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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.
| 501. |
The rms value of potential difference `V` shown in the figure is A. `V_(0)`B. `(V_(0))/(sqrt(2))`C. `(V_(0))/2`D. `(V_(0))/(sqrt(3))` |
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Answer» Correct Answer - B `V_(rms)=[1/Tint_(0)^(T//2) V_(0)^(2) dt]^(1//2)=[(V_(0)^(2))/T [t]_(0)^(T//2)]^(1//2)` `=[(V_(0)^(2))/T(T/2)]^(1//2)` `V_(rms)=(V_(0))/(sqrt(2))` |
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| 502. |
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. `1sqrt2`C. 1D. `sqrt3//2` |
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Answer» Correct Answer - C |
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| 503. |
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 factor of the circuit isA. `1/2`B. `1/(sqrt(2))`C. `1`D. `(sqrt(3))/2` |
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Answer» Correct Answer - C We define phase difference, `tan phi=(X_(L)-X_(C))/R` When `L` is removed, `tan ((pi)/3)=sqrt(3)=(X_(C))/R` `implies X_(C)=sqrt(3)R` When `C` is removed, `tan(pi/3)=sqrt(3)=(X_(L))/R` `implies X_(L)=Rsqrt(3)` Hence, in resonant circuit, `tan phi=(X_(L)-X_(C))/R=0` `implies tan phi=(sqrt(3)R-sqrt(3)R)/R=0` `:.` Power factor `cos phi=1` |
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| 504. |
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 Here, phase difference, `tanphi = (X_(L)-X_(C))/R` `tan pi/3 = (X_(L)-X_(C))/R` When L is removed, `sqrt(3) = X_(C)/R rArr X_(C) = sqrt(3)R` When C is removed, `tanpi/3 = sqrt(3) = X_(L)/R` `X_(L) = Rsqrt(3)` Hence, in resonant |
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| 505. |
Power factor is one forA. (A) pure resistorB. (B) pure inductorC. (C) pure capacitorD. (D) either an inductor or a capacitor |
| Answer» Correct Answer - A | |
| 506. |
An `AC` voltage source `V=V_0siomegat`is connected across resistance `R` and capacitance `C` as shown in figureure. It is given that `R=1/omegaC`. The peak current is `I_0`. If the angular frequency of the voltage source is changed to `omega/sqrt3,` then the new peak current in the circuit is .A. `i_(0)/2`B. `i_(0)/sqrt2`C. `i_(0)/sqrt3`D. `I_(0)/3` |
| Answer» Correct Answer - B | |
| 507. |
An electric appliance draws 3 A current from a 200 V, 50 Hz power supply. (a) Find the average of square of the current. (b) Find the amplitude of the supply voltage. |
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Answer» Correct Answer - `(a) 9A^(2) (b)282 V` |
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| 508. |
For an AC circuit containing only, the applied AC voltage waveform is shown in figure. For this situation , mark the correct stament(s). A. As V increases from a to b, the charging of capacitor rakes placeB. As V increases from a to b ,the current is circuit decreases from maximum to zero valueC. As V decreases from b to c, the capacitor dischargesD. As V decreases from b to c charging of capacitor takes plaece |
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Answer» Correct Answer - A::B::C::D V = `V_o` sin `omega`t q = CV = C`V_o`sin `omega`t = `q_o`sin `omega`t I = dq / dt = `q_o``omega`cos`omega`t = `i_o`cos`omega`t |
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| 509. |
Current in the circuit is wattless, ifA. Inductance in the circuit is zeroB. Resistance in the circuit is zeroC. Current is alternatingD. Resistance and inductance both are zero |
| Answer» Correct Answer - b | |
| 510. |
Choke coil is used to controlA. acB. dcC. Both ac and dcD. Neither ac nor dc |
| Answer» Correct Answer - a | |
| 511. |
The core used in a transformer and other electromagnetic devices is laminated so thatA. ratio of voltage in the primary and secondary may be increasedB. energy loss due to eddy currents may be minimisedC. the weight of the transformer may be reducedD. residual magnetism in the core may be reduced |
| Answer» Correct Answer - B | |
| 512. |
The core of a transformer is laminated to reduceA. eddy current lossB. hysteresis lossC. copper lossD. magnetic loss |
| Answer» Correct Answer - A | |
| 513. |
Choke coil works on the principle ofA. Transient currentB. Self inductionC. Mutual inductionD. Wattless current |
| Answer» Correct Answer - b | |
| 514. |
A transformer works on the principle ofA. self-inductionB. electrical inertiaC. mutual inductionD. magnetic effect of the electrical current |
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Answer» Correct Answer - C A transformer works on the principle of mutual induction |
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| 515. |
A trensformer works on the principle ofA. self inductionB. electrical inertiaC. mutual inductionD. magnetic effect of the electricl current |
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Answer» Correct Answer - C A transformer is based on the principleof mutual induction. |
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| 516. |
The transformer ratio of a transformer is 5. If the primary voltage of the transformer is `400 V, 50 Hz` the secondary voltage will beA. `2000 V, 250 Hz`B. `80 V, 50 Hz`C. `80 V, 10 Hz`D. `2000 V, 50 Hz` |
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Answer» Correct Answer - 4 Frequency remains same. `(V_(s))/(V_(P)) = 5` |
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| 517. |
Statement (A) : In high current low voltage windings of a transformer thick wire is used to minimize energy loss due to heat produced Statement (B) : The core of any transformer is laminated so as to reduce the erergy loss due to eddy currents.A. `A` is true but `B` is falseB. Both `A` and `B` is trueC. `A` is false but `B`d is trueD. Both `A` and `B` are false |
| Answer» Correct Answer - 2 | |
| 518. |
A step-up transformer works on `220 V` and gives `2 A` to an external resistor. The turn ratio between the primary and secondary coils is 2 : 25. Assuming 100% efficiency, find the secondary voltage, primary current and power delivered respectivelyA. `2750 V, 25 A, 5500 W`B. `2750 V, 20 A, 5000 W`C. `2570 V, 25 A, 550 W`D. `2750 V, 20 A, 55 W` |
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Answer» Correct Answer - 1 `(E_(s))/(E_(P)) = (N_(s))/(N_(p)) = (i_(p))/(i_(s)), P = E_(s) i_(s)` |
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| 519. |
To reduce the iron losses in a transformer, the core must be made of a material havingA. low permeability and high resistivityB. high permeability and high resistivityC. low permeability and low resisttivityD. high permeability and low resistivity |
| Answer» Correct Answer - 2 | |
| 520. |
The loss of energy in the form of heat in the iron core of a transformer isA. iron lossB. copper lossC. mechanical lossD. None of these |
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Answer» Correct Answer - A Iron loss is the energy loss in the form of heat due to the formatoin of eddy currents in the iron core of the transformer. |
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| 521. |
Statement (A) : Flux leakage in a transformer can be minimized by winding the primary and secondary coils one over the other. Statement (B) : Core of the transformer is made of soft ironA. `A` is true but `B` is falseB. Both `A` and `B` is trueC. `A` is false but `B`d is trueD. Both `A` and `B` are false |
| Answer» Correct Answer - 4 | |
| 522. |
For an ideal step-down transformer, the quantity which is constant for both the coils isA. current in the coilsB. voltage across the coilsC. resistance of coilsD. power in the coils |
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Answer» Correct Answer - D For an ideal step-down transformer, power is constant for both the coils. i.e., Input power = Output power `therefore V_(p)I_(p) = V_(s)I_(s)` or `(V_(p))/(V_(s))=(I_(s))/(I_(p))` |
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| 523. |
An AC source of frequency `omega` when fed into a RC series circuit, current is recorded to be l. If now frequency is changed to `(omega)/(4)` (keeping voltage same), the current is found to 1/2. The ratio of reactance to resistance at original frequency `omega` isA. 2B. `1/2`C. `(1)/(sqrt(2))`D. None of these |
| Answer» Correct Answer - B | |
| 524. |
The average value of a saw-tooth voltage `V=V_(0) ((2t)/(T)-1)` Over 0 to `(T)/(2)`, (T- time period) is given byA. `(V_(0))/(sqrt(2))`B. `(2V_(0))/(pi)`C. `(V_(0))/(2)`D. `V_(0)` |
| Answer» Correct Answer - C | |
| 525. |
The average value for the saw-tooth voltage of peak value of `V_(0)` over half the cycle as shown in figure is A. `(V_(0))/(sqrt(3))`B. `(V_(0))/(sqrt(2))`C. `(2V_(0))/(3)`D. `(V_(0))/(3)` |
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Answer» Correct Answer - D `V_("mean") = (int v dt)/(int dt), V = (2V_(0))/(T) t - V_(0)` |
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| 526. |
An ac ammeter is used to measure currnet in a circuit. When a given direct current passes through the circuit. The ac ammeter reads 3 A. When another alternating current passes through the circuit, the ac ammeter reads 4A. Then find the reading of this ammeter (inA), if dc and ac flow through the circuit simultaneously.A. 3AB. 4AC. 7AD. 5A |
| Answer» Correct Answer - D | |
| 527. |
A coll has an inductance of `(2.2)/(pi)H` and is joined m series with a resistance of `220Omega`. When an alternating e.m.f. of 220 V at 50 c.p.s. is applied to it, then the wattless component of the runs current in the circuit isA. 5ampereB. 0.5ampereC. 0.7ampereD. 7ampere |
| Answer» Correct Answer - B | |
| 528. |
The natural frequency of the circuit shown in the figure is A. `(1)/(2pi sqrt(LC))`B. `(1)/(pi sqrt(LC))`C. `(2)/(pi sqrt(LC))`D. none |
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Answer» Correct Answer - A `L_(eq) = L + L = 2 L` `C_(eq) = (C_(1) C_(2))/(C_(1) + C_(2)) = (C^(2))/(2C) = (C )/(2)` `f = (1)/(2 pi sqrt(L_(eq)C_(eq))) = (1)/(2 pi sqrt(2 L xx (C )/(2))) = (1)/(2 pi sqrt(LC))` |
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| 529. |
If the phase difference between voltage and current is `pi//6` and the resistance in the circuit is `sqrt(300) Omega`, then the impedance of the circuit will beA. `40 Omega`B. `20 Omega`C. `50 Omega`D. `13 Omega` |
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Answer» Correct Answer - B `cos phi = (R )/(|Z|)` or `(sqrt(3))/(2) = (sqrt(300))/(|Z|)` or `Z = 20 Omega` |
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| 530. |
For the circuit shown in the figure, the current through the inductor is `0.9A` while the current through the condenser is `0.4 A.` hence the current drawn from the generator is A. `I=1.13` ampB. `I=0.9` ampC. `I=0.5` ampD. `I=0.6` amp |
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Answer» Correct Answer - C The current drawn by inductance and capacitor will be in opposite phase. Hence, net current drawn by the generator `=I_(L)-I_(C)=0.9-0.4=0.5` amp |
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| 531. |
A typical light dimmer used to dim the stage lights in a theater consists of a variable induction for L (where inductance is adjustable between zero and `L_(max)`) connected in series with a light bulb B as shown in fig. the mains electrical supply is 220 V at 50 Hz, the light bulb is rated at 220 V, 1100 W. What `(L_(max))` is required if the rate of energy dissipated in the light bulb is to be varied by a factor of 5 from its upper limits of 1100 W? A. `0.69 H`B. `0.28 H`C. `0.38 H`D. `0.56 H` |
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Answer» Correct Answer - B For power to be consumed at the rate of `1100/5=220 W`, we have `P=E_(v)I_(v)cos theta` `220=(220xx220)/(sqrt(R^(2)+L^(2)omega^(2)))xxR/(sqrt(R^(2)+L^(2)omega^(2)))` where `R=(V^(2))/P=(220^(2))/1100=44 Omega` `220=((220)^(2)xx44)/(44^(2)+(Lomega)^(2)) , 44^(2)+(Lomega)^(2)=22xx44` `(Lomega)^(2)=sqrt(220xx44-44^(2))` `=sqrt(44(220-44))=sqrt(44xx176)=88 Omega` `L=88/(2pixxf)=88/(2pixx50)=88/(2xx22)xx7/50=0.28 H` |
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| 532. |
An inductor `L`, a capacitor `C` and `AC` ammeters `A_(1),A_(2)` and `A_(3)` are connected in the circuit as shown. When the frequency of the oscillators is increased, then at resonant frequency, the reading of ammeter ? A. `A_(1)` zeroB. `A_(2)` zeroC. `A_(3)` zeroD. is same in `A_(1),A_(2)` and `A_(3)` |
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Answer» Correct Answer - A At resonant `omegaL=1/(omegaC)`. So, reading of `A_(1)` is zero. |
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| 533. |
A direct current of `5` amp is superimposed on an alternating current `I=10 sin omega t` flowing through a wire. The effective value of the resulting current will be:A. `(15//2)` ampB. `5sqrt(3)` ampC. `5sqrt(5)` ampD. `15`amp |
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Answer» Correct Answer - B Given `1=5+10 sin omegat` `I_(eff)=[(int_(0)^(T)I^(2)dt)/(int_(0)^(T) dt)]^(1//2)` `=[1/Tint_(0)^(T)(5+10 sin omega t)^(2) dt]^(1//2)` `=[1/Tint_(0)^(T) (25+100 sin omegat+100 sin^(2) omegat)]^(1//2)` But as `1/Tint_(0)^(T) sin omega t dt=0` and `1/Tint_(0)^(T) sin^(2) omega t dt=1/2` so `I_(eff)=[25+1/2xx100]^(1//2)` `=5sqrt(3)` amp |
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| 534. |
A direct current of `5` amp is superimposed on an alternating current `I=10 sin omega t` flowing through a wire. The effective value of the resulting current will be:A. `7.5A`B. `2sqrt3A`C. `5sqrt3A`D. `15A` |
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Answer» Correct Answer - C `I =5 + 10 sin omegat` `bari^(2) = 5^(2) + 10^(2)/(2) =75` `i_(rms) = sqrt(bari^(2)) = 5 sqrt3 A` . |
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| 535. |
An arc lamp consumes `10 A` at `40 V`. Calculate the power factor when it is connected with a suitable value of choke coil required to run the arc lamp on `AC` mains of `200 V ("rms")` and `50 Hz`. |
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Answer» Correct Answer - B `R=V/I=40/10=4Omega` `Z=V/I=200/10=20Omega` Power factor, `cosphi=R/Z=4/20` `=0.2` |
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| 536. |
An electric lamp which runs at `100 V DC` and consumes `10 A` current is connected to `AC` mains at `150 V, 50 Hz` cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of choke. |
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Answer» Correct Answer - A::C `V_L=sqrt(V^2-V_R^2)` `=sqrt((150)^2-(100)^2)` `=111.8V` `V_L=IX_L=I(2pifL)` `:. L=V_L/(2pifI)=111.8/(2pixx50xx10)` `=0.036H` |
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| 537. |
In the adjoining figure the impedence of the circuit will be A. `120` ohmB. `50` ohmC. `60` ohmD. `90` ohm |
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Answer» Correct Answer - C `i_(L)=90/30=3A. i_(C)=90/20=4.5A` Net current through circuit `i=i_(C)-i_(L)=1.5A` `:. Z=V/i=90/1.5=60 Omega` |
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| 538. |
A bulb is designed to operate at `12` volts constant direct current.If this bulb is connected to an alternating current source and gives same brightness.What would be the peak voltage of the source? |
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Answer» Correct Answer - `12sqrt2` volts `V_(@)=V_(rms)xxsqrt2=12sqrt2` , `V_(@)=V_(rms)xxsqrt2=12sqrt2` |
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| 539. |
10 kJ/s of heat is produced when a DC flows through a `100 Omega` resistor. Which of the following ac produces nearly same heat per sec through the same resistor ?A. `l = 15 sin omega t`B. `l = 45 sin omega t`C. `l = 30 sin omega t`D. `l=10 sin omega t` |
| Answer» Correct Answer - A | |
| 540. |
Alternating current can be measured byA. moving coil galvanometerB. hot wire ammeterC. tangent galvanometerD. none of the above |
| Answer» Correct Answer - B | |
| 541. |
When a choke is connected in series with a lamp in dc line, the lamp shines brightly. Insertion of an iron core in the choke does not affect the brightness. What happens in case of ac line? |
| Answer» An inductance acts as a simple conductor in a dc, line and reducing its self-inductance by introducing an iron case, has no effect on the brightness of the lamp. In ac line the presence of inductances results in a drop a voltage , So the brightness of the lamp decreases. Introduction of the iron core further reduces the brightness of the lamp. | |
| 542. |
The AC voltage across a resistance can be measured usingA. a potentiometerB. a hot - wire voltmeterC. am moving- coil galvanometerD. a moving - magnet galvanometer |
| Answer» Correct Answer - B | |
| 543. |
An inductor of `10mH` an a capacitor of `16mF` are connected in the circuit as shown in figure The frequency of the power supply is equal to the resonant frequency of the circuit Which ammeter will read will zero ampere .A. `A_(1)`B. `A_(2)`C. `A_(3)`D. none |
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Answer» Correct Answer - C In parallel resonance current in minimum In this case current is zero |
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| 544. |
A choke coil has.A. high inductance and low resistanceB. low inductance and high resistanceC. high inductance and high resistanceD. low inductance and low resistance |
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Answer» Correct Answer - A A choke coil has high inductance and low resistance , due to large inductance L of the coil, the current in the circuit is decreased appreaciably. Due to small resistance of the coil , the power loss in the choke , coil is less. |
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| 545. |
Assertion: Rate of heat generated when resistance is connected with `AC` source depends on time. Reason : `R.M.S.` voltage may be greater than maximum `AC` voltage.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 - C Rate of heat generated depends on time. |
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| 546. |
A small town with a demand of 800 kW of power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is `0.5Omega`.`km^-1`. The line gets power from the line through a 4000-220 V step-down transformer at a substation in the town. How much power must the plant supply, assuming there is negligible power loss due to leakage. |
| Answer» Power supplied by the plant=800+600=1400 kW | |
| 547. |
Assertion: `KVL` rule is also being applied in `AC` circuit shown below. `V_(C)` in the circuit `=2V`.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 - C Voltage will be added vectorially. |
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| 548. |
Two alternating voltage generators produce emfs of the same amplitude`(E_0)` but with a phase difference of `(pi)//3`. The resultant emf isA. `E_(0)sin(omegat+pi//3)`B. `E_(0)sin(omegat+pi//6)`C. `sqrt(3)E_(0)sin(omegat+pi//6)`D. `sqrt(3)E_(0)sin(omegat+pi//2)` |
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Answer» Correct Answer - C `E_(1)=E_(0)sin omegat , E_(2)=E_(0)sin (omegat+pi//3)`, `E=E_(2)+E_(1)` `=E_(0)sin (omegat+pi//3)+E_(0)sin omegat` `=E_(0)(2sin(omegat+pi//6)cos(pi//6)` `=sqrt(3) sin sin (omegat+pi//6)` |
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| 549. |
An alternating e.m.f. is applied to purely capacitive circuit. The phase relation between e.m.f. and current flowing in the circuit is or in a circuit containing capacitance onlyA. e.m.f. is ahead of current by `pi//2`B. Current is ahead of e.m.f. by `pi//2`C. Current lags behind e.m.f. by `pi`D. Current is ahead of e.m.f. by `pi` |
| Answer» Correct Answer - b | |
| 550. |
The phase angle between e.m.f. and current in LCR series ac circuit isA. 0 to `pi//2`B. `pi//4`C. `pi//2`D. `pi` |
| Answer» Correct Answer - a | |