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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.
| 401. |
The process of converting alternating current into direct current is known asA. PurificationB. AmplificationC. RectificationD. Current amplification |
| Answer» Correct Answer - c | |
| 402. |
In a series lags the applied emf. The rate at which energy is dissipated in the resistor, can be increased byA. Decreasing the capacitance and making no other changeB. Increasing the capacitance and making no other changeC. Increasing the inductance and making no other changeD. Increasing the driving frequency and making no other change |
| Answer» Correct Answer - A | |
| 403. |
A coil of resistance `300 Omega` and inductance 1.0 henry is connected across an voltages source of frequency `300//2pi Hz`. The phase difference between the voltage and current in the circuit isA. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(6)` |
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Answer» Correct Answer - B `X_(L) = 2 pi fL = 2pi xx (300)/(2 pi) xx 1 = 300 Omega , R = 300 Omega` `tan phi = (X_(L))/(R ) = 1 implies phi = (pi)/(4)` (lagging) |
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| 404. |
A circuit consists of a noniductive resistor of `50 Omega`, a coil of inductance `0.3 H` and resistance `2 Omega`, and a capacitor of `40 mu F` in series and is supplied with 200 volt rms at 50 cycles `//` sec. ThenA. the current lag or lead by an angle `15^(@) 5^(1)`B. the power in the circuit is `710.4 W`C. the power in th circuit is `640 W`D. the current lag or lead by an angle `12^(@) 5^(1)` |
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Answer» Correct Answer - A,B Total resistance `R = 50 + 2 = 52 Omega` `L = 0.3 H, C = 40 xx 10^(-6) F` `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` , `X_(L) = omega L = 2 xx pi xx 50 xx 0.3 = 30 pi` `X_(c ) = (1)/(omega C) = (1)/(2 xx pi xx 50 xx 40 xx 10^(-6)) = 250 pi` `X_(c ) gt X_(L)` so current leads the applied voltage. `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2)) = sqrt((52)^(2) + (30 pi - 250 pi)^(2))` `I_("rms") = (V_("rms"))/(2)` `P_(av) = E_("rms") I_("rms") cos phi = E_("rms") E_("rms")/(2) xx (R )/(2) = (E_("rms")^(2) R)/(2)` |
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| 405. |
In a series R-L-C circuit, the frequency of the source is half of the resonance frequency. The nature of the circuit will beA. capacitiveB. inductiveC. purely resistiveD. selective |
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Answer» Correct Answer - A `X_(c ) = (1)/(omega C)` and `X_(L) = omega L` At `omega lt omega_("res"), X_(C ) gt X_(L) :.` The circuit capacitive. |
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| 406. |
Instantaneous values of current and e.m.f in an AC circuit are `I=I//sqrt(2)sin 314` tamp and `E=sqrt(2)sin(314t-pi//6)V` respectively. The phase difference between `E` and `I` will beA. `- (pi)/(6)` radB. `- (pi)/(3)`C. `(pi)/(6)`D. `(pi)/(3)` rad |
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Answer» Correct Answer - A `I = (1)/(sqrt(2)) sin 314 t` , `E = sqrt(2) sin (314 t - (pi)/(6))` with respect to current, the emf is lagging. So phase difference is `phi = - (pi)/(6)` |
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| 407. |
The instantaneous values of current and voltage in an AC circuit are `i=100sin314t` amp and `e=200sin(314t+pi//3)V` respectively. If the resistance is `1Omega` then the reactance of the circuit will beA. `-200sqrt(3) Omega`B. `sqrt(3) Omega`C. `-200//sqrt(3) Omega`D. `100sqrt(3) Omega` |
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Answer» Correct Answer - B `V_(0)=i_(0)Z implies 200=100Z implies Z=2 Omega` Also `Z^(2)=R^(2)+X_(L)^(2) implies (2)^(2)=(1)^(2)+X_(L)^(2)impliesX_(L)=sqrt(3)Omega` |
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| 408. |
The current in a circuit containing a capacitance `C` and a resistance `R` in series leads over the applied voltage of frequency `omega/(2pi)` by.A. `tan^(-1)(1)/(omegaCR)`B. `tan^(-1)(omegaCR)`C. `tan^(-1)(omega1)/R`D. `cos^(-1) (omegaCR)` |
| Answer» Correct Answer - A | |
| 409. |
In a circuit containing an inductance of zero resistances, the e.m.f of the applied AC voltage leads the current byA. `90^(@)`B. `45^(@)`C. `30^(@)`D. `0^(@)` |
| Answer» Correct Answer - a | |
| 410. |
Statement 1: In a series LCR circuit at resonance condition power consumed by ciccuit is maximum. Statement 2 : At resonance condition, the effective resistance of circuit is maximum.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 `P_(av)=(VI cos phi)/2` At resonance condition `cos phi=1` But `Z=R` which is minimum. |
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| 411. |
In the figure shown `C_(1)=1F, C_(2)=2F and L=5H`. Initially `C_(1)` is charged 50V and `C_(2)` to 10V. Switch S is closed at time t=D. Suppose at some instant charge on `C_(1)` is 20C with the same polarties as shown in the figure `Current in the circuit at this instant will beA. `10sqrt(2) A`B. `15sqrt(2) A`C. `10 A`D. `20 A` |
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Answer» Correct Answer - B |
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| 412. |
In the figure shown `C_(1)=1F, C_(2)=2F and L=5H`. Initially `C_(1)` is charged 50V and `C_(2)` to 10V. Switch S is closed at time t=D. Suppose at some instant charge on `C_(1)` is 10C with the same polarties as shown in the figure Maximum current in the circuit will beA. `4sqrt(30) A`B. `16sqrt(2) A`C. `20sqrt(3) A`D. `12sqrt(6) A` |
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Answer» Correct Answer - A |
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| 413. |
In the figure shown `C_(1)=1F, C_(2)=2F and L=5H`. Initially `C_(1)` is charged 50V and `C_(2)` to 10V. Switch S is closed at time t=D. Suppose at some instant charge on `C_(1)` is 20C with the same polarties as shown in the figure Energy stored in capacitor `C_(2)` at this instant willA. 10 JB. 15 JC. 25 JD. 40 J |
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Answer» Correct Answer - C |
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| 414. |
In the `AC` network shown in figureure the rms current flowing through the inductor and capacitor are `0.6 A` and `0.8A`, respectively. Then the current coming out of the source is .A. `1.0A`B. `1.4A`C. `0.2A`D. none of the above |
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Answer» Correct Answer - C `I_C` is `90^@` a head of the applied votage and `I_L` lags behind the applied voltage by `90^@`. So, there is a phase difference of `180^@` between `I_L and I_C`. `:. I=I_C=I_L=0.2A` |
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| 415. |
The current in ampere through an inductor is i(10+20t) Here t is in second. The induced emf in the inductor 4V. Total flux linked with the inductor at t= 2 is aA. 10 WbB. 20 WbC. 30 WbD. 40 Wb |
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Answer» Correct Answer - A |
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| 416. |
The ratio of time constant during current growth and current decay of the circuit shown in Figure is A. `1:1`B. `3:2`C. `2:3`D. `1:3` |
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Answer» Correct Answer - B |
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| 417. |
In an L-R circuit connected to a battery of constant emf E, switched is closed at time t=0. If denotes the induced emf across inductor and I the current in the circuit at any time t. Then which of the following graphs shown the variation of e with i?A. B. C. D. |
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Answer» Correct Answer - A |
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| 418. |
Calculate the reading which will be given by a hot-wire voltmeter if it is connected across the terminals of generator whose voltage waveform is represented by `v = 200 sin omega t + 100 sin 3 omega t + 50 sin omega t`A. `110 V`B. `162 V`C. `200 V`D. `220 V` |
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Answer» Correct Answer - B Since hot wire voltmeter reads only rms value we will have to find rms value of the given voltage. Considering one complete cycle, `v_("rms") = sqrt((1)/(2 pi) int_(0)^(2 pi) v^(2) d theta)` `v_("rms") (1)/(2 pi) int_(0)^(2 x) (200 sin theta + 100 sin 3 theta + 50 sin 5 theta)^(2) d theta` `= (1)/(2 pi) int_(0)^(2pi) (200 sin theta + 100 sin 3 theta + 50^(2) sin^(2) 5 theta)^(2)` `+ 2 xx 200 xx 100 sin theta sin 3 theta` `+ 2 xx 100 xx 50 sin 3 theta sin 5 theta` `+ 2 xx 50 + 200 sin 5 theta sin theta) d theta` `= (1)/(2 pi) ((200^(2))/(2) + (100^(2))/(2) + (50^(2))/(2)) 2 pi = 26,250` `v_("rms") = sqrt(2650) = 162 V` |
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| 419. |
If an arc is represented by `I=100sin200pit` ampere, determine the peak value of current and time period. |
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Answer» Comparing with the general equation of ac, `I=I_0sin(omegat+alpha)` we get, the peak value of current `I_0=10A` and angular frequency , `omega=200piHz` `therefore` Time period, `T=(2pi)/omega=(2pi)/(200pi)=0.01s` |
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| 420. |
If the current in an `AC` circuit is represented by the equation, `i=5sin(300t-pi/4)` Here `t` is in second and in an ampere, calculate (a) peak and rms value of current (b) frequecne of `AC` (c) average current. |
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Answer» i) Comparing the given equation with general equation, we have `I= I_(0)sin(omegat+-phi)` `therefore` The peak value, `I_(0)`=5A and `I_(rms) = I_(0)/sqrt(2)= 5/sqrt(2) = 3.535`A ii) Angular frequency, `omega = 300 rads^(-1)` `therefore` `v= omega/(2pi) = 300/(2pi) ~~ 47.75 Hz` iii) Average current, `I_(av) = (2/pi)I_(0) = (2/pi)(5) = 3.18`A |
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| 421. |
In an `A.C` circuit having resistance and capacitanceA. emf leads the currentB. current lags behind the emfC. both the current and emf are in phaseD. current leads the emf. |
| Answer» Correct Answer - 4 | |
| 422. |
If the frequency of alternating e.m.f. is in `L-C-R` circuit, then the value of impedance `Z` will change with log (frequency) asA. increaseB. increases and then becomes equal to resistance, then it will start decreasingC. decreases and when it becomes minimum equal to the resistance then it will start increasingD. go on decreasing |
| Answer» Correct Answer - 3 | |
| 423. |
The voltage supplied to a circuit is given by `V=V_(0)t^(3/2)`, where t is time in second. Find the rms value of voltage for the period, t=0 to t=1s. |
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Answer» The mean square voltage, `barV^(2)=1/tint_(0)^(t)V^(2)dt=1/tint_(0)^(t)t^(3)dt` `=V_(0)^(2)/4[t^(4)/4]_(0)^(1)=V_(0)^(2)/4` `therefore = sqrt(barV^(2)) = V_(0)/2` |
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| 424. |
An `AC` source rated `100 V (rms)` supplies a current of `10 A(rms)` to a circuit. The average power delivered by the sourceA. must be 1000WB. may be less than 1000WC. may be greater than 1000WD. may be 1000W |
| Answer» Correct Answer - B::D | |
| 425. |
An `AC` source rated `100 V (rms)` supplies a current of `10 A(rms)` to a circuit. The average power delivered by the sourceA. must be `1000 W`B. may be less than `1000 W`C. may be greater than `1000 W`D. may be `1000 W` |
| Answer» Correct Answer - B,D | |
| 426. |
One application of `L-R-C` series circuit is in high pass or low pass filter, which out either the low or high frequency components of a signal. A has pass filter is shown in figure where the output voltage is taken across the `L-R` where `L-R` combination represents and inductive coil that also has resistance due to the large length of the wire in the coil. which of the following statementds is correct when `omega` is samll in the case of `V_("out")//V_(s)`A. `omega RC`B. `(omega R)/(L)`C. `omega R L`D. `(omega R)/(C )` |
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Answer» Correct Answer - A As `omega to 0`, `(V_("out"))/(V_(s)) = omega C R` |
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| 427. |
One application of `L-R-C` series circuit is in high pass or low pass filter, which out either the low or high frequency components of a signal. A has pass filter is shown in figure where the output voltage is taken across the `L-R` where `L-R` combination represents and inductive coil that also has resistance due to the large length of the wire in the coil. Find the ratio `V_("out") //V_(s)` as a function of the angular frequency `omega` of the sourceA. `sqrt((R^(2) + omega L^(2))/(R^(2) + (omega L - (1)/(omega C))^(2)))`B. `sqrt((R^(2) + omega^(2) L^(2))/(R^(2) + (omega L - (1)/(omega C))^(2)))`C. `sqrt((R^(2) + omega^(2) L)/(R^(2) + (omega C - (1)/(omega L))^(2)))`D. 1 |
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Answer» Correct Answer - B `V_("out") = (V_(s))/(sqrt(R^(2) + (omega L - (1)/(omega C))^(2))) xx sqrt(R^(2) + omega^(2) L^(2))` `(V_("out"))/(V_(s)) = (sqrt(R^(2) + omega^(2) L^(2)))/(sqrt(R^(2) + (omega L - (1)/(omega C))^(2))) = sqrt((R^(2) + omega^(2) L^(2))/(R^(2) + (omega L - (1)/(omega C))^(2)))` |
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| 428. |
Which statement(s) is False for the series resonant conditionA. current maximum and phase difference between `E` and `i` is `pi//2`B. current maximum and phase difference between `E` and `i` is zeroC. voltage maximum and phase difference between `E` and `i` is zeroD. voltage maximum and phase difference between `E` and `i` is `pi//2` |
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Answer» Correct Answer - A,C,D At resonance `X_(L) = X_(C )` and `Z = R` (minimum) `tan phi (X_(L) - X_(C ))/(R ) = 0 implies phi = 0` `I = (E)/(sqrt(R^(2) + (X_(L) - X_(C ))^(2))) = (E)/(R )` (maximum). So current is maximum and in phase with applied voltage. |
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| 429. |
The average power dissipated in a pure inductor `L carrying an alternating current of rms value I is .A. `(1)/(2) LI^(2)`B. `LI^(2)`C. `(1)/(4) LI^(2)`D. zero |
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Answer» Correct Answer - D Average power dissipated in purely inductive capacitive circuit is zero `(phi = 90^(@))` The power is dissipated only in resistance . |
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| 430. |
In an `LCR` circuit the energy is dissipated inA. R onlyB. R and L onlyC. R and C onlyD. R, L and C |
| Answer» Correct Answer - A | |
| 431. |
A charged `30muF` capacitor is connected to a 27mH inductor. What is the angular frequency of free oscillations of the circuit? |
| Answer» `omega_r=1/sqrt(LC)=1/sqrt(27times10^-3times32times10^-6)=1.1times10^3s^-1` | |
| 432. |
Suppose the initial charge on the capactor in the above question is 6 mC. What is the total energy stored in the capactor intially ? What is the total energy at later time ? |
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Answer» Capacitance of the capacitor, C=`30muF=30xx10^(-6)F` Inductance of the inductor, L`=27mH=27xx10^(-3)H` Charge on the capacitor, Q=6mC=6`xx10^(-3)C` Total energy stored in the capacitor can be calculated by the relation `E=(1)/(2)(Q ^(2))/(C)` `=(1)/(2) xx ((6xx10^(-3))^(2))/(30xx10^(-6)) ` `=(6)/(10)=0.6J` Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor. |
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| 433. |
A charged `30muF` capacitor is connected to a 27mH inductor. If the initial charge on the capacitor is 6mC then What is the total energy stored in the circuit initially. What is the total energy at a later time? |
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Answer» `E=1/2Q^2/C=1/2((6times10^-3)^2)/(30times10^-6)=0.61` There will be no charge in total energy. |
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| 434. |
Power dissipated in an `L-C-R` series circuit connected to an `AC` source of emf `epsilon` isA. `(E^(2)R)/[[R^(2) + (Lomega - (1)/(Comega))^(2)]`B. `(E^(2) sqrt(R^(2) + (Lomega - (1)/(Comega))^(2)))/R`C. `(E^(2) [R^(2) + (Lomega - (1)/(Comega))^(2)])/R `D. `(E^(2)R)/(sqrt(R^(2) + (Lomega (1)/(Comega))^(2))` |
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Answer» Correct Answer - A `P = V_(rms) I_(rms) cos phi = V_(rms) V_(rms)/(Z) R /(Z)` `(V_(rms)^(2)R)/(Z^(2)) = (E^(2)R)/(R^(2) + (Lomega- (1)/(Comega))^(2))` . |
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| 435. |
The current in series `LCR` circuit will be the maximum when `omega` isA. As large as possibleB. Equal `o` natural frequency of `LCR` systemC. `sqrt(LC)`D. `sqrt(1//LC)` |
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Answer» Correct Answer - D At resonant frequency current in series `LCR` circuit is maximum. |
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| 436. |
For a series `RLC` circuit `R=X_(L)=2X_(C)`. The impedence of the current and phase different (between) `V` and `i` will beA. `(sqrt(5)R)/2, tan^(-1)(2)`B. `(sqrt(5)R)/2, tan^(-1)(1/2)`C. `sqrt(5) X_(C), tan^(-1)(2)`D. `sqrt(5) R, tan^(-1)(1/2)` |
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Answer» Correct Answer - B `X_(L)=R, X_(C)=R//2` `:. tan varphi=(X_(L)-X_(C))/R=(R-R/2)/R=1/2` `implies varphi=tan^(-1)(1//2)` Also `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt(R^(2)+(R^(2))/4)=(sqrt(5))/2 R` |
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| 437. |
When an ideal choke is connected to an ac source of 100V and 50Hz, a current of 8 A flows through the circuit, A current of 10 A flows throught the circuit when a pure resistor is connected instead of the choke coil. IF the two are connected in series with an ac supply of 100 V and 40 Hz, then the current in the circuit isA. `10A`B. `8A`C. `5sqrt2A`D. `10sqrt2A` |
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Answer» Correct Answer - C `X_L=omegaL=100/8` `thereforeL=100/(8omega)=1/(8pi)``R=100/10=10Omega` Where R and L are connected in series, `Z=sqrt((1/(8pi)times2pitimes40)^2+10^2)=10sqrt2` `thereforeI=V/Z=100/(10sqrt2)=5sqrt2A` |
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| 438. |
A virtual current of `4A` and `50Hz` flows in an `AC` circuit contaning a coil. The power consumed in the coil is `240W`.If the virtual voltage across the coil is 100v then its inductance will beA. `1/(3pi) H`B. `1/(5 pi) H`C. `1/(7 pi) H`D. `1/(9 pi) H` |
| Answer» Correct Answer - b | |
| 439. |
A coil of inductance 0.50 H and resistance `100Omega` is connected to a 240 V, 50Hz ac supply. what is the time lag between the voltage maximum and current maximum? |
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Answer» `tanphi=X_L/R=(2pifL)/R=(2pitimes50times0.5)/100=1.571` `thereforephi=tan^-1 1.571=57.5^@` `therefore` Time interval `=T/360timesphi=phi/(360f)` `=57.5/(360times50)=3.19times10^-3s` |
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| 440. |
The reading of ammeter in the circuit shown will be A. `2 A`B. `2.4 A`C. ZeroD. `1.7 A` |
| Answer» Correct Answer - c | |
| 441. |
The reading of ammeter in the circuit shown will be A. 2AB. 2.4AC. ZeroD. 1.7A |
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Answer» Correct Answer - C Given `underset(L)(X)` = `underset(C )(X)` = `5Omega` , this is the condition of resonance . So `underset(L)(V)` = `underset(C )(V)` , so net voltage across L and C combinations will be zero. |
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| 442. |
A box `P` and a coil `Q` are connected in series with an `AC` source of variable frequency. The emf of the source is constant at `10 V`. Box `P` contains a capacitance of `1 mu F` in series with a resistance of `32 Omega`. Coil `Q` has a self-inductance 4.9 mH and a resistance of `68 Omega` in series. The frequency is adjusted to that the maximum current flows in `P` and `Q`. At this frequency (a) The impedance of `P` is `77 Omega` (b) The impedance of `Q` is `85 Omega` (c ) Voltage across `P` is `7.7 V` (d) Voltage across `Q` is `9.7 V` A. Only (a),(c ) are correctB. Only (a),(d) are correctdC. Only (c ), (d) are correctD. (a),(c ),(d) are correct |
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Answer» Correct Answer - D `i_(0) = (V_(0))/(R ) , Z_(p) = sqrt(R_(1)^(2) + X_(c )^(2)) , Z_(Q) = sqrt(R_(2)^(2) + X_(L)^(2))` `V_(P) = i_(0) Z_(p)` and `V_(Q) = i_(0) Z_(Q)` |
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| 443. |
A fully charged capacitor C with initial charge `q_0` is connected to a coil of self-inductance L at t=0. The time at which the energy is stored equally between the electric and the magnetic field isA. `pi/4sqrt(LC)`B. `2pisqrt(LC)`C. `sqrt(LC)`D. `pisqrt(LC)` |
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Answer» Correct Answer - A During discharging of capacitor C through inductance L, let at any instant, charge in capacitor be Q. `thereforeQ=Q_0sinomegat` Maximum energy stored in capacitor-`1/2Q_0^2/C` Let at an instant t, the energy be stored equally between electric and magnetic field. Then energy stored in electric field at instant t is `1/2Q^2/C=1/2[1/2Q_0^2/C]`or,`Q^2=Q_0^2/2` or,`Q=Q_0/sqrt2`or,`sinomegat=Q_0/sqrt2` or,`omegat=pi/4`or,`t=pi/(4omega)=(pisqrt(LC))/4[becauseomega1/sqrt(LC)]` |
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| 444. |
The resistance of a coil for DC is `5Omega` . In case of AC, the resistance willA. remain `5Omega`B. decreaseC. increaseD. be zero |
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Answer» Correct Answer - C In case of AC, Z = `sqrt R^(2) + omega(L)^(2)` , while in case of DC it is only R. |
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| 445. |
when the speed of a dc motor increase the armature currentA. increasesB. decreases does not changeC. increases and decreases continuoslyD. increases and decreases continuously |
| Answer» Correct Answer - B | |
| 446. |
Assertion (A) : The average value of `lt sin^(2) omega t gt` is zero. Reason (R ) : The average value of function `F (t)` over a period `T` is `lt F (t) gt = (1)/(T) int_(0)^(T) F (t) dt`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 |
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Answer» Correct Answer - 4 `lt sin^(2) omega t gt = 1//2` |
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| 447. |
A coil of self-inductance `((1)/(pi)) H` is connected is series with a `300 Omega` resistance. A voltage of `200 V` at frequency `200 Hz` is applied to this combination. The phase difference between the voltage and the current will beA. `tan^(-1) ((4)/(3))`B. `tan^(-1) ((3)/(4))`C. `tan^(-1) ((1)/(4))`D. `tan^(-1) ((5)/(4))` |
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Answer» Correct Answer - 1 `tan theta = (2 pi f L)/(R )," "f = (1)/(2 pi sqrt(LC))` |
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| 448. |
Assertion: When a ferromagnetic rod is inserted inside an inductor, then current in L-C-R, alternating circuit will decrease. Reason: By inserting the ferromagnetic rod inside the inductor, coefficient of self induction and hence the net impedance will increases.A. If both Assertion and Reason are true and Reason 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 is true but Reason is false.D. If both Assertion and Reason are false. |
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Answer» e) `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))` By inserting the ferromagnetic rod L, and therefore `X_(L)` will increase. But value of Z and therefore I may increase, decrease or remain same. |
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| 449. |
Assertion : In series `L-C-R` circuit, if a ferromagnetic rod is inserted inside an inductor, incr current in the circuit may ease or decrease. Reason : By doing so `X_L` will increase.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A::B `Z=sqrt(R^2+(X_C-X_L)^2)` `X_L` will increase. So `Z` may increase or decrease depending on the value of `X_C` therefore, current may decrese or increase. |
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| 450. |
A reacangular loop if size `(2m xx 1m)` is placed in x-y plane. A uniform but time varying magnetic field of strength T where t is the time elsapsed in second exists in sosace. The magnitude of induced emf (in V) at time t isA. 20=20iB. 20C. 20iD. zero |
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Answer» Correct Answer - D |
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