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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.
| 451. |
In an LCR series ac circuit, the current isA. always in phase with the voltageB. lags the generator voltageC. leads the generator voltageD. none of these |
| Answer» Correct Answer - D | |
| 452. |
In resonant circuits, at resonance, the phase difference between current and emf is |
| Answer» Correct Answer - A | |
| 453. |
The voltage across inductor and condenser at resonance isA. zeroB. of equal magnitude and in phaseC. of equal magnitude and out of phase by `pi`D. of difference magnitudes and out of phase by `pi` |
| Answer» Correct Answer - C | |
| 454. |
Voltage magnification factor of a series resonance circuit isA. `Q=omegaL//R`B. `Q=1//omegaCR`C. both a and bD. neither a nor b |
| Answer» Correct Answer - C | |
| 455. |
In parallel resonant circuit, the current through condenser leads the current through inductor byA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `270^(@)` |
| Answer» Correct Answer - C | |
| 456. |
In a series LCR circuit, voltage applied is `V=3sin(314t+(pi)/(6))` and current from the supply is `l=2sin(315t+(pi)/(3))`. Which of the following is correct ?A. Impendance of circuit is `1.5Omega`B. Reactance of circuit is `(4)/(3)Omega`C. Resistance of circuit is `(3sqrt3)/(4)Omega`D. Wattless component of current is `(1)/(sqrt2)` |
| Answer» Correct Answer - A::C::D | |
| 457. |
In a series LCR circuit, at resonance theA. total inpedance is `Lomega-(1//Comega)`B. toal impedance is RC. voltage across C and L are in phaseD. the voltage across C lags the source voltage by `pi//2` |
| Answer» Correct Answer - B | |
| 458. |
The parallel resonance circuit is called asA. acceptor circuitB. rejector circuitC. rectifier circuitD. transfer circuit |
| Answer» Correct Answer - B | |
| 459. |
With increase in frequency of ac supply the impedance of the parallel resonant circuitA. remains constantB. increasesC. decreasesD. increases at first, becomes maximum and then decreases |
| Answer» Correct Answer - D | |
| 460. |
A squre loop of side a= 12 cm with its sides parallel to x, and y-axis is moved with velocity, V=8 cm/s in the positive x direction in a magnetic field along the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient `delB//delx=-10^(-3) T//cm` along the x-direction, and it is changing in time at the rate `delB//delt=7T//sec` in the loop if its resistance is R`=4.5 Omega`. Find the current . |
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Answer» Correct Answer - A::B::D `I=(EMF)/R=-(a^(2)deltaB)/(RdeltaX)v-(a^(2)deltaB)/(Rdeltat)=22.4mA` Method II: Take `B=-10^(-3)x+7t+B_(0)` where `B_(0)` is constant `phi=underset(y=x)overset(x+a)intadyB, epsilon=|(ddelta)/(dt)|,i=epsilon/R` |
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| 461. |
In the figure shown a long conductor carries constant current `I`. A rod `PQ` of length `l` is in the plane of the rod.The rod is rotated about point `P` with constant angular velocity `omega` as shown in the figure.Find the `e.m.f` induced in the rod in the position shown.Indicate which points is at high potential. |
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Answer» Correct Answer - A::B::C Magnetic field `B` at point of differential element is `B=(mu_(0)i)/((2pi(a+x cos theta)) emf` induced in this element is `da= B V dx` `rArr depsilon=(mu_(0)i omega)/(2pi).(xdx)/((1+x cos theta))` `rArr epsilon=underset(0)overset(epsilon)int depsilon=(mu_(0)iomega)/(2pi cos theta)underset(0)overset(l)int (x/((a/cos theta)+x))dx` `=(mu_(0)iomega)/(2pi cos theta)[l-a/costhetaln((a+l cos theta)/a)]` |
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| 462. |
A current of 3A, in one coil, causes a flux in the second coil of 2000 turns to change by `6xx10^(-4)Wb` per turns of the secondary coil. The mutual inductance of the pair of coils isA. 6 HB. 2 HC. 0.4 HD. 4 H |
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Answer» Correct Answer - C `nphi=MI` `therefore M=(nphi)/(I)=(2000xx6xx10^(-4))/(3)=0.4H` |
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| 463. |
A coil with air inside it has a self inductance of 0.05 H. A soft iron rod of relative permeabiilty 100 is introduced inside the coil. The value of self inductance isA. 5 HB. 0.05 HC. 2.5 HD. 10 H |
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Answer» Correct Answer - A `L_(m)=mu_(r)L_(a)` `=100xx0.05=5H` |
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| 464. |
Does the current in an A.C. circuit lag, lead or remain in phase with the voltages of frequency v applied to the circuit when (i) `v = v_(r)` (ii) `v lt v_(r)` (iii) `v gt v_(r)` where `v_(r)` is the resonance frequency. |
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Answer» (i) When `v = v_(r), X_(L) = X_(C)`. The circuit is purely resistive, i.e., it is non-inductive. Therefore, current and voltage in the circuit are in the same phase. (ii) When `v lt v_(r)` As `X_(L) = omega L = 2 pi v L` and `X_(C ) = (1)/(omega C) = (1)/(1 pi v C)` `:.` when v is small, `X_(L)` is small, `X_(C )` is large. The circuit is capacitance dominated. Therefore, current in the circuit leads the voltage by phase angle `phi`. `X_(L)` becomes large and `X_(C )` becomes small. The circuit is inductance dominated. The current legs behind the voltage, by phase angle `phi`. |
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| 465. |
In an alternaitng current circuit, resistor, inductor and capacitor are connected in series in diagram shown in Fig. Find the value of current and power foctor of the circuit |
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Answer» Here, `R = 50 Omega, L = 10 H` `C = 2 mu F = 2 xx 10^(-6) F, E_(v) = 250 V, v = 50 Hz` `X_(L) = omega L = 2 pi L = 2 xx 3.14 xx 50 xx 10 = 3140 Omega`. `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `= (1)/(2 xx 3.14 xx 50 xx 2 xx 10^(-6)) = (10^(4))/(6.28)` `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` ` = sqrt(50^(2) + (3140 - 1.6 xx 10^(3))) = 1540 Omega` `I_(v) = (E_(v))/(Z) = (250)/(1540.8) = 0.162 A` Power factor, `cos phi = (R )/(Z) = (50)/(1540.8) = 0.032` |
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| 466. |
For the situation described in figure, the magnetic field changes with time according to `B=(2.00t^3-4.00t^2+0.8)t` and `r_2=2R=5.0cm` (a) Calculate the force on an electron located at `P_2` at `t = 2.00 s`(b) What are the magnitude and direction of the electric field at `P_1` when `t = 3.00 s` and `r_1= 0.02 m`. |
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Answer» a. At `P_2` `El=(dphi)/(dt)` `E(2pir_2)=piR^2.(dB)/(dt)` `:. E=R^2/(2r_2)((dB)/(dt))` |
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| 467. |
A solenoid of 50 cm length and 8 cm diameter is would with 500 turns of wire . Another coil of 20 insulated wire is colsely wound over it at its middle region . Calculate the coefficient of mutual induction [Hint : Use the formula `M=(mu_(0)mu_(r)N_(1)N_(2)S)/(l)henry`] |
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Answer» Correct Answer - 0.126 mH |
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| 468. |
A 50 mH coil carries a current of 2 ampere. The energy stored in joules isA. 1B. 0.1C. 0.05D. 0.5 |
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Answer» Correct Answer - B Energy stored in the coil, `E=(1)/(2)Li^(2)=(1)/(2)xx50xx10^(-3)xx4=0.1J` |
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| 469. |
Half of the core of a solenoid of `2xx 10^(-3) m^(2)` cross -section , is made up of air and the other half iron `(mu_(r)=500)`. The length of the solenoid is 2 m . If the number of turns is 1000, calculate its coefficient of self induction . |
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Answer» Correct Answer - 0.315 H |
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| 470. |
A solenoid has 2000 turns would over a length of 0.30 m. The area of its cross-section is `1.2xx10^(-2)m^(2)`. If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil isA. `6xx10^(-4)V`B. `4.8xx10^(-3)V`C. `6xx10^(-2)V`D. `32.1xx10^(-2)V` |
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Answer» Correct Answer - D Induced emf, `e=(Li)/(dt)=(mu_(0)N^(2)S)/(l)(di)/(dt)` `=(4pixx10^(-7)xx(2000)^(2)xx1.2xx10^(-3))/(0.30)xx((4)/(0.25))=32.1xx10^(-2)V` |
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| 471. |
The current in an ideal, long solenoid is varies at a uniform rate of `0.01 As^(-1)`. The solenoid has 2000 turns/m and its radius 6.0 cm. (a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. write the change in the magnetic flux through this circle in 2.0 seconds. (b) find the electric field induced at a point on the circumference of hte circle. (c) find the electric field induced at a piont outside the solenoid at a distance 8.0cm from its axis. |
| Answer» Correct Answer - A::B::C::D | |
| 472. |
A circular loop of radius `1m` is placed in a varying magnetic field given as `B=6t` Tesla, where `t` is time in sec. (a)Find the `emf` induced in the coil if the plane of the coil is perpendicular to the magnetic field. (b) Find the electric field in the tangential directin, induced due to the changing magnetic field. (c)Find the current in the loop if its resistance is `1Omega//m`. |
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Answer» Correct Answer - A::B::C (a)`epsilon=A(dB)/(dt)=pi(1)^(2) 6=6pi V` (b)`Exx2pir=A(dB)/(dt)=6pi` `E=3/r=3/1=3` volt/meter. (c)Find the current in the loop if its resistance is `1Omega//m` `i=epsilon/R=(6pi)/(1xx2pir)=3/r=3/1=3 amp.` |
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| 473. |
An a.c. voltage of 200 V is applied to the primary of a transformer and voltage of 2000 V is obtained from the secondary . Calculate the ratio of currents through primary and secondary coils. |
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Answer» Here, `E_(P) = 200 V, E_(s) = 2000 V, (I_(P))/(I_(s)) = ?` `(I_(P))/(I_(s)) = (E_(s))/(E_(P)) = (2000)/(200) = 10` |
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| 474. |
State the condition under which resonance occurs in LCR circuit ? |
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Answer» In LCR circuit, resonance occurs, when `X_(L) = X_(C )` or `V_(L) = V_(C )`. The resonance frequency is `v_(r) = (1)/(2 pi sqrt(LC))`. |
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| 475. |
A circular loop of radius 0.3 cm lies parallel to amuch bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isA. `9.1xx10^(11) weber`B. `6xx10^(11) weber`C. `3.3xx10^(11) weber`D. `6.6xx10^(11) weber` |
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Answer» Correct Answer - A (a) As we know, magnetic flux , `(phi)=B.A` `(mu_(0)(2)(20xx10^(-2))^(2))/(2[(0.2)^(2)+(0.15)^(2)]xx pi (0.3xx10^(-2)^(2)` `=9.216xx10^(-11)=9.2xx10^*-11) weber`. |
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| 476. |
A solenoid of length `l` metre has self-inductance `L` henry. If number of turns are doubled, its self-inductanceA. remains sameB. becomes`2L`henryC. becomes `4L` henryD. becomes `(L)/(sqrt(2))` henry |
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Answer» Correct Answer - C `LpropN^(2)` |
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| 477. |
The resistance of a coil is 5 ohm and a current of 0.2A is induced in it due to a varying magnetic field. The rate of change of magnetic flux in it will be-A. 0.5 W/bB. 0.05 Wb/sC. 1 W b/sD. 20 Wb/s |
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Answer» Correct Answer - C `R= 5Omega , i=0.2A`, `V=-(dphi)/(dt)=ixxR=5xx0.2 =1` volt Rate of change of magnetic flux =1 volt `=(1wb)/(mu)` |
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| 478. |
If the length and area of cross-section of an inductor remain same but the number of turns is doubled, its self-inductance will become-A. halfB. four timesC. doubleD. one-fourth |
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Answer» Correct Answer - B `L alpha n^(2)` |
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| 479. |
The current carrying wire and the rod AB are in the same plane. The rod moves parallel to the wire with a velocity v . Which one of the following statements is true about induced emf in the rod A. End A will be at lower potential with respect to BB. A and B will be at the same potentialC. There will be be no induced emf in the rodD. Potential at A will be higher than that at B |
| Answer» Correct Answer - D | |
| 480. |
A car moves up on a plane road. The induced emf in the axle connecting the two wheel is maximum when itA. moves at the polesB. moves at equatorC. remains stationaryD. no emf is induced at all |
| Answer» Correct Answer - B | |
| 481. |
The induced emf across the secondary coil depends upon (i) the number of turns `n_(1) and n_(2)` in primary and secondary coil respectively. (ii) the permeability of the medium between the two coils (iii) area of cross section of two coils (iv) resistance of wire of two coils Which of the above is/are true?A. All are trueB. `i`,ii,iiiC. iii,ivD. ii,iii,iv |
| Answer» Correct Answer - B | |
| 482. |
A resistor `R` an inductance `L` and a capacitor `C` are all connected in series with an ac supply The resistance of `R` is `16ohm` and for the given frequency the inductive reactance of `L` is `24` ohm and the capacitive reactance of `C` is `12ohm` If the current in the circuit is `5A` find (a) the potential difference across `R,L` and `C` (b) the impedance of the circuit (c) the voltage of the ac supply and (d) the phase angle . |
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Answer» Correct Answer - `80 V, 120 V, 60 V (b) 20 Omega ( c) 100 V (d) 37^(@)` |
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| 483. |
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)` is |
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Answer» Correct Answer - d In case of ocillatory discharge of a apacitor through an inductor, charge at instant t is given by `q=q_(0)cos omegat` where `omega=(1)/(sqrt(LC))" "....(i)` `:. cos omegat=(q)q_(0))=(CV_(2))/(CV_(1))=(V_(2))/(V_(1))" " ( :. Q=CV)` Current through the inductor `I=(dq)/(dt)=(d)/(qt)(q_(0)cos omegat)=-q_(0)omega sin omegat` `|I|=CV_(1)(1)/(sqrt(LC))[1-cos^(2)omega]^(1//2)` `=V_(1)sqrt((C)/(L))[-((V_(2))/(V_(1)))^(2)]^(1//2)=[(C(V_(1)^(2)-V_(2)^(2)))/(L)]" "` (Using (i)) |
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| 484. |
In figure-5.54 a wire ring of radius R is in pure rolling on a surface. Find the EMF induced across the top and bottom points of the ring at any instant. |
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Answer» Correct Answer - `2BomegaR^(2)` |
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| 485. |
From the two e.m.f. Equation `e_(1)=E_(0) sin (100 pi t)` and `e_(2)=E_(0)sin (100 pi t+(pi)/3)`, we find thatA. `e_(1) "leads" e_(2) "by" 60^(@)`B. `e_(2) "lags behind" e_(1) "by" 60^(@)`C. `e_(2)` achieves its maximum value `1/300` second before `e_(1)`D. `e_(1)` achieves its maximum value `1/300` second before `e_(2)` |
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Answer» Correct Answer - C For both, `omega t = 2 pi f t = 100 pi t` `:. F = 50 Hz :. T = 1/50 s` For a phase diff. = `2 pi`, time required = `T = 1/50 s` `:.` For a phase diff. of `(pi)/3`, time = `(pi)/3 xx 1/50 xx 1/(2 pi) = 1/300 s`. |
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| 486. |
The length l of a wire is shaped to form a coil of 1 turn. This coil has self inductance L. if the same length is bend more sharply to form 3 turns, the self inductance will become/remainA. LB. 2LC. 3LD. 4L |
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Answer» Correct Answer - C `L=(mu_(0)n^(2)pia)/(2)" "therefore (L_(1))/(L_(2))=((n_(1))/(n_(2)))^(2)(a_(1))/(a_(2))` `therefore L_(2)=3L` |
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| 487. |
Two coils have the mutual inductance of 0.05 H. The current changes in the first coil as `I=I_(0)sin omegat`, where `I_(0)=1A` and `omega=100pi"rad/s"`. The maximum emf induced in secondary coil isA. `5piV`B. `7piV`C. `2.5piV`D. `piV` |
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Answer» Correct Answer - A `e=M(dI)/(dt)=MI_(0)omegacos omegat` `therefore e_(0)=MI_(0)omega=5piV` |
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| 488. |
If the rate of change of current is 1A per second in one coil induces an emf of 2 V in the neighbouring coil, the mutual inductance of two coils isA. 2 HB. 1.5 HC. 1 HD. 2.5 H |
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Answer» Correct Answer - A `e=M(dI)/(dt)` `2=Mxx1` `therefore M=2H` |
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| 489. |
A thin beam of n identical positively charged particle are constrained to move in a circular orbit of radius R in a particle accelerator. Each particle has charge q and mass m and the current in the circular orbit is `I_(0)`. The magnetic flux through the circular path is made to increases at a constant rate of `beta Wb s^(–1)`. Calculate the current after the particles complete one turn. |
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Answer» Correct Answer - `I = sqrt(I_(0)^(2)+(n^(2)q^(3)beta)/(2pi^(2)R^(2)m))` |
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| 490. |
There is a long cylinder of radius R having a cylindrical cavity of radius `R//2` as shown in the figure. Apart from the cavity, the entire space inside the cylinder has a uniform mag- netic field parallel to the axis of the cylinder. The magnetic field starts changing at a uniform rate of `(dB)/(dt) = kT//s`. Find the induced electric field at a point inside the cavity. |
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Answer» Correct Answer - `(kR)/(4)` perpendicular to `OO_(1)` |
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| 491. |
What inductance would be needed to store `1.0 kWh` of energy in a coil carrying a `200 A` current. `(1kWh=3.6xx10^6J)` |
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Answer» We have `i=200A` and `U=1kWh=3.6xx10^6J` `:. L=(2U)/i^2 (U=1/2Li^2)` `=(2(3.6xx10^6))/((200)^2)=180H` |
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| 492. |
A sensitive electronic device of resistance `175Omega` is to be connected to a source of emf by a switch. The device is designed to operate with a current of `36mA` but to avoid damage to the device, the current can rise to no more than `4.9mA` in the first `58mus` after the switch the closed. To protect the device it is connected in series with an inductor. a. What emf must the source have?b. What inductance is required? c. What is the time constant? |
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Answer» a. Given `R=175Omega` and peak value current `i_0=36xx10^-3A` Applied voltage `V=i_0R=(175)(36xx10^-3)"volt"=6.3V` b. From the relation `i=i_0(1-e^(-t//tau_L))` We have, `(4.9)=(36)[1-e^(-t//tau_L)]` or `e^(-t//tau_L)=1-4.9/36=0.864` `:. t/tau_L=ln(0.864)=0.146` or `t/(L//R)=0.146` `:. (Rt)/L=0.146` or `L=(Rt)/0.146=((175)(58xx10^-6))/0.146` `=7.0xx10^-2H` c.Time constant of the circuit `tau_L=L/R=(7.0xx10^-2)/175` `=4.0xx10^-4` |
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| 493. |
An inductor coil stores 32 J of magnetic field energy and dissiopates energy as heat at the rate of 320 W when a current of 4 A is passed through it. Find the time constant of the circuit when this coil is joined across on ideal battery.A. `0.1sec`B. `0.2sec`C. `0.3sec`D. `0.4sec` |
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Answer» Correct Answer - B `(1)/(2)Li^(2)=32 implies (1)/(2)Lxx(4)^(2)=32` `L=4H` `i^(2)R=320 implies (4)^(2)R=320` `R=20QOmega` `tau=(L)/(R)=(4)/(20)=0.2sec` |
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| 494. |
In an a.c circuit, value of voltage and current change every instant. Therefore, power of an a.c. circuit at any instant is the product of instantaneous voltages `(E)` and instantaneous current `(I)`. The average power supplied to a pure resistance `R` over a complete cycle of a.c is `P = E_(upsilon). I_(upsilon)`. When circuit is inductive, average power/cycle `= E_(upsilon) I_(upsilon) cos phi`, where `phi` is the phase angle between alternating voltage an altenating current in the circuit. In an a.c. circuit, `800 mH` inductor and a `60 mu F` capacitor are connected in series with `15 ohm` resistance. The a.c. supply to the circuit is `230 V, 50 Hz` The electrial energy spent in running the circuit for one hour isA. `20.0.5 J`B. `20.05 k w h`C. `7.22 xx 10^(4) J`D. Zero |
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Answer» Correct Answer - C As energy spent = power `xx` time `:.` energy spent ` = 20.05 xx 60 xx 60` `= 7.22 xx 10^(4)` Joule |
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| 495. |
In an a.c circuit, value of voltage and current change every instant. Therefore, power of an a.c. circuit at any instant is the product of instantaneous voltages `(E)` and instantaneous current `(I)`. The average power supplied to a pure resistance `R` over a complete cycle of a.c is `P = E_(upsilon). I_(upsilon)`. When circuit is inductive, average power/cycle `= E_(upsilon) I_(upsilon) cos phi`, where `phi` is the phase angle between alternating voltage an altenating current in the circuit. In an a.c. circuit, `800 mH` inductor and a `60 mu F` capacitor are connected in series with `15 ohm` resistance. The a.c. supply to the circuit is `230 V, 50 Hz` The total power abosrbed per cycle by all the three circuit elements isA. `230 W`B. `690 W`C. `60.15 W`D. `20.05 W` |
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Answer» Correct Answer - D Total power absorbed/cycle `P = P_(R ) + P_(L) + P_(C ) = 20.05 + 0 + 0 = 20.05` watt |
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| 496. |
In an a.c circuit, value of voltage and current change every instant. Therefore, power of an a.c. circuit at any instant is the product of instantaneous voltages `(E)` and instantaneous current `(I)`. The average power supplied to a pure resistance `R` over a complete cycle of a.c is `P = E_(upsilon). I_(upsilon)`. When circuit is inductive, average power/cycle `= E_(upsilon) I_(upsilon) cos phi`, where `phi` is the phase angle between alternating voltage an altenating current in the circuit. In an a.c. circuit, `800 mH` inductor and a `60 mu F` capacitor are connected in series with `15 ohm` resistance. The a.c. supply to the circuit is `230 V, 50 Hz` The average power transferred per cycle ot capacitor isA. `230 W`B. `220 W`C. `40 W`D. Zero |
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Answer» Correct Answer - D `P_(C ) = E _(upsilon) I_(upsilon) cos phi` In a capacitor, phase difference, `phi = 90^(@)` `:. P_(C ) = E_(upsilon) I_(upsilon) cos 90^(@) = `zero |
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| 497. |
In an a.c circuit, value of voltage and current change every instant. Therefore, power of an a.c. circuit at any instant is the product of instantaneous voltages `(E)` and instantaneous current `(I)`. The average power supplied to a pure resistance `R` over a complete cycle of a.c is `P = E_(upsilon). I_(upsilon)`. When circuit is inductive, average power//cycle `= E_(upsilon) I_(upsilon) cos phi`, where `phi` is the phase angle between alternating voltage an altenating current in the circuit. In an a.c. circuit, `800 mH` inductor and a `60 mu F` capacitor are connected in series with `15 ohm` resistance. The a.c. supply to the circuit is `230 V, 50 Hz` This average power transferred per cycle to resistance isA. `2.05 W`B. `20.05 W`C. ZeroD. `230 W` |
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Answer» Correct Answer - B Here, `L = 500 mH = 800 xx 10^(-3) H = 0.8 H` `C = 60 mu F = 60 xx 10^(-6) F` ` R = 15 ohm, E_(upsilon) = 230 V` and `v = 50 Hz`. `X_(L) = omega L = 2 pi v L` `2 xx 22//7 xx 50 xx 0.8 = 251.43 Omega` `X_(C ) = (1)/(omega C) = (1)/( 2 pi v C) = (1)/(2 xx 22//7 xx 50 xx 60 xx 10^(-6))` `= 53.03 ohm` `Z =sqrt(R^(2) + (X_(L) - X_(C ))^(2)) = sqrt(15^(2) + (251.43 - 52.03)^(2))` `= 98.98 ohm` `I_(upsilon) = (E_(upsilon))/(Z) = (230)/(198.97) = 1.156 A` `:.` Average power transferred/cycle to resistance `P_(R ) = I_(upsilon)^(2) xx R = (1.156)^(2) xx 15 = 20.05` watt |
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| 498. |
In an a.c circuit, value of voltage and current change every instant. Therefore, power of an a.c. circuit at any instant is the product of instantaneous voltages `(E)` and instantaneous current `(I)`. The average power supplied to a pure resistance `R` over a complete cycle of a.c is `P = E_(upsilon). I_(upsilon)`. When circuit is inductive, average power/cycle `= E_(upsilon) I_(upsilon) cos phi`, where `phi` is the phase angle between alternating voltage an altenating current in the circuit. In an a.c. circuit, `800 mH` inductor and a `60 mu F` capacitor are connected in series with `15 ohm` resistance. The a.c. supply to the circuit is `230 V, 50 Hz` The average power transferred per cycle t inductors isA. zeroB. `40.10 W`C. `20.05 W`D. `20.5 W` |
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Answer» Correct Answer - A `P_(L) = E_(upsilon) I_(upsilon) cos phi` In an inductor, phase difference `phi = 90^(@)` `:. P_(L) = E_(upsilon) I_(upsilon) cos 90^(2) =` zero |
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| 499. |
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) |
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Answer» (i) As the town is 10 km away, length of pair of Cu-wires used, `l = 20 km = 20000 m`. Resistance of wire, `R = (rho l)/(a) = (rho l)/(pi r^(2)) = (1.7 xx 10^(-8) xx 20000)/(3.14 (0.5 xx 10^(-2))^(2)) = 4 Omega` When Power `P = 10^(6)` watt is transmitted at V = 220 V. Current drawn, `I = (P)/(V) = (10^(6))/(220) = 0.45 xx 10^(4) A` Power loss, `I^(2) R = (0.45 xx 10^(4))^(2) xx 4 = 0.81 xx 10^(8) W`, which is so large. Therefore, this method of transmission of power is not feasible. (ii) When power `P = 10^(6)` wattis transmitted at 11000 V, Current drawn, `I = (P)/(V) = (10^(6))/(11000) = (10^(3))/(11) A` , Power loss, `I^(2) R = ((10^(3))/(11))^(2) xx 4 = 3.3 xx 10^(4)` watt Fraction of ohmic losses to power transmitted `= (3.3 xx 10^(4))/(10^(6)) xx 100% = 3.3%` |
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An average emf of 25 V is induced in an inductor when the current in it is changed from 2.5 A in one direction ot the same value in the opposite direction in 0.1s. Find the self-inductance of the inductor. |
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Answer» Correct Answer - 0.5 H |
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