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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.
| 1501. |
An electric motor operating on a `60 V` dc supply draws a currrent of `10A`. If the effeciency of the motor is `50%`, the resistance of its winding isA. `3Omega`B. `6Omega`C. `15Omega`D. `30Omega` |
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Answer» Correct Answer - A Efficiency `eta=50%` so `e=E//2` and `i=(E-e)/(R ) impliesi=(E-E//2)/(R )=(E)/(2R)` `impliesR=(E)/2i)=(60)/(2xx10)=3Omega` |
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| 1502. |
A pure resistive circuit element X when connected to an ac supply of peak voltage 400 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y, when connected to the same ac supply also gives the same value of peak current but the current lags behind by `90^(@)`. If the series combination of X and Y is connected to the same suply, what will be the rms value of current?A. 2.5AB. 2AC. 1AD. 0A |
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Answer» Correct Answer - A For X, current is in phase `therefore R=(e)/(I)=40Omega` For Y curret lags by is `90^(@)` `therefore X_(L)=(e)/(I)=40Omega` `therefore Z=sqrt(R^(2)+X_(L)^(2))=40sqrt(2)Omega` `I_(rms)=(e_(rms))/(Z)=2.5A` |
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| 1503. |
In the circuit shown in Fig. currrent through the battery at `t = 0` and `t = oo` is A. `1.5 A, 1.5 A`B. `0,0`C. `1 A, 1.5 A`D. `1.5 A, 0 ` |
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Answer» Correct Answer - A In Fig. at `t = 0`, current will flow from the capacitor ans at `t = oo`, inductor provides the path of zero resistance to the flow of current. Therefore, in each case, current through battery. `I = (V)/(R ) = (15 V)/(10 Omega) = 1.5 A` |
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| 1504. |
A pure resistive circuit element `X` when connected to an a.c. supply of peak voltage `200 V` gives a peak current of `5 A` . A second current element `Y` when connected to same a.c. supply gives the same value of peak current but the current lags behind by `90^(@)`. If series combination of `X` and `Y` is connected to the same supply, what is the impedance of the circuit ?A. `40 sqrt2 ohm`B. `40 Omega`C. `80 Omega`D. `2 sqrt40 ohm` |
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Answer» Correct Answer - A `R = (E_(0))/(I_(0)) = (200)/(5) = 40 Omega` As current lags behind tha applied voltage by `90^(@)` therefore, element `Y` must be a pure inductor. `X_(L) = (E_(0))/(I_(0)) = (200)/(5) = 40 Omega` `Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(40^(2) + 40^(2)) = 40 sqrt2 ohm` |
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| 1505. |
In a series LCR circuit, the resonance frequency is 800 Hz. The half power points are obtained at frequencies 745 and 855 Hz. Calcualte the Q factor of the circuit and the band width. |
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Answer» Here, `v_(r) = 800 Hz, v_(1) = 745 Hz, v_(2) = 855 Hz` `Q = (v_(r))/(v_(2) - v_(1)) = (800)/(855 - 745) = (800)/(110) = 7.27` Band width ` = v_(2) - v_(1) = 855 - 745 = 110 Hz` |
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| 1506. |
In a series LCR circuit, at resonance, power factor is …….. .A. 0.1B. 0C. 1D. infinite |
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Answer» Correct Answer - C At resonance, for the series LCR circuit, the impedance Z=R and the circuit behaves as a purely resistive circuit, in which e and I are in phase. `:.` The power factor is one. |
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| 1507. |
A long solenoid of cross-sectional area `5.0 cm^2` is wound with 25 turns of wire per centimetre. It is placed in the middle of a closely wrapped coil of 10 turns and radius 25 cm as shown. (a) What is the emf induced in the coil when the current through the solenoid is decreasing at a rate `-0.20 A//s` ?(b) What is the electric field induced in the coil? |
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Answer» In the theory we have already derived mutual inductance between solenoid and coil `M=(mu_0N_1N_2(piR_1^2))/l_1` `=(mu_0N_1N_2S_1)/l_1` `|e|=|M(di)/(dt)|=|(mu_0N_1N-2S_1)/l_1 (di_1)/(dt)|` `=((4pixx10^-7)(25)(10)(5xx10^-4)(0.2))/10^-2` `=3.14xx10^-6V` b. `El=|(dphi)/(dt)|=e` `:. E=e/l=e/(2piR_2)` `=(3.14xx10^-6)/((2pi)(0.25))` `=2xx10^-6V//m` |
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| 1508. |
A very small circular loop of radius `a` is initially coplanar & concentric with a much larger circular loop of radius `b( lt lt a)`.A constant current `I` is passed in the large loop which is kept fixed in space & the small loop is rotated with constant angular velocity `omega` about a diameter.The resistance of the small loop is `R` & its inductance is negligible The induced `emf` in the large loop due to current induced in smaller loop as a function of time is equal to `1/x((pia^(2)mu_(0)omega)/b)^(2) (I cos 2omegat)/R`.Find out value of `x`. |
| Answer» Correct Answer - D | |
| 1509. |
In the circuit shows in Fig, the coil has inductance and resistance. When `X` is joined to `Y`, the time constant is `tau` during the growth of current. When the steady state is reached, heat is produced in the coil at a rate `P`. `X` is now joined to `Z`. After joining `X` and `Z`: A. the total heat produced in the coil is `P tau`B. the total heat produced in the coil is `(1)/(2)P tau`C. the total heat produced in the coil is `2Ptau`D. the data given is not sufficient to reach a conclusion |
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Answer» Correct Answer - B Power, `P=(i_(0))^(2)R, i.e.,(i_(0))^(2)=(P)/(R) and I=(L)/(R)` Total heat produced, `U=(1)/(2)Li_(0)^(2)=(1)/(2)(tauR)((P)/(R))=(1)/(2)P tau` |
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| 1510. |
Assertion Coefficient of self-induction of an inductor depends upon the rate of change of current passing through it. Reason From, `e=-L(di)/(dt)` We can see that, `L=(-e)/((di//dt))or L prop (1)/((di//dt))`A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is ture. |
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Answer» Correct Answer - D Self-inductance does not depend upon current flowing or change in current flowing but it depends upon turns N. Area of cross-section (A) and permeabillity of medius `(mu)`. We have, `=(di)/(dt)`. |
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| 1511. |
The inductor shown in figure has inducance `0.54 H` and carries a current in the direction shown thast is decreasing at a uniform rate `(di)/(dt)=0.03A//s`. a. Find the self induced emf b. Which emf of the inductor a ro b is at a higher potential? |
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Answer» (i) Given, inductance, L= 0.54H `(di)/(dt)=-0.03` Self-induced emf is givne as `e=-L(di)/(dt)=(-0.54)(-0.03)V` `=1.62xx10^(-2)V` (ii) We know that, `V_(ba)=L(di)/(dt)=-1.62xx10^(-2)V` Since, `V_(ba)(=V_(b)-V_(a))` is negative, It implies that `V_(a)gtV_(b)` or a is at higher potential. |
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| 1512. |
An inductive coil has a resistance of `100 Omega`. When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by `45^(@)`. What is the inductance of the coil?A. `1/(40 pi)H`B. `1/(20 pi)H`C. `1/(60 pi)H`D. `1/(10 pi)H` |
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Answer» Correct Answer - B In the L-R circuit, the current lags behind the voltage and phase difference `phi = 45^(@)` `:. tan 45^(@) = 1 = (L omega)/R ` `:. L = R/(omega) = R/(2 pif) = (100)/(2 pi xx 1000) = 1/(20 pi) H`. |
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| 1513. |
What would you do to obtain a large deflection of the galvanometer? |
| Answer» To obtain a large deflection, one or more of the following steps can be taken : (i) Use a rod made of soft iron inside the coil `C_2` (ii) Connect the coil to a powerful battery, and (iii) Move the arrangement rapidly towards the test coil `C_1`. | |
| 1514. |
In the figure, galvanometer G gives maximum deflection when A. magnet is pushed into the coilB. magnet is rotated into the coilC. magnet is stationary at the center of the coilD. number of turns in the coil is reduced |
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Answer» Correct Answer - A When the magent is pushed into the coil, magnetic flux linked with the coil change. An emf is induced in the coil, which produces maximum deflection. |
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| 1515. |
The power factor in the circuit is unity, when the circuit contains an idealA. resistanceB. conductanceC. capacitanceD. reactance |
| Answer» Correct Answer - A | |
| 1516. |
In a series LCR circuit, at resonance, power factor is …….. .A. zeroB. 0.5C. 1D. `oo` |
| Answer» Correct Answer - C | |
| 1517. |
The system in which energy of the system continuously changes between electric field of capacitor and magnetic field of oscillator to produce oscilaltions isA. acepter circuitB. tank circuitC. rectifier circuitD. amplifier circuit |
| Answer» Correct Answer - B | |
| 1518. |
In series LCR circuit at resonance,A. current is maximum and voltage is minimumB. current is maximum and voltage is maximumC. current is minimum and voltage is maximumD. current is minimum and voltage is minimum |
| Answer» Correct Answer - A | |
| 1519. |
At resonance the peak value of current in a series LCR circuit isA. `e_(0)Z`B. `e_(0)//Z`C. `e_(0)`D. `e_(0)//R` |
| Answer» Correct Answer - D | |
| 1520. |
In an LCR series circuit the current isA. in phase with applied emfB. lags the applied voltageC. lead the applied voltageD. may lead or lag behind the applied voltage |
| Answer» Correct Answer - D | |
| 1521. |
Power in an AC circuit is rated per second at whichA. charge flowsB. work is doneC. energy is spentD. current alternates |
| Answer» Correct Answer - B | |
| 1522. |
An AC source is connected in parallel with an L-C-R circuit as shown.Let `l_(s), l_(L), l_(C) " and " l_(R)` denote the currents through and `V_(s), V_(L), V_(C) " and "V_(R)`the voltage across the corresponding componts.Then,A. `V_(S)=V_(R)+V_(L)+V_(C)`B. `I_(S)=I_(R)+I_(L)+I_(C)`C. `(I_(R),I_(L),I_(C)) lt I_(S)`D. `I_(L),I_(C)` may be `gtI_(S)` |
| Answer» Correct Answer - D | |
| 1523. |
In an `LCR` series ac circuit the voltage across `L,C` and `R` are `V_(1), V_(2)` and `V_(3)` respectively The voltage of the source is .A. `e_(R)+e_(L)+e_(C)`B. `sqrt(e_(R)^(2)+(e_(L)-e_(C))^(2))`C. `e_(R)+e_(C)-e_(L)`D. `sqrt((e_(R)+e_(L))^(2)+e_(C)^(2))` |
| Answer» Correct Answer - B | |
| 1524. |
The graph between inductive reactance and frequency isA. parabolaB. straight lineC. hyperbolaD. an arc of a circle |
| Answer» Correct Answer - B | |
| 1525. |
The average power in LCR series circuit isA. `e_(rms)I_(rms)cosphi`B. `e_(rms)I_(rms)sinphi`C. `e_(rms)I_(rms)`D. `e_(rms)I_(rms)tanphi` |
| Answer» Correct Answer - A | |
| 1526. |
In a purely resistive `AS` circuit,A. current leads emf by a phase angle of `pi` radiansB. current leads emf by a phase angle of `pi//2` radiansC. current and emf are in phaseD. current lags behind emf by a phase angle of `pi//2` radians |
| Answer» Correct Answer - C | |
| 1527. |
The product `e_(rms)I_(rms)` is called asA. true powerB. apparent powerC. power factorD. Q factor |
| Answer» Correct Answer - B | |
| 1528. |
Which increase in frequency of an AC supply , the impedance of an L-C-R series circuitA. remains constantB. increasesC. decreasesD. decreases at first becomes minimum and then increases |
| Answer» Correct Answer - D | |
| 1529. |
Power factor of the AC circuit varies betweenA. 0 to 0.l5B. 0.5 to 1C. 0 to 1D. 1 to 2 |
| Answer» Correct Answer - C | |
| 1530. |
For purely reactive circuit containing only L or C, power factor is |
| Answer» Correct Answer - A | |
| 1531. |
The ratio of tru epower to apparent power, isA. phase factorB. power factorC. frequency factorD. Q factor |
| Answer» Correct Answer - B | |
| 1532. |
An AC of frequency f is flowing in a circuit containing a resistance R and capacitance C in series. The impedance of the circuit is equal toA. `R+f`B. `R+2pifC`C. `R+(1)/(2pifC)`D. `sqrt(R^(2)+X_(C)^(2))` |
| Answer» Correct Answer - D | |
| 1533. |
In an `AC` circuit with voltage `V` and current `I`, the power dissipated isA. `VI//2`B. `VI//sqrt(2)`C. `VI`D. depends on phase voltage V and current I |
| Answer» Correct Answer - D | |
| 1534. |
The term `cosphi` in an `AC` circuit is calledA. phase factorB. power factorC. frequency factorD. resonance factor |
| Answer» Correct Answer - B | |
| 1535. |
When the frequency of AC is doubled, the impedance of an RC circuit isA. doubledB. halvedC. increasesD. decreases |
| Answer» Correct Answer - D | |
| 1536. |
The average power dissipated in an AC circuit containing a resistance along isA. `e_(rms)I_(rms)`B. `e_(rms)I_(rms)cosphi`C. 0D. none of these |
| Answer» Correct Answer - A | |
| 1537. |
The inductive reactance of an inductor of inductance L isA. `(1)/(2pifC)`B. `(1)/(2pifL)`C. `2pifC`D. `2pifL` |
| Answer» Correct Answer - D | |
| 1538. |
When the frequency of AC is doubled, the impedance of an LCR, circuit isA. halvedB. doubledC. increasesD. decreases |
| Answer» Correct Answer - C | |
| 1539. |
In a purely capacitive circuit, the currentA. lead the applied emf by `pi//2`B. lags behind of applied emf by `pi//2`C. in same phase of applied emfD. none of these |
| Answer» Correct Answer - A | |
| 1540. |
The capacitivie reactance of a condenser of capacitance C isA. `(1)/(2pifC)`B. `(1)/(2pifL)`C. `2pifC`D. `2pifL` |
| Answer» Correct Answer - A | |
| 1541. |
What do you mean by the impedance of `LCR`-circuitA. `sqrt(R^(2)(X_(L)-X_(C))^(2))`B. `sqrt(R^(2)+(X_(L)+X_(C))^(2))`C. `sqrt(R+(X_(L)+X_(C))^(2))`D. `sqrt(X_(L)-X_(C)+R)` |
| Answer» Correct Answer - A | |
| 1542. |
In a purely inductive circuit, the current isA. lead the applied emf by `pi//2`B. lags behind of applied emf by `pi//2`C. in same phase of applied emfD. none of these |
| Answer» Correct Answer - B | |
| 1543. |
What will be the phase difference between current and emf when 220 V, 50 Hz ac source is connected to a circuit containing pure resistor?A. current lags behind voltage in phaseB. current and voltage are in same phaseC. current leads ahead of voltage in phaseD. current and voltage are in opposite phase |
| Answer» Correct Answer - B | |
| 1544. |
In a simple circuit with resistance, phase between AC emf and AC current isA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `270^(@)` |
| Answer» Correct Answer - A | |
| 1545. |
The working of dynamo is based on principle ofA. electromagnetic inductionB. self inductionC. mutual inductionD. none of these |
| Answer» Correct Answer - A | |
| 1546. |
A rectangular coil of 100 turns and size `0.1mxx0.05m` is placed perpendicular to a magnetic field of 0.1 T. If the field drops to 0.05 T in 0.05 s, the magnitude of the emf induced in the coil isA. 2B. 3C. `05`D. 6 |
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Answer» Correct Answer - C Here, Area of coil, `A=0.1mxx0.05m=5xx10^(-3)m^(2)` Number of turns, N=100 Initial flux linked with the coil `phi_(1)=BAcostheta=0.1xx5xx10^(-3)costheta=5xx10^(-4)Wb` Find flux linked with the coil `phi_(2)=0.05xx5xx10^(-3)cos0^(@)=25xx10^(-5)Wb=2.5xx10^(-4)Wb` The magnitude of induced emf in the coil is `epsi=(N|Deltaphi|)/(Deltat)=(N|phi_(2)-phi_(1)|)/(t)=(100|2.5xx10^(_4)-5xx10^(-4)|)/(0.05)` `=(100xx2.5xx10^(-4))/(0.05)V=0.05` |
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| 1547. |
A rectangular coil of 100 turns and size `0.1mxx0.05m` is placed perpendicular to a magnetic field of 0.1 T. If the field drops to 0.05 T in 0.05 s, the magnitude of the emf induced in the coil isA. `sqrt2`B. `sqrt3`C. `sqrt(0.6)`D. 0.5 |
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Answer» Correct Answer - D Given, n= 100 turns, A = `0.1xx0.05m^(2)` `B_(1)=0.1T, B_(2)=0.05T, dt=0.05s` We know that, `e=|(-dphi)/(dt)|=|(d)/(dt)("n B A cos"theta)|=|(nAdBcos theta)/(dt)|` Here, `theta=0` `therefore" "e=|(nAdB)/(dt)|=|(100xx0.1xx0.05xx(0.1-0.05))/(0.05)|` `rArr" e = 0.5 V` None of the option is correct. |
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| 1548. |
A coil having an inductance of 50 mH and a resistance of `10 Omega` is connected in series with a `25 muF` capacitor across a 200 V supply. What is the Q factor of the circuit at resonance?A. 3.5B. 4.47C. 5.5D. 7 |
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Answer» Correct Answer - B Q factor =`1/R sqrt(L/C) = 1/10sqrt((50xx10^(-3))/(25 xx 10^(-6)))` `1/10sqrt(2000) = sqrt(20) = 4.47`. |
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| 1549. |
A capacitor of capacitance `100 muF` and a coil of resistance `50 Omega` and inductance `0.5 H` are connected in series with `a 110 V, 50 Hz AC` source. Find the rms value of the current. |
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Answer» Here, `I_(v) = ? R = 50 ohm, L = 0.5 H, C = 100 mu F = 100 xx 10^(-6) F = 10^(-4) F,` `E_(v) = 110 V, v = 50 Hz, X_(L) = omega v L = 2 pi xx 50 xx 0.5 = 157 ohm` `X_(C ) = 1//2 pi v xx 10^(-4) = (10^(4))/(2 xx 3.14 xx 50) = 31.85 ohm` `Z = sqrt(R^(2) + (X_(L) - X_(C )^(2))) = sqrt(50^(2) + (157 - 31.85)^(2)) = 134.8 ohm` `I_(v) = (E_(v))/(Z) = (110)/(134.8) A 0.816 A` |
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| 1550. |
A resistance of `100 Omega`, a coil inductance 5 mH and a capacitor of `10 muF` are joined in series. When this L-C-R circuit is joined to a suitable frequency a.c., the circuit resonates. What is the effect on the resonant frequency, if the resistance is made `50 Omega`?A. It is halvedB. it is doubledC. it does not changeD. it is tripled |
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Answer» Correct Answer - C The resnant frequency, `f = 1/(2pisqrt(LC))` It does not depend upon the value of the resistance. Hence f will remain the same, evem if R is made `50 Omega or 200 Omega`. |
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