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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.
| 751. |
STATEMENT-1 : A resistor opposes the flow of current . and STATEMENT-2 : An inductor opposes change in strength of current.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 752. |
An ac having a peak value 1.41 A is used to heat a wire. A dc producing the same heating rate will be approximatelyA. 1.41 AB. 2.0 AC. 0.705 AD. 1.0 A |
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Answer» Correct Answer - D The rms value of the alternating current, `I_(rms)=I_0/sqrt2=1.41/1.41=1A` Hence, the dc required to produce the same heating rate =1 A. |
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| 753. |
Find the reactance of a capacitor `(C = 200 muF)` when it is connected to (a) `a 10 Hz AC source`, (b) `a 50 Hz AC` source and ( c) `a 500 Hz AC` source. |
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Answer» `(a) X_(c)=(1)/(omegaC)=(1)/(2pifC)cong 80Omega "For" f=10Hz "AC source"` `(b) X_(c)=(1)/(omegaC)=(1)/(2pifC)cong 16Omega "For" f=50Hz and` `(c) X_(c)=(1)/(omegaC)=(1)/(2pifC)cong 1.6Omega "For" f=500Hz` |
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| 754. |
A series L-C-R circuit is connected across a variable frequency source of emf. `V_(L), V_(C), V_(R)` are instantaneous voltage drops across inductor capacitor and resistor and V is instantaneous voltage across the circuit. `{:(" Column I",," Column II"),((A) |V|,,(p) "Can be greater than zero"),((B) |V|-|V_(L)|,,(q) "Can be less than zero"),((C) |V|-|V_(C)|,,(r)"Can be equal to zero"),((D) |V|-|V_(R)|,,):}` |
| Answer» Correct Answer - A(p,r), B(p,q,r), C(p, q, r), D(p, q, r) | |
| 755. |
An ac voltage `e=E_0sinomegat` is applied across an ideal inductor of self-inductance [L]. Write down the peak current. |
| Answer» Peak value of the current = `E_0/(omegaL)` [Inductive reactance =`omegaL`] | |
| 756. |
Find the rms value of current from t=0 to t=`(2pi)/(omega)` if the current avries as `i=I_(m)sin omegat`. |
| Answer» `i_(rms) =sqrt((underset((pi)/(2omega))overset((2pi)/(omega))int I_(m)^(2) sin omegatdt)/((2pi)/(omega)))= sqrt((i_(m)^(2))/(2))=(I_(m))/(sqrt2)` | |
| 757. |
In the series LCR circuit , the voltmeter and ammeter reading are: A. `V = 100` volt, `I = 2A`B. `V = 100` volt, `I = 5 A`C. `V = 1000` volt, `I = 2A`D. `V = 300` volt, `I = 1 A` |
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Answer» Correct Answer - 1 `I_("r.m.s") = (V_("r.m.s"))/(Z) = (V_("r.m.s"))/(R ) = (100)/(50) = 2 A` `V = sqrt(V_(R)^(2) + (V_(L) - V_(C ))^(2))` |
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| 758. |
In the series LCR circuit , the voltmeater and ammeter reading are: A. `V=100V, I=2A`B. `V=100V, I=5A`C. `V=1000V, I=2A`D. `V=300V, I=1A` |
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Answer» Correct Answer - A Here `(V_L) = (V_C)` , they are in opposite phaes. Hence, they will cancel each other. Now the resultant potential difference is equal to the applied potential difference =100V `Z=R(:. X_(L)=X_(C ))` `:. I_(rms)=(V_(rms))/(Z) =(V_(rms))/(R ) =(100)/(50) = 2A`. |
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| 759. |
For the circuit shown A. Current in circuit in 10AB. Voltage across inductor is 100VC. Voltage acros capacitor is less than that of supply voltageD. Voltage across capacitor is more than that of supply voltage |
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Answer» Correct Answer - D |
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| 760. |
Current in an ac circuit is given by `i=2 sqrt(2) sin [(pi t+((pi)//4)]`. Then find the average value of current during time t=0 to t=1 s. |
| Answer» `ltigt =(underset(0)overset(1)int idt)/1=2sqrt2 underset(0)overset(1)int sin (pit+pi/4)=4/pi` | |
| 761. |
In the series LCR circuit as shown in figure, the heat developed in 80 seconds and amplitude of wattless current is : A. 4000J,5AB. 8000J, 3AC. 4000J, 4AD. 8000J,5A |
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Answer» Correct Answer - A |
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| 762. |
The current flowing in a wire fluctuates sinusoidally as shown in the diagram. The root mean square valuie of the current is A. `i_(0)((1)/(2)+1)^(2)`B. `i_(0)(sqrt2+1)^(t//2)`C. `2sqrt2 t_(0)`D. `i_(0)((2sqrt2+1)/(2))^(t//2)` |
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Answer» Correct Answer - A |
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| 763. |
There is a conducting ring of radius `R`. Another ring having current `i` and radius `r (r lt lt R)` is kept on the axis of bigger ring such that its center lies on the axis of bigger ring at a distance `x` from the center of bigger ring and its plane is perpendicular to that axis. The mutual inductance of the bigger ring due to the smaller ring isA. `(mu_(0)piR^(2)r^(2))/((R^(2)+x^(2))^(3//2)`B. `(mu_(0)piR^(2)r^(2))/(4(R^(2)+x^(2))^(3//2)`C. `(mu_(0)piR^(2)r^(2))/(16(R^(2)+x^(2))^(3//2)`D. `(mu_(0)piR^(2)r^(2))/(2(R^(2)+x^(2))^(3//2)` |
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Answer» Correct Answer - D |
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| 764. |
A resistance (R), inductance (L) and capacitance (C) are connected in series to an ac source of voltage V having variable frequency. Calculate the energy delivered by the source to the circuit during one period if the operating frequency is twice the resonance frequency. |
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Answer» Correct Answer - `(piRsqrt(LCV)^(2))/(R^(2)+2.25(L)/(C))` |
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| 765. |
In the circuit shown the source voltage is given as` v = V_(0)` sin `omegat`. Find the current through the source as a function of time. |
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Answer» Correct Answer - `i=V_(0)[(1)/(R)sinomegat+(omegaC-(1)/(omegaL))cosomegat]` |
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| 766. |
A box has a large electric circuit inside it. When it was connected to an ac generator it was found that it was putting a lot of load on the generator and the power factor of the box was `(1)/(sqrt(2)).` A capacitor of capacitance C was connectedin series with the box and the power factor of the circuit became equal to the ideal value. Find the impedance of the box. The generator has an angular frequency of`omega.` |
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Answer» Correct Answer - `(sqrt(2))/(omegaC)` |
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| 767. |
Following figure shows an `AC` generator to a "block box" through a pair of terminals. The box contains possible `R,L,C` or their combination, whose elements and arrangements are not known to us. Measurements, outside the box reveals that `e=75 sin(omega t) `volts, `i=1.5 sin(omegat+45^(@))` amp then. the wrong statements isA. There must be a capacitor in the boxB. There must be an inductor in the boxC. There must be a resistance in the boxD. The power factor is 0.707 |
| Answer» Correct Answer - b | |
| 768. |
For an ac circuit `V=15 sin omega t` and `I=20 cos omega t` the average power consumed in this circuit isA. 300 wattB. 150 wattC. 75 wattD. zero |
| Answer» Correct Answer - d | |
| 769. |
A bulb is connected first with `DC` and the then `AC` of same voltage then it will shine brightly withA. ACB. DCC. Brightness will be in ratio 1/1.4D. Equally with both |
| Answer» Correct Answer - d | |
| 770. |
The instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance are given by `I = sqrt(2)sin (100 t-pi//4)A` and `upsilon = 40 sin(100t)` V. Calculate the resistance ? |
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Answer» `i=sqrt(2)sin (100t-pi//4)A" " (because i=i_(0)sin(omega t-phi))` `upsilon = 40 sin (100 t)V " " (because V = V_(0)sin(omega t))` `i_(0)=sqrt(2), V_(0)=40, omega = 100, phi = pi//4` `R = (V_(0))/(i_(0))cos phi = (40)/(sqrt(2))cos.(pi)/(4)` `R=(40)/(sqrt(2))xx(1)/(sqrt(2))` `R = 20 omega` |
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| 771. |
The resonant frequency of a circuit is `f`. If the capacitance is made `4` times the initial values, then the resonant frequecy will becomeA. f/2B. 2fC. fD. f/4 |
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Answer» Correct Answer - A Resonant frequency , f = 1/`2pi``sqrt LC` f `propto` 1/`sqrt C` or `underset(1)(f)` / `underset(2)(f)` = `sqrt underset(2)(C )/underset(1)(C )` `implies` f/`underset(2)(f)` = `sqrt 4C/C ` `implies` `underset(2)(f)` = f/2 |
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| 772. |
The resonant frequency of a circuit is `f`. If the capacitance is made `4` times the initial values, then the resonant frequecy will becomeA. `f//2`B. `2f`C. `f`D. `f//4` |
| Answer» Correct Answer - a | |
| 773. |
The potential differences `V` and the current `i` flowing through an instrument in an `AC` circuit of frequency `f` are given by `V=5 cos omega t` and `I=2 sin omega t` amperes (where `omega=2 pi f`). The power dissipated in the instrument isA. zeroB. `10W`C. `5W`D. `2.5W` |
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Answer» Correct Answer - A `V =5 cos omega t = 5 sin (omega t + pi//2)` `I = 2 sin omegat` phase different between `V` and I `phi = pi//2` `P =0` . |
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| 774. |
Which quantity in an ac circuit is not dependent on frequency?A. resistanceB. impedanceC. inductive reactanceD. capacitative reactance |
| Answer» Correct Answer - A | |
| 775. |
The series resonant frequency of an LCR circuit is f. IF the capacitance is made 4 times the initial value, then the resonant frequency will comeA. `f//2`B. 2fC. fD. `f//4` |
| Answer» Correct Answer - A | |
| 776. |
The condition of getting maximum current in an LCR series circuit isA. `X_L=0`B. `X_C=0`C. `X_L=X_C`D. `R=X_L-X_C` |
| Answer» Correct Answer - C | |
| 777. |
In an `LR` circuit, `R = 100 Omega` and `L = 2 H`. If an alternating voltage of `120 V` and `60 Hz` is connected in this circuit, then the value of current flowing in it will be ______ `A` nearlyA. 0.32B. 0.16C. 0.48D. `0.8` |
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Answer» Correct Answer - 2 `I = (E)/(Z) = (E)/(sqrt(R^(2) 4 pi^(2) f^(2) L^(2)))` |
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| 778. |
The frequency of ac mains in India isA. 30 c/s or HzB. 50 c/s or HzC. 60 c/s or HzD. 120 c/s or Hz |
| Answer» Correct Answer - b | |
| 779. |
A series `R-C` circuit is connected to an alternating voltage source. Consider two situations (a) When capacitor is air filled. (b) When capacitor is mica filled. current through resistor is `i` and voltage across capacitor is `V` thenA. `V_(a)ltV_(b)`B. `V_(a)gtV_(b)`C. `i_(a)gti_(b)`D. `V_(a)=V_(b)` |
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Answer» Correct Answer - B |
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| 780. |
A transformer having efficiency of `90%` is working on `200 V` and `3 kW` power supply. If the current in the secondary coil is `6A`, the voltage across the secondary coil and current in the primary coil respectively areA. `300 V,15 A`B. `450 V,15 A`C. `450 V,13.5 A`D. `600 V,15 A` |
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Answer» Correct Answer - B We know efficiency `eta=(Output)/(Input)=(V_(s)I_(s))/(V_(p)I_(p))` `implies eta=(V_(s)I_(s))/(V_(p)I_(p)) implies 0.9=(V_(s)(6))/(3xx10^(3))` `implies V_(s)=450 V` As `V_(p)I_(p)=3000` `I_(p)=3000/(V_(p))=3000/200 A=15 A` |
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| 781. |
r.m.s. value of current i=3+4sin `(omegat+pi//3) isA. 5AB. `sqrt2A`C. `(5)/(sqrt2A)`D. `(7)/(sqrt2)A` |
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Answer» Correct Answer - B |
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| 782. |
When the current in the portion of the circuit shown in the figure is `2A` and increasing at the rate of `1A//s`,the measured potential difference `V_(a)-V_(b)=8V`.However when the current is `2A` and decreasing at the rate of `1A//s`, the measured potential difference `V_(a)-V_(b)=4V`.The values of `R` and `L` are: A. `3Omega` and 2H, respectivelyB. `2Omega` and 3H, respectivelyC. `3Omega` and 2H, respectivelyD. |
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Answer» Correct Answer - A |
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| 783. |
Which of the following plots may represent the reactance of a series `LC` combination? A. aB. bC. cD. d |
| Answer» Correct Answer - d | |
| 784. |
A choke coil of resistance `R` and inductance `L` is connected to `A.C` sourceof frequency `f` and maximum voltage `V_(0)`. Then, the average power dissipated in the choke is proportional to:A. `f^(2)`B. `f^(-2)`C. `f^(1)`D. `f^(0)` |
| Answer» Correct Answer - D | |
| 785. |
In given LR circuit the switch S is closed at time `t=0` thenA. The ratio of induced emfs in the inductors of inductances L and 2L will be correctB. The ratio of indued emfs in the inductor of inductances L and 2L will decrease with timeC. The potential difference `V_(A)-V_(B)` increase with timeD. The potential difference `V_(A)-V_(B)` will be constant |
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Answer» Correct Answer - A::D |
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| 786. |
A power outlet puts out 60 Hz AC. Which of the following statements is true ?A. Voltage goes to zero only 120 times per secondB. Current goes to zero only 60 times per secondC. Power goes to zero, only 60 times per secondD. Voltage , current and power stay constant |
| Answer» Correct Answer - A | |
| 787. |
In the circuit as shown in the figure, switch S is added at t=0. Then: A. after a long time interval potential differences across capacitor and inductor will be equalB. after a long time interval charge on a capacitor will be ECC. after a long time interval curent in the inductory will be E/RD. after a long time interval current through battery willl be same as the curren through it initially |
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Answer» Correct Answer - D |
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| 788. |
Electric circuit is composed of three conducting rods MO, ON and PQ as shown in the figure. The resistance of the rods per unit length is known to be 1. The rod PQ slides as shown in the figure. At t=0, rod PQ is at O. The whole system is embledded ina uniform magnetic field B, which is directed perpendicularly into page. The induced electric current is: A. Proportional to time tB. Inversibly proportional to time tC. Proportional to square at time tD. Independent of time t |
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Answer» Correct Answer - D |
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| 789. |
In the shown AC circuit phase different between current `I_(1)` and `I_(2)` is A. `(pi)/(2)-tan^(-1).(x_(L))/(R)`B. `tan^(-1)-(X_(L)-X_(C))/(R)`C. `(pi)/(2)+tan^(-1).(x_(L))/(R)`D. `tan^(-1).(X_(L)-X_(C))/(R)+(pi)/(2)` |
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Answer» Correct Answer - C |
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| 790. |
In series LCR AC circuit, the voltage of the source at any instant is equal toA. The sum of the maximum voltage across the elementsB. The voltage drop across the resistorC. The sum of the instantaneous voltages across the elementsD. The sum of the rms voltages across the elements |
| Answer» Correct Answer - C | |
| 791. |
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. `100Omega`B. `25Omega`C. `sqrt(125xx75)"Omega"`D. `400Omega` |
| Answer» Correct Answer - B | |
| 792. |
A Capacitor and a coil in series are connected to a 6volt ac source. By varying the frequency of the source, maximum current of 600mA is observed. If the same coil is now connected toa cell of emf 6volt dc and internal resistance of 2ohm, the current h through it will beA. 0.5AB. 0.6AC. 1.0AD. 2.0A |
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Answer» Correct Answer - A |
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| 793. |
An inductor L, a resistanc eR and two identical bulbs `B_(1) and B_(2)` are connected to a battery through a switch S as shown in the figure . The resistance of coil having inductance L is also R. Which of the following statement gives the corrrect description of the happenings when the switch S is closed? A. `B_(2)` lights earlier than `B_(1)` and finally both the bulbs shine equally bright.B. `B_(1)` light up earlier and finally both the bulbs acquire equal brightnesss,C. `B_(2)` lights up earlier and finally `B_(1)` shines brighter than `B_(2)`D. `B_(2)` lights up together with equal brightness all the time. |
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Answer» Correct Answer - A |
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| 794. |
In the circuit shown in figure, if both the bulbs `B_(1)` and `B_(2)` are identical: A. their brightness will be the sameB. `B_(2)` will be brighter than BC. as frequency of supply voltage is increased the brightness of bulb BD. only `B_(2)` will glow because the capacitor has infinite impedance |
| Answer» Correct Answer - B::C | |
| 795. |
An ac voltage is represented by `E=220 sqrt(2) cos (50 pi) t` How many times will the current become zero in 1 s?A. (A) 50 timesB. (B) 100 timesC. (C) 30 timesD. (D) 25 times |
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Answer» Correct Answer - A `E=(E_0) cos omega t` `:. Omega =50 pi` `2 pi f = 50 pi implies f=25 Hz` In one cycle ac current becomes zero twice. therefore, 50 times the current becomes zero in 1 s. |
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| 796. |
A pure inductance of 1.0H is connected across a 110V, 70Hz Source. Find the (a) reactanace, (b) current and (c ) peak value of current.A. reactance of the circuit is `440 Omega`B. current of the circuit is `0.25 Omega`C. reactance of the circuit is `880 Omega`D. current of the circuit is `0.5 Omega` |
| Answer» Correct Answer - A,B | |
| 797. |
In an AC series circuit, the instanctaneous current is zero when the instantaneous voltage is xamimum. Connected to the source may be aA. pure capacitorB. pure inductorC. combination of pure an inductor and pure capacitorD. pure resistor |
| Answer» Correct Answer - A::C | |
| 798. |
A pure inductance of 1.0H is connected across a 110V, 70Hz Source. Find the (a) reactanace, (b) current and (c ) peak value of current.A. reactance of the circuit is `440Omega`B. for `omega=500s^(-1)"peak current" is 4AC. reactance of the circuit is`880Omega`D. for `omega=1000s^(-1)"peak current" is 4A |
| Answer» Correct Answer - A::B::C::D | |
| 799. |
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=100s^(-1)"peak current is 20A"`B. `"for" omega=500s^(-1)"peak current is 4A"`C. `"for" omega=500s^(-1)"peak current is 2A"`D. `"for" omega=1000s^(-1)"peak current is 4A"` |
| Answer» Correct Answer - A::B::C::D | |
| 800. |
An alternating emf is applied across a parallel combination of a resistance `R`, capacitance `C` and an inductance `L`. If `I_(R),I_(L),I_(C)` are the currents through `R,L` and `C` respectively. Then the diagram which correctely represents, the phase relationship among `I_(R),I_(L),I_(C)` and source emf `E`m is given byA. B. C. D. |
| Answer» Correct Answer - C | |