit. The impedence of this combination is very nearlyA. 30 ohmB. 40 ohmC. 50 ohmD. 60 ohm |
| Answer» Correct Answer - c | |
| 571. |
For high frequency, a capacitor offersA. More reactanceB. Less reactanceC. Zero reactanceD. Infinite reactance |
| Answer» Correct Answer - b | |
| 572. |
The r.m.s current in an `AC` circuit is `2A`. If the wattless current be `sqrt(3)A`, what is the power factor?A. `1/sqrt(3)`B. `1/sqrt(2)`C. `1/2`D. `1/3` |
| Answer» Correct Answer - c | |
| 573. |
In an `LCR` a.c. circuit at resonance, the currentA. Is always in phase with the voltageB. Always leads the voltageC. Always lags behind the voltageD. May lead or lag behind the voltage |
| Answer» Correct Answer - 1 | |
| 574. |
An inductor `20xx10^(-3)` a capacitor `100(mu)F` and a resistor `50 Omega` are connected in series across a source of emf `V=10 sin 314 t`. Then the energy dissipated in the circuit in 20 mim isA. 960 JB. 900 JC. 250JD. 500J |
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Answer» Correct Answer - A `Z^(2)=(X_(C )-X_(L))^(2) + R^(2)=(31.85 - 6.28)^(2) + (50)^(2) = 3154` `P=((E_(rms)^(2))/(Z^(2))) R= ((10//sqrt(2))^(2))/(3154) xx 50 = 0.8 W` Heat produced in 20 min `=(0.8)(20xx60)=960 J` `X_(C )-X_(L) = 31.85 - 2(6.28)=19.29` `I_(m)=(10)/(19.29)=0.52` Hence, `I=0.52 sin (314t+(pi)//(2))=0.52 cos 314 t`. |
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| 575. |
Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values `1000 Omega, 1muF` and 2.0 H respectively . Given emf as, V = 100sqrt2 sin 1000 t V` |
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Answer» The rms value of voltage across the source, `V_(rms) = V_(0)/sqrt(2)` comparing the given equation with general equation, `V_(0) = 100sqrt(2),omega=1000rads^(-1)` Also, R=`1000Omega`, C=`1muF= 1xx10^(-6)`F L=2H `V_(rms)=(100sqrt(2))/(sqrt(2)) = 100V` `therefore I_(rms) = V_(rms)/|Z| = (V_(rms))/(sqrt(R^(2) + (X_(L)-X_(C))^(2)))` = `V_(rms)/(sqrt(R^(2)+(omegaL-1/(omegaC))^(2)))` `=100/sqrt((1000)^(2)+(1000xx2-1/(1000 xx 1xx 10^(-6))))` The current will be same everywhere in the circuit, therefore, PD across resistor, `V_(R) = I_(rms)R=0.0707xx1000 = 70.7V` PD across inductor `V_(L) = I_(rms)X_(L) = 0.0707 xx 1/(1xx1000xx10^(-6)) = 70.7` V Note:" The rms voltages do not add directly as, `V_(R)+V_(L)+V_(C)=282.8`V Which is not the source voltage 100V. The reason is that these voltages are not in phase and cna be added by vector or by phasor algebra. |
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| 576. |
An inductor `20xx10^(-3)` a capacitor `100(mu)F` and a resistor `50 Omega` are connected in series across a source of emf `V=10 sin 314 t`. If resistance is removed from the circuit and the value of inductance is doubled, the variation of current with time in the new circuit is |
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Answer» Correct Answer - `951.52 J;0.52 cos 314 t` Given that `L=20xx10^(-3) H` `C=100xx10^(-6) F` `R=50 Omega , omega=314` `V=10 sin 314 t , Deltat=20xx60` second (i)`DeltaH=I_(rms)^(2)R(Deltat)=(V_(0)/(sqrt2 Z))^(2) RxxDeltat` here `z=sqrt(R^(2)+(omegaL-1/(omegac))^(2))=sqrt((50)^(2)+(6.28-31.8)^(2))=56.15 Omega` (ii)`I=I_(0)sin (314t+pi//2)` where `I_(0)=V_(0)/((omega(2L)-1/(omegac)))` |
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| 577. |
A bulb and a capacitor are in series with an `AC` source. On increasing frequency how will glow of the bulb changeA. increaseB. decreaseC. remain unchangedD. sometimes increases and sometimes decreases |
| Answer» Correct Answer - A | |
| 578. |
An inductance `L` and capacitance `C` and resistance `R` are connected in series across an `AC` source of angular frequency `omega`. If `omega^(2)gt (1)/(LC)` thenA. emf leads the currentB. both the emf and the current are in phaseC. current lead the emfD. emf lags behind the current |
| Answer» Correct Answer - 1 | |
| 579. |
A bulb and a capacitor are in series with an `AC` source. On increasing frequency how will glow of the bulb changeA. The glow decreaseB. The glow increaseC. The glow remains the sameD. Bulb quenches |
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Answer» Correct Answer - B |
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| 580. |
A bulb and a capacitor are in series with an `AC` source. On increasing frequency how will glow of the bulb changeA. The glow decreasesB. The glow increasesC. The glow remain the sameD. The bulb quench |
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Answer» Correct Answer - B This is because, when frequency `v` is increased, the capacitive reactance `X_(C)=1/(2pivC)` decrease and hence the current through the bulb increases |
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| 581. |
A capacitor of capacitance 250 pF is connected in parallel with a choke coil having inductance of `1.6 xx 10^(-2) H` and resistance `20 Omega`. Calculate (a) the resonance frequency and (b) the circuit impedance at resonance. |
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Answer» a) The resonance frequency of a rejector L-C-R circut is given by `f=1/(2pi)sqrt(1/(LC) - R^(2)/L^(2))` `=1/(2pi)sqrt(1/((1.6xx10^(-2))(250xx10^(-12)))-(20)^(2)/(1.6xx10^(-2))^(2))`= `7.96 xx 10^(4) Hz` b) the circuit impedance at resonance is given by, `Z=L/(CR) = (1.6 xx 10^(-2))/(250 xx 10^(-12)xx(20))=3.2 xx 10^(6) Omega` |
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| 582. |
In a series `LCR` circuit with an `ac` source of `50 V,R=300 Omega`,frequency `v=50/piHz`.The average electric field energy, stored in the capacitor and average magnetic energy stored in the coil are `25 mJ` and `5 mJ` respectively.The `RMS` current in the circuit is `0.10 A`.Then find |
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Answer» Correct Answer - (a)`C=20 muF` , (b)`1 H` , (c )`35.36 V` Av. Electric field energy =`(1/2CV_(rms)^(2))=25xx10^(-3) J` `therefore 1/2xxc.I_(rms)^(2)xx1/(2pi^(2)v^(2)c^(2))=25xx10^(-3)J thereforeC=20 muF` Av.magnetic energy `(1/2Li_(rms)^(2))=5xx10^(-3)` `therefore L=(2xx5xx10^(-3))/(0.10)^(2) rArrL=1` henry `V_(R)=I_(rms)R , V_(C)=I_(rms)X_(C) , V_(L)=I_(rms)xxomegal.` `=0.10xx300 , =(0.10)xx1/(2pi(50/pi)xx20xx10^(-6)) ,=0.10xx2pixx50/pi`(1) =`V_(R)=30 V , V_(C)=50 V , V_(L)=10 V` `rms` voltage of source `E_(rms)=50 V` (given in the equation) |
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| 583. |
Transformer is used toA. convert AC to DC voltageB. convert DC to AC voltageC. obtain desired DC powerD. obtain desired AC voltage and current |
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Answer» Correct Answer - D Transformer is used to obtain desired AC voltage and current because a transformer is a device based on the principle of mutual induction, which is used for converting large AC at low voltage into small current at high voltage. |
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| 584. |
Consider the following two statements `A` and `B` and identify the correct answer. `{:A)` At resonance of `L-C` series circuit, the reactance of circuit is minimum. `{:B)` The reactance of a capacitor is an `A.C` circuit is similar to the reactance of a capacitor in a `D.C` circuitA. `A` is true but `B` is falseB. Both `A` and `B` are tureC. `A` is false but `B` is trueD. Both `A` and `B` are false |
| Answer» Correct Answer - 1 | |
| 585. |
Choose the wrong statement of the followingA. The peak voltage acorss the indcutor can be less than the peak voltage of the source in an `LCR` circuitB. In a circuit containing a capacitor and an ac soruce the currentsd is zero at the instant source voltage is maximumC. When an `AC` source is connected to a capacitor, then the rms current in the circuit gets increased if a dielectic slab is inserted into the capacitorD. In a pure inductive circuitd emf will be in phase with the current. |
| Answer» Correct Answer - 4 | |
| 586. |
A dynamo convertsA. mechanical energy into thermal energyB. electrical energy into thermal energyC. thermal energy into electrical energyD. mechanical energy into electrical energy |
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Answer» Correct Answer - D A dynamo convert mechanical energy into electrical energy. |
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| 587. |
A series LCR circuit with a resistance of `100 sqrt(%) Omega` is connected to an ac source of 200 V. When the capacitor is removed from the circuit, current lags behind emf by `45^(@)`. When the inductor is removed from the circuit keeping the capacitor and resistor in the circuit, current leads by an angle of `tan^(-1)((1)/(2))`. Calculate the current and power dissipated in LCR circuit. |
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Answer» When capacitor is removed, `tan phi = (X_(L))/(R) ` (It is L - R circuit) `tan phi = tan 45^(@) = (X_(L))/(R) rArr X_(L)= R` When inductor is removed, `tan phi = (X_(C))/(R) = (1)/(2) = (X_(C))/(R) rArr X_(C)=(R)/(2)` Impendances `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt(R^(2)+(R-(R)/(2))^(2))=(Sqrt(5))/(2) R` Now, `l_("rms")=(E_("rms"))/(Z) = (200)/(250) = 0.8A, cos phi = (R)/(Z) = (2) /(sqrt(5))` `P_(av)=V_("rms")l_("rms")cos phi = 64 sqrt(5) W` |
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| 588. |
The essential difference between a `d.c` dynamo and an `a.c` dynamo is thatA. a.c. has an electromagent but d.c. has a permanent magnetB. a.c. will generate a higher voltageC. a.c. has slip rings but the d.c has a commutatorD. a.c. dynamo has a coil wound on soft iron, but the d.c. dynamo has a coil wound on copper |
| Answer» Correct Answer - 3 | |
| 589. |
Find the potential difference across resistance, capacitance and inductance in series LCR circuit where `L=2.0 H, C = 1 mu F and R = 100 Omega`. The applied emf has the form `V= 100 sqrt(2) sin (1000 t)` |
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Answer» Impendace ` Z= sqrt(R^(2)+(X_(L)-X_(C))^(2)` `=sqrt((1000)^(2)+[1000xx2-(1)/(1000xx1xx10^(-6))]^(2))=1000sqrt(2)Omega` As `l_("rms") X_(L) = 0.0707 xx 1000xx2= 141.4 V` rms voltage across capacitor, `V_(C)=l_("rms")X_(C) = 70.7 V` rms voltage across resistor `V_(R) = l_("rms")R=70.7 V` |
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| 590. |
The unit of impedence isA. ohmB. mhoC. ampereD. volt |
| Answer» Correct Answer - 1 | |
| 591. |
The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity `omega` isA. `(sqrt(R^(2) + omega^(2) L^(2)))/(R )`B. `(R )/(sqrt(R^(2) + omega^(2) L^(2)))`C. `(omega L)/(R )`D. `(R )/(omega L)` |
| Answer» Correct Answer - 2 | |
| 592. |
Assertion : Resonance is exhibited by a circuit only if both L and C are present in the circuit. Reason : Only then the voltage across L and C cancel each other, both being out of phase.A. If both assertion ans reason are true ans reaason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion istrue but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A The resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then the voltages across L and C cancel each other since both being out of phase and current amplitude is `V_(m//R)`, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit. |
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| 593. |
Assertion : The power in ac circuit is minimum if the circuit has only a resistor. Reason : Power of a circuit is independent of the phase angle.A. If both assertion ans reason are true ans reaason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion istrue but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D Power in a series ac circuit consisting of L,C and R is given by `P = I_("rms")V_("rms")cos phi` where `phi = tan^(-1)((|X_(L)-X_(C)|)/(R ))` For a purely resistive circuit `X_(L) = 0` and `X_(C) = 0` Therefore, `tan phi = 0` or `phi = 0` and thereby `cos phi = 1` and P = IV. The power is maximum as `cos phi` is maximum. Power depends on the phase angle through the power factor `cos phi`. |
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| 594. |
Assertion: When `AC` circuit contain resistor only, its power is minimum. Reason: Power of a circuit is independent of phase angle.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the a ssertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - d | |
| 595. |
In an `AC` circuit, peak value of voltage is `423` volts. Its effective voltage isA. `400` voltsB. `323` voltsC. `300` voltsD. `340` volts |
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Answer» Correct Answer - C Effective voltage `V_(rms)=(V_(0))/(sqrt(2))=423/(sqrt(2))=300 V` |
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| 596. |
An air core coil and an electric bulb are connected in series with an AC source. If an iron rod is put in the coil, then the intensity of light of the bulb willA. remains sameB. increasesC. decreaseD. first decrease then increase |
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Answer» Correct Answer - C If an iron rod is puts in the coil, then the intensity of light is of bulb will decrease because the iron rod increase the resistance. |
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| 597. |
For a series L-C-R circuit with L=1.00 mH, C=10`muF` and `R=50Omega`, is driven with 5V AC voltage. At resosance, the current through the circuit isA. 0.2 AB. 0.25 AC. 0.15 AD. 0.1 A |
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Answer» Correct Answer - D We know that for series L-C-R in circuit. `V_(rms) = sqrt(V_(R)^(2) + (V_(L)-V_(C)^(2)))` Given, `V_(L)`= 90V, `V_(C)` = 60V and `V_(R)` = 40 V Now, `V_(rms) = sqrt((40)^(2)+(90-60)^(2))`=50 V |
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| 598. |
An electron lamp is conected to `220V, 50Hz` supply. Then the peak value of voltage isA. 210 VB. 211 VC. 311 VD. 320 V |
| Answer» Correct Answer - c | |
| 599. |
The voltage of domestic ac is 220 volt . What does this representA. Mean voltageB. Peak voltageC. Root mean voltageD. Root mean square voltage |
| Answer» Correct Answer - d | |
| 600. |
In a circuit, the value of the alternating current is measured by hot wire ammeter as `10` ammeter. Its peak value will beA. `10 A`B. 20 AC. 14.14 AD. 7.07 A |
| Answer» Correct Answer - c | |