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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.
| 351. |
A series L-C-R circuit contains inductancle 5 mH, capacitor `2muF` and resistance `10 Omega`. If a frequency AC source is varied, then what is the frequency at which maximum power is dissipated?A. `(10^(5))/(pi)` HzB. `(10^(5))/(pi)`HzC. `2/3 xx 10^(5)` HzD. `5/pi xx 10^(3)` Hz |
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Answer» Correct Answer - D We know that, Resonant frequency is given by `f_(0) = 1/(2pisqrt(LC)) = 1/(2pisqrt(5 xx 10^(-3) xx 2 xx 10^(-6)))` where, L`=5 xx 10^(-3)` H and `C= 2 xx 10^(-6)`F `rArr f_(0) = 1/(2pisqrt(10^(-8))) = 1/(2pixx 10^(-4))` `=10^(4)/(2pi)` Hz = `5/pi xx 10^(3)` |
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| 352. |
The alternating current in a circuit is given by `I = 50sin314t`. The peak value and frequency of the current areA. `I_(0)`=25 A and f=100HzB. `I_(0)`=50 A and f=50 HzC. `I_(0)` = 50 A and f=100 HzD. `I_(0)`=25 A and f=50 Hz |
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Answer» Correct Answer - B From standard equation, we have `I= I_(0)sinomegat`...............(i) Given, I=50sin314t ..............(ii) Comparing Eqs. (i) and (ii), we get `I_(0)`= 50 A, `omega = 2pif` = 314 `rArr f=314/(2 xx 3.i14) = 50 Hz` |
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| 353. |
What is Q-factor? |
| Answer» Q-factor is a dimensionless parameter that describes how underdampede an oscillator or a resonator is. | |
| 354. |
State the condition under which the phenomenon of resonance occurs in series LCR circuit when ac voltage is applied. In a series LCR circuit, the current is in same phase with voltage. Calculate the value of self-inductance if the capacitor used in `20muF` and resistance used in 10 ohm with the ac source of frequency 50 Hz. |
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Answer» Condition of resonance in a series LCR circuit: `omegaL=1/(omegaC)` where `omega`= angular frequency of the source, L=self inductance of the coil and C=capacitance. Accordingly to the question, electric current and emf are in the same phase in the LCR circuit, i.e., the circuit is in the resonance condition. Here, `omega=2pitimes50=2times3.14times50=314Hz` `C=20muF=20times10^-6F=2times10^-5F` Now, `omegaL=1/(omegaC)` or, `L=1/(omega^2C)=1/((314)^2times(2times10^-5))=0.507 H` |
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| 355. |
If the readings `V_(1)` and `V_(3)` are 10. volt each, then reading of `V_(2)` is: A. 0 voltB. 100 voltC. 200 voltD. cannot be determined by given information |
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Answer» Correct Answer - C Resultant voltage = 200 volt Since `v_(1)` and `v_(3)` out of phase, the resultant voltage is equal to `v_(2)` `:. V_(2) = 200` volt |
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| 356. |
If the frequency of the source e.m.f. in an AC circuit is n, the power varies with a frequency:A. nB. 2nC. n/2D. zero |
| Answer» Correct Answer - B | |
| 357. |
By what percentage the impedance in an AC series circuit should be increased so that the power factir changes from `(1//2) "to" (1//4)` (when R is constant)?A. 2B. 1C. 0.5D. 4 |
| Answer» Correct Answer - B | |
| 358. |
In a circuit containing R and L , as the frequency of the impressed AC increase, the impedance of the circuitA. decreasesB. increasesC. remains unchangedD. first increases and then decreases |
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Answer» Correct Answer - B Impedance , Z = `sqrt R^(2) + (2`pi`fL)^(2)` As f increases, Z will increase. |
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| 359. |
Assertion (A) : The electrostatic energy stored in capacitor plus magnetic energy stored in inductor will always be zero in series `LCR` circuit driven by ac voltage source under condition of resonance. Reason (R ) : The complete voltage of ac source appears across the resistor in a series `LCR` circuit driven by ac voltage source under condition of resonance.A. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D In resonance condition when energy across capacitor is maximum, energy stored in inductor is zero, vice versa is also true. Hence statement 1 is false. |
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| 360. |
Resonance occures in a series LCR circuit when the frequency of the applied emf is 1000 Hz.A. when frequency = 900Hz then the current through the voltage source will be ahead of emf of the sourceB. the impedance of the circuit is minimum at `f=1000 Hz`C. at only resonance the voltages across L and C differ in phase by `180^(@)`D. if the value of C is double, resonance occurs at `f=2000 Hz`. |
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Answer» Correct Answer - A::B `(f=900Hz) lt (f_(res)=1000Hz), so (X_C)gt(X_L)`, circuit becomes capacitive and current leads emf. At resistance, impedance is minimum. Not only at resonance, but at all conditions voltage across L and C differs in phases by `180^(@)` when connected in series. |
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| 361. |
An inductor and a resistor are connected in series with an ac source. In this circuit.A. the current and `P.d` across the resistance lead `P.d` across the inductance by `pi//2`B. the current and `P.d` acorss the resistance lags behind the `P.d` across the inductance by angle `pi//2`C. The currentd across resistance leads and the `P.d` across resistance lags behind the `P.d` across the inductance by `pi//2`D. the current across resistance lags behind and the `P.d` across the resistance leads the `P.d` across the inductance by `pi//2` |
| Answer» Correct Answer - 2 | |
| 362. |
Assertion (A) : The electrostatic energy stored in capacitor plus magnetic energy stored in inductor will always be zero in series `LCR` circuit driven by ac voltage source under condition of resonance. Reason (R ) : The complete voltage of ac source appears across the resistor in a series `LCR` circuit driven by ac voltage source under condition of resonance.A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. Assertion is true but Reason is falseD. Assertion is false but Reason is true |
| Answer» Correct Answer - 4 | |
| 363. |
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to consitute the circuit ?A. Only resistorB. Resistor and an inductorC. Resistor and a capacitorD. Only a capacitor |
| Answer» Correct Answer - 3,4 | |
| 364. |
The value of current at resonance in a series L-C-R circuit is affected by the value ofA. R onlyB. C onlyC. L onlyD. L, C and R |
| Answer» Correct Answer - A | |
| 365. |
If the value of potential in an `ac`, circuit is `10 V`, then the peak value of potential isA. `10/(sqrt(2))`B. `10sqrt(2)`C. `20sqrt(2)`D. `20/(sqrt(2))` |
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Answer» Correct Answer - B `V_(0)=sqrt(2)V_(rms)=10sqrt(2)` |
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| 366. |
The maximum value of `AC` voltage in a circuit is `70V`. Its `r.m.s.` value isA. `70.7 V`B. `100 V`C. `500V`D. `707 V` |
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Answer» Correct Answer - C `E_(rms)=(E_(0))/(sqrt(2))=707/1.41=500 V` |
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| 367. |
An LC circuit contains a 20 mH inductor and `25 mu F` capacitor with an initial charge of 5 mC. The total energy stored in the circuit initially isA. 5 JB. `0.5 J`C. 50 JD. 500 J |
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Answer» Correct Answer - B Here, `C = 25 mu F = 25xx10^(-6)F`, `L=20 mH=20xx10^(-3)H, q_(0)=5 mC=5xx10^(-3)C` `therefore` Total energy stored in the circuit initially is `U = (q_(0)^(2))/(2C)=((5xx10^(-3))^(2))/(2xx25xx10^(-6))=(25xx10^(-6))/(2xx25xx10^(-6))=(1)/(2)=0.5 J` |
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| 368. |
A condenser of capacity `C` is charged to a potential difference of `V_(1)`. The plates of the condenser are then connected to an ideal inductor of inductance `L`. The current through the inductor wehnn the potential difference across the condenser reduces to `V_(2)` isA. `((C(V_(1)-V_(2))^(2))/(L))^(1/2)`B. `(C(V_(1)^(2)-V_(2)^(2)))/(L)`C. `(C(V_(1)^(2)+V_(2)^(2)))/(L)`D. `((C(V_(1)^(2)-V_(2)^(2)))/(L))^(1/2)` |
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Answer» Correct Answer - D In case of oscillatory discharge of a capacitor through an inductor, charge at instant t is given by `q = q_(0)cos omega t` where, `omega = (1)/(sqrt(LC))` `therefore cos omega t=(q)/(q_(0))=(CV_(2))/(CV_(1))=(V_(2))/(V_(1)) " "(because q=CV)` .....(i) Current through the inductor `I=(dq)/(dt)=(d)/(dt)(q_(0)cos omega t)=-q_(0)omega sin omega t` `|I|=CV_(1)(1)/(sqrt(LC))[1-cos^(2)omega t]^(1//2)` `=V_(1)sqrt((C)/(L))[1-((V_(2))/(V_(1)))^(2)]^(1//2)=[(C(V_(1)^(2)-V_(2)^(2)))/(L)]^(1//2)` (using (i)) |
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| 369. |
The variation of the instantaneous current `(I)` and the instantaneous e.m.f `(E)` in a circuit is as shown in figure. Which of the following statement is correct? A. The voltage lags behind the current by `pi//2`B. The voltage leads the current by `pi//2`C. The voltage and the current are in phaseD. The voltage leads the current by `pi` |
| Answer» Correct Answer - b | |
| 370. |
The variation of the instantaneous current `(I)` and the instantaneous e.m.f `(E)` in a circuit is as shown in figure. Which of the following statement is correct? A. the voltage lags behind the current by `pi//2`B. The voltage leads the current by `pi//2`C. The voltage and the current are in phaseD. The voltage leads the current by `pi` |
| Answer» Correct Answer - B | |
| 371. |
Current and voltage in AC are I = `underset(o)(I)` sin (`omega`t + `pi`/4), ThenA. `underset(L)(X)` gt `underset( C)(X)`B. R = 0C. Both are correctD. Both are wrong |
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Answer» Correct Answer - C Phase difference between current and voltage is `pi`/4 , with voltage leading. |
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| 372. |
1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transimitted if (i) power is transformer is used to boost the voltage to 11000 V, power transmitted, then a step down transformer is used to bring voltages to 220 V. `(rho_(Cu) = 1.7 xx 10^(-8) SI`unit)A. `1.8%`B. `1.5%`C. `3.6%`D. `7.2%` |
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Answer» Correct Answer - C Given, `l=20 km = 20xx10^(3)m` `r=0.5 cm =0.5xx10^(-2)m, rho = 1.7xx10^(-8)Omega m` `R=(rho l)/(A)=(1.7xx10^(-8)xx20xx10^(3))/(3.14xx(0.5xx10^(-2))^(2))=4.3 Omega` `P=10^(6)W, V=11000 V` `I=(P)/(V)=(10^(6))/(11000)=(10^(3))/(11)A` Power loss, `I^(2)R=((10^(3))/(11))^(2)xx4.3=3.6xx10^(4)W` `(P_("LOST"))/(P_("TRANSMITTED"))xx100%=(3.6xx10^(4))/(10^(6))xx100=3.6%` |
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| 373. |
An L-R circuit has R = 10 `Omega` and L = 2H . If 120 V , 60 Hz AC voltage is applied, then current in the circuit will beA. 0.32 AB. 0.16 AC. 0.45 AD. 0.80 A |
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Answer» Correct Answer - B `l = V/Z = V / sqrt R^(2) + (2pifL)^(2)` `120/sqrt 10^(2) + (2pi xx 60 xx 2 )^(2)` = 0.16 A |
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| 374. |
An alternating current is given by the equation `I=i_1sinomegatcosomegat`.The rms current is given byA. `(i_2+i_1)//sqrt2`B. `(i_2-i_1)//sqrt2`C. `sqrt({(i_1^2+i_2^2)//2})`D. `sqrt({(i_1^2+i_2^2)//(sqrt2)})` |
| Answer» Correct Answer - C | |
| 375. |
In a resonant LCR circuit,A. power factor is zeroB. power factor is oneC. power dissipated in the resistor is zeroD. power dissipated in the capacitor is zero |
| Answer» Correct Answer - B::D | |
| 376. |
A coil having an inductance of `1//pi` henry is connected in series with a resistance of `300 Omega`. If 20 volt from a 200 cycle source are impressed across the combination, the value of the phase angle between the voltage and the current is :A. `"tan"^(-1) (5)/(4)`B. `"tan"^(-1) (4)/(5)`C. `"tan"^(-1) (3)/(4)`D. `"tan"^(-1) (4)/(3)` |
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Answer» Correct Answer - D `X_(L) = 2 pi f L = 2 pi xx 200 xx (1)/(pi) = 400 Omega` `R = 300 Omega` `tan phi = (X_(L))/(R ) = (400)/(300) = (4)/(3)` `implies phi = tan^(-1) ((4)/(3))` |
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| 377. |
Which of the following options is correct with respect to the circuit diagram given below? A. R = `400 Omega` , C = 0.5 `mu`FB. R = 500 `Omega` , C = `1mu`FC. R = 500 `Omega` , C = 1 `mu`FD. R = 400 `Omega` , C = 0.1 `mu`F |
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Answer» Correct Answer - B Circuit is in resonance, `therefore` V = `V_R` = IR `or` R = V / I = 100 / 0.2 = 500` Omega` `X_L` = `V_L` / I = 400 / 0.2 = `2000 Omega` = `omega`L `omega` = `1000 rads^(_1)` Further, `X_L` = `X_C` = `2000 Omega` = 1 / `omega`C `therefore` C = 1 / 2000`omega` = 0.5 `mu`F |
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| 378. |
An AC source is connected with a resistance ( R) and an unchanged capacitance C, in series. The potential difference across the resistor is in phase with the initial potential difference across the capacitor for the first time at the instant (assume that at t =0 , emf is zero)A. `pi/Omega`B. `2pi/Omega`C. `pi/2Omega`D. `3pi/2Omega` |
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Answer» Correct Answer - D Let V = `V_o sin Omegat ` [as V = 0 at t = 0] Then `V_R = V_o sin omega`t and `V_C = V_0 sin(omegat - pi / 2)` C and `V_R` are in same phase . While `V_C` lags V (or `V_R) by 90^(@)` . Now `V_R` is in same phase with initial potential difference across the capacitor for the first time when, `omegat = pi / 2 + 2pi = 3pi/ 2` `therefore` `t = 3pi/2omega` |
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| 379. |
In an ac circuit the instantaneous values of current and applied voltage are respectively `i=2(Amp) sin (250pis^(-1)t and epsi =(10V) sin [(250pis^(-1))t+(pi)/(3)`, Find the instantaneous power drawn from the source at `t=(2)/(3)ms` and its average value. |
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Answer» Correct Answer - `10 W, 5W` `P=iepsilon=[2 sin250pit][10sin(250pit+pi//3)]` at `t=2/3xx10^(-3)rArr P=10 watt` `ltPgt = i_(rms)epsilon_(rms)cos phi=2/sqrt2xx10sqrt2xxcos pi//3` `P=iepsilon=[2sin 250pi t][10 sin (250pit+pi//3)] , t=2/3 xx10^(-3) rArr P=10` watt `ltPgt = i_(rms)epsilon_(rms)cos phi=2/sqrt2xx10sqrt2xxcos pi//3` |
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| 380. |
The instantaneous values of current and voltage in an ac circuit are `i=100 sin 314 r amp` and `e=200 sin (314 t+pi//3) V` respectively. If the resistance is `1 Omega` then the reactance of the circuit will beA. `-200sqrt(3) Omega`B. `sqrt(3) Omega`C. `-200//sqrt(3) Omega`D. `100 sqrt(3) Omega` |
| Answer» Correct Answer - b | |
| 381. |
If `A` and `B` are indentical bulbs, which bulb glows brighter? A. AB. BC. Both equally brightD. Cannot say |
| Answer» Correct Answer - a | |
| 382. |
A circuit operating at `360/(2pi) Hz` contains a `1muF` capacitor and a `20 Omega`. resistor. How large an inductor must be added in series to make the phase angle for the circuit zero? Calculate the current in the circuit if the applied voltage is `120 V`.A. `7.7 H`B. `10 H`C. `3.5 H`D. `15 H` |
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Answer» Correct Answer - 1 `omega L = (1)/(omega C)` |
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| 383. |
The equation of an alternating current is `I = 50 sqrt(2) sin 400 pi t A`, then the frequency and the root mean square value of current are respectively.A. `200 Hz, 50 A`B. `400 Hz, 50 sqrt(2) A`C. `200 Hz, 50 sqrt(2) A`D. `500 Hz, 200 A` |
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Answer» Correct Answer - 1 `I = 50 sqrt(2) sin 400 pi t`, `l = l_(0) sin omega t` Comparing two equation, we get `omega = 2 pi f = 400 pi , f = 200 Hz` and `I_(0) = 50 sqrt(2)` `I_("rms") = (I_(0))/(sqrt(2)) = 50 A` |
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| 384. |
A box `P` and a coil `Q` are connected in series with an ac source of variable freguency The emf of the source is constant at `28V` The frequency is so adjusted that the maximum current flows in `P` and `Q` Find (a) impedance of `P` and `Q` at this frequency (b) voltage across `P` and `Q` . |
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Answer» Current is maximum I e of series resonance `omega_(r) = (1)/(sqrt(LC)) =(1)/(sqrt( 1.2 xx 10^(-3) xx (25)/(3) xx 10^(-6))) =10^(4) rad//sec` `X_(C) = (1)/(omega,_(r)C) = (1)/(10^(4) xx (25)/(3) xx 10^(-6)) =12Omega =X_(L)` `I = (V)/(R_(1) + R_(2)) = (28)/(16 + 12) =(28)/(28) =1A` `Z_(p) = sqrt((16)^(2) + (X_(L))^(2)) = sqrt((16)^(2) + (12)^(2)) =20Omega` `Z_(Q) = sqrt((12)^(2) + (X_(L)^(2)) = sqrt((12)^(2) + (12)^(2))) =12sqrt2Omega` `V_(P) = iZ_(p) = 1xx 20 =20V` `V_(Q) = iZa = 1 xx 12 sqrt2 = 12sqrt2V` . |
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| 385. |
A box `P` and a coil `Q` are connected in series with an ac source of variable freguency The emf of the source is constant at `28V` The frequency is so adjusted that the maximum current flows in `P` and `Q` Find (a) impedance of `P` and `Q` at this frequency (b) voltage across `P` and `Q` . |
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Answer» Current is amximum i.e. Case of series resonance, `omega_(0)= 1/(sqrt(LC)` = `1/(sqrt(1.2 xx 10^(-3) xx 25/3 xx 10^(-6)) = 10^(4)rads^(-1)` At resonance, `X_(C) = X_(L)` `X_(C) = 1/(omega_(0)C) = 1/(10^(4)xx25/3xx10^(-6)) = 12 Omega=X_(L)` Resistance `R_(1)` and `R_(2)` are in series, so `R_(net)= R_(1) + R_(2)` `therefore I= V/(R_(1)+R_(2)) = 28/(16+12) = 28/28 = 1A` a) `Z_(P) = sqrt((16)^(2) + (X_(C))^(2)) = sqrt(16^(2)+(12)^(2))=20Omega` `Z_(Q) = sqrt(12^(2)_X_(L)^(2)) = sqrt((12)^(2)+(12)^(2))` = `20Omega` b) `V_(P) = IZ_(P)= 1 xx 20 = 20V` `V_(Q) = IZ_(Q) = 1 xx 12sqrt(2) = 12sqrt(2)V` |
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| 386. |
Why is the use of ac voltage preferred over dc voltage? Given two reasons. |
| Answer» The main reason for preferring use of `ac` voltage over `dc` voltage is that `ac` voltages can be easily and effieciently converted from one voltage to the other by means of transfomer.Further, electrical energy can be transmitted economically over long distances using `AC`. | |
| 387. |
IF the rotating speed of a dynamo is doubled, the induced electromotive force will beA. doubledB. halvedC. four times as muchD. unchanged |
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Answer» Correct Answer - A `e=omegaBANsin(omegat+alpha)`, if `omega` is doubled then e is also doubled. |
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| 388. |
Why is the use of ac voltage preferred over dc voltage? |
| Answer» Ac voltage can be stepped up or stepped-down using a transformer, which is essential for power transmission ans power consumption in daily life. In addition, any capacitor or inductor can be used as an active component in an ac circuit. These advantages are not provided by dc voltages. | |
| 389. |
For the `AC` circuit shown, the reading of ammeter and voltmeter are `5A` and `50 sqrt(5)` volts respectively, then A. average power deliverd by the source is `250 W`B. rms value of `AC` source is 50 voltsC. voltage gain is 2D. frequency of ac source is 1000rad/s |
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Answer» Correct Answer - A,B,C,D `i_("rms") = 5A` Reading of voltmeter `= i_("rms") sqrt(R^(2) + X_(C )^(2))` From this `X_(C ) = 20 Omega implies omega = 10^(3) "rad"//s` `:. X_(L) = omega L = 20 Omega` Therefore the circuit is in resonance `E_("rms") - i_("rms") R = 50 V` also power factor = 1 Av, power `= E_("rms") i_("rms") cos phi = 250 W` voltage gain ` = (X_(L))/(R ) = 2` |
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| 390. |
A sinusoidal voltage `V (t) = 100 sin (500 t)` is applied across a pure inductance of `L = 0.02 H`. The current through the coil is :A. `10 cos (500 t)`B. `- 10 cos (500 t)`C. `10 sin (500 t)`D. `- 10 sin (500 t)` |
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Answer» Correct Answer - 2 `i = i_(0) sin (omega t - (pi)/(2)), i_(0) = (V_(0))/(X_(L))` |
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| 391. |
The energy of the electromagetic wave is of the order of `15` keV. To which part of the spectrum dose it belong?A. `X-rays`B. Infrared raysC. Ultraviolet raysD. `gamma-rays` |
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Answer» Correct Answer - A |
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| 392. |
In the circuit shown in Fig. A conducting wire HE is moved with a constant speed v towards left. The complete circuit is placed in a uniform magnetic field `vec(B)` perpendicular to the plane of circuit inwards. The current in HKDE is A. clockwiseB. anticlockwiseC. altenatingD. ZERO |
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Answer» Correct Answer - D |
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| 393. |
The transformer ratio of transformer is `10 : 1`. The current in the primary circuit if the secondary current required is `100 A` assuming the transformer be ideal, isA. `500 A`B. `200 A`C. `1000 A`D. `2000 A` |
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Answer» Correct Answer - 3 `(N_(s))/(N_(p)) = (I_(P))/(I_(S))` |
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| 394. |
Find rms value in the following cases (a) `I = 5 +3 sin omegat` (b) ` I = a sin omegat + b cos omegat` (c ) `I =i_(1) sin omegat + i_(2) cos omegat + i_(3) sin 2 omegat` . |
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Answer» `bari^(2)= (int_(0)^(T) i^(2)dt)/(int_(0)^(T) dt) = (1)/(T) int_(0)^(T) (5 + 3sin omegat)^(2)dt` `= (1)/(T) int_(0)^(T) (25 + 30sin omegat + 9sin^(2) omegat)dt` `= (1)/(T) [25 T + 30 int_(0)^(T) sin omegat dt + 9 int_(0)^(T) sin^(2) omega dt]` `(1)/(T) [25T + 0 + 9 .(T)/(2)] =25+ (9)/(2) =5^(2) + (3^(2))/(2)` `i_(rms) = sqrt(bari^(2)) = sqrt(5^(2) + (3^(2))/(2) ) =sqrt59/(2)` `bar(i^(2))=(a^(2))/(2) + (b^(2))/(2)` `i _(rms) = sqrt((a^(2) +b^(2))/(2))` `bar(i^(2))=(i_(1)^(2))/(2) + (i_(2)^(2))/(2) + (i_(3)^(2))/(2)` `i_(mas) = sqrt((i_(1)^(2) + i_(2)^(2) +i_(3)^(2))/(2))` . |
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| 395. |
The transformer ratio of transformer is `10 : 1`. If the primary voltage is 440 V`, secondary emf isA. `44 V`B. `440 V`C. `4400 V`D. `44000 V` |
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Answer» Correct Answer - 3 `(N_(s))/(N_(p)) = (V_(S))/(V_(p))` |
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| 396. |
A setup transformer develops `440 V` in secondary coil for an input of `200 V A.C` Then the type of transformer isA. Steped downB. steped upC. SameD. Same but with reversed direction |
| Answer» Correct Answer - 2 | |
| 397. |
The electric current in AC circuit is given by the relation i =`3sinomegat + 4cosomegat`. The rms value of the current in the circuit in ampere isA. `5/sqrt(2)`B. `5sqrt(2)`C. `sqrt(2)/5`D. `1/sqrt(2)` |
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Answer» Correct Answer - A The current through the AC circuit is given by i = `3sinomegat + 4cosomegat` The maximum value of current is `I_(rms) = sqrt(3^(2)+4^(2))= sqrt(25)` = 5unit The rms value of current, `I_(rms) = I_(max)/sqrt(2) = 5/sqrt(2)` unit |
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| 398. |
What is the r.m.s. value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of `2` amperes in the same resistor?A. 6ampereB. 2ampereC. `2sqrt3ampere`D. 0.65ampere |
| Answer» Correct Answer - C | |
| 399. |
What is the r.m.s. value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of `2` amperes in the same resistor?A. 6 ampB. 2 ampC. 3.46 ampD. 0.66 amp |
| Answer» Correct Answer - c | |
| 400. |
In an `AC` circuit with voltage `V` and current `I`, the power dissipated isA. VIB. `1/2 VI`C. `1/sqrt(2) VI`D. Depends on the phase between V and I |
| Answer» Correct Answer - d | |