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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.
| 451. |
In the given circuit, let `i_(1)` be the current drawn battery at time `t=0` and `i_(2)` be steady current at `t=oo` then the ratio `(i_(1))/(i_(2))` is A. 0.6B. 0.8C. 1.2D. 1.5 |
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Answer» Correct Answer - B |
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| 452. |
The time required for a 50Hz alternating current to increase from zero to `70.7%` of its peak value is-A. 2.5 msB. 10 msC. 20 msD. 14.14 ms |
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Answer» Correct Answer - A `1/(50xx8)=1/400 sec =2.5 ms` |
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| 453. |
An `L-C-R` series circuit with `100Omega` resistance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the current and the power dissipated in the `L-C-R` circuitA. `200 W`B. `100 W`C. `50 W`D. `400 W` |
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Answer» Correct Answer - D Av.Electric field energy = `((1)/(2) CV_("rms")^(2)) = 25 xx 10^(-3) J` `:. (1)/(2) C (I_("rms") X_(C ))^(2) = 25 xx 10^(-3) J` `:. (1)/(2) C I_(0)^(2) (1)/(2 pi^(2) v^(2) c^(2)) = 25 xx 10^(-3) J` `:. C = 20 mu F` Average magnetic energy `((1)/(2) LI_("rms")^(2)) = L = 2 H` `:. L = (2 xx 5 xx 10^(-3))/((.10)^(2)) L = 1` henry `V_(R ) = I_("rms") R = V_(C ) = 50 V = V_(L) = I_("rms") L` `= (0.10) .300 = (.10) xx (1)/(2 pi ((50)/(sqrt(t))).20 xx 10^(-6))` `= (.10) xx 2 pi (50)/(sqrt(2)) (I)` `V_(R ) = 30 V = V_(C ) = 50 V = V_(L) = 10 W` rsm voltageof source `E_("rms") = (50)/(sqrt(2))` `:. E_("rms") = 35.36 V` |
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| 454. |
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 findA. capacitance `C` of capacitor is `20 mu F`B. inductance `L` of inductor is `2 H`C. peak voltage of source is `50 V`D. the sum of rms voltage across the three elements is `35.4 V` |
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Answer» Correct Answer - A,B,C,D Let `I_(r)` be the rms current through the circuit then `I_(r ) = 2A, (I_(r ))/(omega C) = 20 V, I_(r ) omega C = 20 V` and `I_(r ) R = 10 V` solving we get `R = 50 Omega, C = (1)/(pi) xx 10^(-3) F` and `L = (1)/(10 pi) H` `:. V_(s)` = source voltage = `I_(r ) = sqrt(R^(2) + (omega L - (1)/(omega C))^(2))` `= sqrt((I_(r ) R)^(2) + (I_(r) omega L - (I)/(omega C))^(2))` `= sqrt(10^(2) + (20 - 20)^(2)) = 10` volts After in inductor is shorted `I_(r ) = (V_(s))/(sqrt(R^(2) + (1)/(omega^(2) C^(2)))) = (10)/(sqrt(25 + 10)) = (2)/(sqrt(5)) A` `v_(1) = I_(r ) R = 2 sqrt(5)` volts, `v_(2) = (I_(r ))/(omega C) = 4 sqrt(5)` volts |
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| 455. |
A square loop of side and a straight infinity conducor are placed in the same plane with two sides of the squre parallel to the conductor. The resistance of the loop is R. The loop is turned through `180^(@)` about the axis . The electric charge that flows in the square loop is `(mu_(0)I_(a))/(2piR)1m |(na+b)/(b)|`. Find the value of n. |
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Answer» Correct Answer - 2 |
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| 456. |
An `AC` voltage is applied acrss a series combination of `L` and `R`. If the voltage drop across the resistor and inductor are `20 V` and `15 V` respectiely, then applied peak voltage isA. `25V`B. `35V`C. `25sqrt2V`D. `5sqrt7V` |
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Answer» Correct Answer - C `V=sqrt(V_R^2+V_L^2)` `=sqrt((20)^2(15)^2)=25V` But this is the rms value. `:.` Peak value of `=sqrt2 V_("rms")=25sqrt(2)V`. |
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| 457. |
The potential difference between the ends of a resistance `R` is `V_(R)`, between the ends of capacitor is `V_(C ) = 2V_(R)` and between the ends of inductance is `V_(L) = 3V_(R )`. Then the alternating potential of the source in terms of `V_(R )` will beA. `sqrt(2) V_(R)`B. `V_(R)`C. `(V_(R))/(sqrt(2))`D. `5 V_(R )` |
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Answer» Correct Answer - 1 `overline(V_(s)) = overline(V_(B)) + overline(V_(C )) + overline(V_(L)) = V_(R ) hat(i) - 2 V_(R ) hat(j) + 3 V_(R ) hat(j)` `= V_(R ) hat(i) + V_(R ) hat(j) , |overline(V)| = sqrt(2) V_(R)` |
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| 458. |
Choose correct statement if capcitance increases from zero `(0)` to inifinity `oo` A. Current increases from 0 (Zero) to maximum then decreases to zeroB. Reading of voltmeter first increases and it will be maximum when `X_(L) = X_(C )`C. Power factor of circuit first increases then decreasesD. `V_(1)` may be greater than `V, V_(1)` may be equal `V, V_(1)` maybe less than `V`, where `V_(1)` is reading of volmeter and `V` is source voltage. |
| Answer» Correct Answer - A,B,C,D | |
| 459. |
In the given series `R-L-C` circuit, `R = 100 Omega, L = 10^(-3) H, C= 0.1 mu F, V_(0) = 200 V` (a) The resonant frequency is `15924 Hz` (b) The current at resosnance is `1A` (c ) The power dissipated in the circuitd at resonance is `100 W` A. (a),(b),(c ) are correctB. (a),(b),(c ) are wrongC. Only (a),(b) are correctD. Only (b),(c ) are correct |
| Answer» Correct Answer - A | |
| 460. |
A steady `4 A` flows in an inductor coil when connected to a `12 V` source as shown in figure. If the same coil is connected to an ac source of `12 V`, `50 "rad"//s` a currentd of `2.4 A` flows in the circuit as shown in figure2. Now after these observations , a capacitor fo capacitance `(1)/(50) F` is connected in series with the coil and with same `AC` source as shown in figure 3. The resistance of the coil is:A. `1 Omega`B. `2 Omega`C. `3 Omega`D. `4 Omega` |
| Answer» Correct Answer - C | |
| 461. |
A steady `4 A` flows in an inductor coil when connected to a `12 V` source as shown in figure. If the same coil is connected to an ac source of `12 V`, `50 "rad"//s` a currentd of `2.4 A` flows in the circuit as shown in figure2. Now after these observations , a capacitor fo capacitance `(1)/(50) F` is connected in series with the coil and with same `AC` source as shown in figure 3. The resistance of the coil is:A. `0.01 H`B. `0.02 H`C. `0.04 H`D. `0.08 H` |
| Answer» Correct Answer - A | |
| 462. |
A steady `4 A` flows in an inductor coil when connected to a `12 V` source as shown in figure. If the same coil is connected to an ac source of `12 V`, `50 "rad"//s` a currentd of `2.4 A` flows in the circuit as shown in figure2. Now after these observations , a capacitor fo capacitance `(1)/(50) F` is connected in series with the coil and with same `AC` source as shown in figure 3. The resistance of the coil is:A. `24 W`B. `72 W`C. `144 W`D. `18.2 W` |
| Answer» Correct Answer - D | |
| 463. |
In a series L-R circuit, connected with a sinusoidal ac source, the maximum potential difference across L and R are respectivaly 3 volts and 4 volts. At an instant the potemtial difference across resistor is 2 volts. The potential difference in volt, across the inductor at the same instant will be:A. `3 cos 30^(@)`B. `3 cos 60^(@)`C. `6 cos 45^(@)`D. 6 |
| Answer» Correct Answer - A | |
| 464. |
In a series `L-R` circuit, connected with a sinusoidal ac source, the maximum potential difference acrosssd `L` and `R` are respectively 3 volts and 4 volts At the same instant, the magnitude of the potential difference in volt, across the ac source will beA. `3 cos 67^(@)`B. `5 sin 37^(@)`C. `4 cos 97^(@)`D. 0 |
| Answer» Correct Answer - B | |
| 465. |
In a series `L-R` circuit, connected with a sinusoidal ac source, the maximum potential difference acrosssd `L` and `R` are respectively 3 volts and 4 volts At the same instant, the magnitude of the potential difference in volt, across the ac source will beA. `3 cos 67^(@)`B. `5 cos 83^(@)`C. `6 cos 97^(@)`D. 0 |
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Answer» Correct Answer - B |
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| 466. |
A uniform magnetic field existsin region given by `vec(B) = 3 hat(i) + 4 hat(j)+5hat(k)`. A rod of length `5 m` is placed along `y`-axis is moved along `x`- axis with constant speed `1 m//sec`. Then the magnitude of induced `e.m.f` in the rod is :A. zeroB. 25VC. 20VD. 15V |
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Answer» Correct Answer - B |
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| 467. |
In an ac circuit the current is given by `i=0.5 sin (314 t + 60^@)` milliampere. Then peak to peak value of current is-A. 0.5AB. 1.0AC. 0.5mAD. 1.0 mA |
| Answer» Correct Answer - D | |
| 468. |
In an electric circuit applied ac emf is `e=20 sin 300t` volt and the current is `i=4 sin 300t` ampere. The reactance of the circuit is-A. 5 ohmB. 80 ohmC. `1.8 k Omega`D. zero |
| Answer» Correct Answer - D | |
| 469. |
STATEMENT-1 : In series LC(AC) circuit, the current is infinite at resonance. STATEMENT-2 : In parallel LC(AC) circuit, the current is zero at resonance. STATEMENT-3 : Series LCR(AC) circuit is used at resonance for turning radio receivers.A. T F TB. F F TC. T T TD. T T F |
| Answer» Correct Answer - C | |
| 470. |
Show that average heat produced during a cycle of `AC` is same as produced by `DC` with `i=i_("rms")`. |
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Answer» For an `AC,i=i_0sinomegat` Therefore, instantaneous value of heat produced in time `dt` across a resistance `R` is `:. dH-i^2Rdt=i_0^2Rsin^2omegatdt` `:.` Averge value of the produced during as cycle, `H_(av) =(int_0^TdH)/(int_0^Tdt)=(int_0^(2pi/omega)(i_0^2Rsin^2omegat)dt)/(int_0^(2pi//omega)dt)` `i_0^2/2R((2pi)/omega)=i_(rms)i^2RT` i.e. `AC` produces same heating efects as `DC` of value `i=i_(rms)` |
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| 471. |
A steady dc current of 2.83A is flowing in a resistance wire. The rms value of this current is-A. 2.83 AB. about 2AC. about 4AD. undefined for dc |
| Answer» Correct Answer - A | |
| 472. |
STATEMENT-1 : The rms value of ac is that value of steady current,which would generale the same amount of heat in a given resistance in a given time interval. STATEMENT-2 : In parallel AC circuit, impedance of LCR circuit is equal to R. STATEMENT-3 : A magnetic field can increase the speed of the charged particle, when the field is time varyingA. T F TB. F F TC. T T FD. T T T |
| Answer» Correct Answer - D | |
| 473. |
A capacitor of capacitance `1 mu F` is charged to potential 2 V and is connected to an inductor of 1 mH. At an instant when potential difference across. The capacitor is 1 V, the current in the circuit is `10^(-2) sqrt((10x)/(3))` ampere. Find out the value of x. |
| Answer» Correct Answer - 9 | |
| 474. |
In the L-R circuit as shown in figure, potential difference across the resistance at some instant is 4 V. Then A. current is incresin at a rate of 8A/s at this instantB. power supplied by the battery at this instant is 20WC. power stored in the magnetic field at this instant is 16WD. current in the circuit at this instant is 1A |
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Answer» Correct Answer - A::B::C::D |
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| 475. |
In LCR circuit as shown in figure A. current will lead the voltageB. rms value of current is 20AC. power factor of circuit is `(1)/(sqrt2)`D. voltage drop across resistance is 100V |
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Answer» Correct Answer - A::C |
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| 476. |
In LCR circuit during resonanceA. power factor is zeroB. power factor is oneC. power developed acrtoss resistance is zeroD. power developed across capacitance is zero |
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Answer» Correct Answer - B::D |
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| 477. |
In an L-R circuit, if an iron core is inserted inside the coilA. steady state current will increaseB. steady state current will increase remain unchangedC. time constant will increaseD. time constant will increase |
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Answer» Correct Answer - B::D |
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| 478. |
Current in `R_(3)` A. just after closing the switch is zeroB. long after closing the switch is zeroC. just after closing the switch is `(E)/(R_(3))`D. long after clolsing the switch is `(E)/(R_(3))` |
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Answer» Correct Answer - A::B |
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| 479. |
A V-shaped conducting wire is moved inside a magnetic field as shown in figure. Magnetic fiel d is perpendicular to paper inwards. Then A. `V_(a)=V_(c)`B. `V_(a)gtV_(c)`C. `V_(a)gtV_(b)`D. `V_(c)gtV_(b)` |
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Answer» Correct Answer - A::C::D |
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| 480. |
A series LCR circuit has `R=5 Omega, L=40 mH` and `C=1mu F`, the bandwidth of the circuit isA. 10 HzB. 20 HzC. 30 HzD. 40 Hz |
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Answer» Correct Answer - B Resonant angular frequency `omega_(r ) = (1)/(sqrt(LC))` …(i) Quality factor `Q=("Resonant angular frequency")/("Bandwidth")` Bandwidth `=upsilon_(2)-upsilon_(1)=(upsilon_(r ))/(Q)` ….(ii) where `upsilon_(r )` = resonant frequency `=(1)/(2pi sqrt(LC))` Q = quality factor. Also, `Q = (omega_(r )L)/(R )` `therefore upsilon_(2)-upsilon_(1)=(upsilon_(r )R)/(2pi upsilon_(r )L)=(R )/(2pi L)` (Using (i) and (ii)) `upsilon_(2)-upsilon_(1)=(R )/(2pi L)=(5)/(2pi xx 40xx10^(-3))=20 Hz` |
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| 481. |
Figure shows a series LCR circuit connected to a variable frequency 230 V source. The source frequency which drives the circuit the circuit in resonance isA. 4 HzB. 5 HzC. 6 HzD. 8 Hz |
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Answer» Correct Answer - D Here, `L=5 H,C = 80 mu F = 80xx10^(-6)F, R = 40 Omega` `V_("rms")=230 V` The resonannt angular frequency is `omega_(r )=(1)/(sqrt(LC))=(1)/(sqrt(5xx80xx10^(-6)))=50" rad " s^(-1)` `therefore upsilon_(r )=(omega_(r ))/(2pi)=(50)/(2pi)=8 Hz` |
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| 482. |
An inductance of `(200)/(pi) mH` a capacitance of `(10^(-3))/(pi)` and a resistance of `10 Omega` are connected in series with an `AC` source of `220 V, 50Hz`. The phase angle of the circuit is |
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Answer» Here, `L = (200)/(pi) mH = (200 xx 10^(-3))/(pi) H = (0.2)/(pi) H` `C = (10^(-3))/(pi) F, R = 10 Omega , E_(v) = 220 V, n = 50 Hz` `X_(L) = omega L = 2 pi n L = 2 pi xx 50 xx (0.2)/(pi) = 20 Omega` `X_(C )= (1)/(omega C) = (1)/(2 pi n C) = (pi)/(2 pi xx 50 xx 10^(-3)) = 10 Omega` `tan phi = ((X_(L) - X_(C )))/(R ) = (20 - 10)/(10) = 1 , phi = (pi)/(4)` |
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| 483. |
A series L-C-R circuit is operated at resonance . ThenA. voltage across R is minimumB. impedance is minimumC. impedance is maximumD. current amplitude is minimum |
| Answer» Correct Answer - B | |
| 484. |
An L-C-R series is under resonance . If `underset(m)(l)` is current amplitude `underset(m)(V)` is voltage amplitude, R is the resonance , Z is the impedance , `underset(L)(X)` is the inducitve reactance and `underset( C )(X) ` is the capacitive reactance , thenA. `underset(m)(l)` = `underser(m)(V)`/ ZB. `underset(m)(V)`/ `underset(L)(X)`C. `underset(m)(l)` = `underset(m)(V)` / `underset( C )(X)`D. `underset(m)(l)` = `underset(m)(V)` / R |
| Answer» Correct Answer - D | |
| 485. |
In an L-C-R series, AC circuit at resonanceA. the capacitive reactance is more than the inductiveB. the capacitive reactance equals the inductive reactanceC. the capactive reactance is less than the inductive reactanceD. the power dissipated is minimum |
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Answer» Correct Answer - B At resonance (series resonant circuit), `underset(L)(X)` = `underset(C )(X)` `implies` `underset(min)(Z)` = R , i.e., circuit behaves as resistive circuit. |
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| 486. |
In the circuit shown in the figure `R = 50 Omega, E_(1) = 25 sqrt(3)` volt and `E_(2) = 25 sqrt(6) sin omega t` volt where `omega = 100 pi s^(-1)`. The switch is closed at time `t = 0` and remains closed for 14 minutes, then it is opened. If total heat produced is usedd to raise the temperature of 3 kg of water at `20^(@)C`, what would be the final temperature of water ?A. `15^(@)C`B. `25^(@)C`C. `45^(@)C`D. `75^(@)C` |
| Answer» Correct Answer - B | |
| 487. |
In the circuit shown in the figure `R = 50 Omega, E_(1) = 25 sqrt(3)` volt and `E_(2) = 25 sqrt(6) sin omega t` volt where `omega = 100 pi s^(-1)`. The switch is closed at time `t = 0` and remains closed for 14 minutes, then it is opened. Find the amount of heat produced in the resistor A. `64000 J`B. `56000 J`C. `63000 J`D. `75000 J` |
| Answer» Correct Answer - C | |
| 488. |
A `20 V` 5 watt lamp is used in ac main `220 V` and frequency 50 c.p.s. What pure resistance should be included in place of the above passive elements so that the lamp can run on its rated voltage?A. `120 Omega`B. `240 Omega`C. `800 Omega`D. `720 Omega` |
| Answer» Correct Answer - C | |
| 489. |
A `20 V` 5 watt lamp is used in ac main `220 V` and frequency 50 c.p.s. Capacitance of capacitor, to be put in series to run the lampA. `2 * F`B. `4 * F`C. `6 * F`D. `8 * F` |
| Answer» Correct Answer - B | |
| 490. |
Which of the following curves correctly represent the variation of capacitve reactance (`underset(C )(X)`) with frequency (f) ?A. B. C. D. |
| Answer» Correct Answer - C | |
| 491. |
What is step up transformer ? How it differs from step down transformer ? |
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Answer» The ratio of number of turns in the secondary coil to the number of turns in the primary coil is called transformer ratio. `T = (N_(S))/(N_(P))=("No of turns in the secondary")/("No of turns in the primary")` If `N_(S) gt N_(P)`, then the transformer is called step up transformer. If `N_(S)lt N_(P)`, then the transformer is called step down transformer. |
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| 492. |
In series LCR circuit at resonanceA. Power factor is zeroB. Power developed across the resistor is maximumC. Power developed across the inductor is zeroD. Both (2) & (3) |
| Answer» Correct Answer - D | |
| 493. |
The capacitance of a pure capacitance is `1` farad. In DC circuits, its effective resistance will beA. zeroB. infinteC. `1Omega`D. `1/2Omega` |
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Answer» Correct Answer - B In DC circuits , f = 0 `therefore` `underset(C )(X)` = 1/`2pi`(0)C= 1/0 = `oo` |
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| 494. |
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 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` |
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Answer» Correct Answer - B For purely capacitive circuit `e=e_(0) sin omegat` `i=i_(0)sin (omegat+(pi)/2)` i.e., current is ahead of emf by `pi//2` |
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| 495. |
A capacitor becomes a perfect insulator forA. Alternatinc currentsB. Direct currentC. Both `AC` and `DC`D. None of these |
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Answer» Correct Answer - B `X_(C)=1/(omegaC)=1/(2pivC) , for DC v=0, :. X_(C)=oo` |
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| 496. |
A capacitor becomes a perfect insulator forA. Alternating currentsB. Direct currentsC. Both ac and dcD. None of these |
| Answer» Correct Answer - b | |
| 497. |
A capacitor becomes a perfect insulator forA. DCB. ACC. DC as well as ACD. neither AC nor DC |
| Answer» `X_(C)=(1)/(omegaC)"for" DComega=0, "So",X_(C)=oo` | |
| 498. |
A capacitor becomes a perfect insulator forA. constant direct currentB. alternating currentC. direct as well as alternating currentD. variable current |
| Answer» Correct Answer - A | |
| 499. |
The current (I) in the inductance is varying with time according to the plot shown in figure. Which one of the following is the correct variation of voltage with time in the coil?A. B. C. D. |
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Answer» Correct Answer - D For inductor as we know induced voltage leads the current For t=0 to `t=T//2` `V= L(dl)/(dt) = Ld/(dt)(2I_(0)t)/T`=constant For t`=T//2` to t=T `V=(Ldl)/(dt)(-2I_(0)t)/T=-"constant"` Therefore, answer will represents by graph (d) |
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| 500. |
An alternating current `I` in an inductance coil varies with time `t` according to the graph as shown: Which one of the following graph gives the variation of voltage with time? A. B. C. D. |
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Answer» Correct Answer - D For inductor as we know induced voltage For `t=0` to `t=-T//2` `V=L(dI)/(dt)=Ld/(dt)((2I_(0)t)/T)=` constant, For `t=T//2` to `t=T` `V=(LdI)/(dt)((-2I_(0)t)/T)=-` constant therefore, answer will be represented by graph `(d)`. |
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