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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.
| 751. |
Which of the two curves shown below has lesser time constant. |
| Answer» Correct Answer - Curve 1 | |
| 752. |
A capacitor of `10 mu F` is connected to an a.c. source of e.m.f. `E = 220 sin 100 pi t`. Write the equaiton of instananeous current through the circuit. What will be the reading of a.c. ammeter connected in the circuit ? |
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Answer» Here, `C = 10 mu F = 10 xx 10^(-6) F` `E = 220 sin 100 pi t = E_(0) sin omega t` `:. E_(0) = 220 V, omega = 2 pi v = 100 pi, v = 50 hz`. `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` ` =(1)/(2 xx 3.14 xx 50 xx 10^(-5)) = 318.5 Omega` `I_(0) = (E_(0))/(X_(C )) = (220)/(318.5) = 0.691 A` Reading of a.c. ammeter, `I_(v) = (I_(0))/(sqrt2) = (0.691)/(1.414) = 0.489 A` Instantaneous current in the circuit, `I = I_(0) sin (omega t + pi//2)` `I = 0.691 sin (100 pi t + pi//2)` |
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| 753. |
Following readings were obtained in the experiment to determine self inductance of a coil. If frequency of a.c. is 50 Hz, what is the value of self inductance ? |
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Answer» From D.C. Part. `R_(1) = (V_(1))/(I_(1)) = (0.5)/(0.1) = 5 Omega, R_(2) = (0.6)/(0.12) = 5 Omega` `:.` Average value of `R = (R_(1) + R_(2))/(2) = (5 + 5)/(2) = 5 Omega` `|{:(,"D.C.Part",),("S.No."," V"," I"),(1,0.5V,0.1A),(2,0.6V,0.12A):}| " " |{:(,"A.C.Part",),("S.No."," V"," I"),(1,1.0V,0.1A),(2,1.6V,0.16A):}|` From A.C. Part `Z_(1) = (V_(1))/(I_(1)) = (1.0)/(0.1) = 10 Omega , Z_(2) = (1.6)/(0.16) = 10 Omega` Average value of `Z = (Z_(1) + Z_(2))/(2) = (10 + 10)/(2) = 10 Omega` As `X_(L) = omega L = 2 pi L = sqrt(Z^(2) - R^(2))` `L = (sqrt(Z^(2) - R^(2)))/(2 pi v) = (sqrt(10^(2) - 5^(2)))/(2 xx 3.14 xx 50) = (sqrt 75)/(314)` `= 0.0276 H` |
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| 754. |
An electric lamp which runs at 80 volt d.c. and consumes 10 ampere is connected to 100 volt, 50 Hz a.c. mains. Calculate the inductance of the choke required. |
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Answer» Here, `V = 80 V, I = 10 A`, `R = (V)/(I) = (80)/(10) = 8 Omega` `E_(v) = 100 V, v = 50 Hz, I_(v) = I = 10 A, L = ?` If Z is impedance of lamp and choke coil, then `Z = (E_(v))/(I_(v)) = (100)/(10) = 100 Omega` As `R^(2) + X_(L)^(2) = Z^(2)` `:. X_(L)^(2) = Z^(2) - R^(2) = 10^(2) - 8^(2) = 36, X_(L) = 6 Omega` Now `X_(L) = omega L = 2 pi v L` `:. L = (X_(L))/(2 pi v) = (6 xx 7)/(2 xx 22 xx 50) = 1.9 xx 10^(-2) H` |
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| 755. |
In a pure LC circuit, what is the energy stored when peak current is `I_(0)`? |
| Answer» Total energy stored in inductor `= (1)/(2) LI_(0)^(2)`. | |
| 756. |
An electrical element X when connected to an alternating voltage source has current through it leading the voltage by `pi//2` radian. Identify X and write an expression for its reactance. |
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Answer» As current through element X leads the alternating voltage applied by `pi//2` radian, therefore, X is a pure capacitance. Its reactance is `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` |
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| 757. |
At resonance in an a.c. circuit, what is the value of power factor ? |
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Answer» At resonance. `Z = R`, `:.` Power factor, `cos phi = (R )/(Z) = (R )/(R ) = 1 =` maximum. |
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| 758. |
A wire is sliding as shown in Figure. The angle between the acceleration and the velocity of the wire is A. (a) `30^(@)`B. (b) `40^(@)`C. ( c) `120^(@)`D. ( d) `90^(@)` |
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Answer» Correct Answer - C ( c) `vec(F)_(m) _|_ vec(v)` ``:.` theta = 90^(@) + 30^(@) = 120^(@)` |
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| 759. |
A `0.1 m` long conductor carrying a current of `50 A` is perpendicular to a magentic field of `1.25 mT`. The mechanical power to move the conductor with a speed of `1ms^(-1)` isA. (a) `0.25 mW`B. (b) `6.25mW`C. ( c) `0.625W`D. (d) `1W` |
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Answer» Correct Answer - B (b) `P = Fv = Bilv = 1.25 xx 10^(-3) xx 50 xx 0.1 xx 1 W` `= 6,25 xx 10^(-3) W = 6.25m W` An iterating alternating `P = EI = (Blv) I` |
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| 760. |
A rectangular copper coil is placed in a uniform magnetic field of induction 40 mT with its plane perpendicular to the field. The area of the coil is shrinking at a constant rate of `0.5m^(2)s^(-1)`. The emf induced in the coil isA. 10 mVB. 20 mVC. 80 mVD. 40 mV |
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Answer» Correct Answer - B Given, B = 40 mT = `40xx10^(-3)T` Rate of shrinking of area = `0.5m^(2)s^(-1)` `rArr" "(dA)/(dt)=0.5m^(2)s^(-1)` As, `phi=BA cos theta=BA" "[because cos theta=cos 0^(@)=1]` On differentiating with w.r.t. to t, `(dphi)/(dt)=B.(dA)/(dt)=e` `rArr" "e= "induced emf" = 40xx10^(-3)xx0.5=20xx10^(-3)V` = 20 mV |
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| 761. |
Changing magnetic fields can set up current loops in nearby metal bodies and the currents are called asA. eddy currentsB. flux currentsC. alternating currentsD. leaking currents |
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Answer» Correct Answer - A Changing magnetic field can set up current loop in nearly metal bodies and the current is called as eddy current. |
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| 762. |
Electric field induced by changing magnetic fields areA. ConservativeB. Non-conservativeC. May be conservatie or non-conservative depending on the conditionD. Nothing can be said |
| Answer» Correct Answer - B | |
| 763. |
Statement 1: Electric field produced by changing magnetic field is nonconservative. Statement 2: For the electric field `vec E` induced by a changing magnetic field which has closed lines of force, `oint vec E . vec(dl) =0`A. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 764. |
Show that if two inductors with equal inductance `L` are connected in parallel then the equivalent inductance of combination is `L//2`.The inductors are separated by a large distance. |
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Answer» Correct Answer - A::B `1/L_(eq)=1/L_(1)+1/L_(2)=1/L+1/L=2/L` or `L_(eq)=1/2` |
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| 765. |
A long cylinder of radius `a` carrying a uniform surface charge rotates about its axis with an angular velocity `omega`. Find the magnetic field energy per unit length of the cylinder if the linear charge density equals `lambda` and `mu = 1`. |
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Answer» Correct Answer - `(mu_(0)lamda^(2)omega^(2)a^(2))/(8pi)` |
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| 766. |
Express the induced emf as scalar triple product of `vecv, vecB and vecl`. |
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Answer» `e=intde=int(vec v xx vec B)vec(dl)` If `vecv and vecB` are uniform upon the entire length l of the `e=(vec v xx vecB)vecJ` |
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| 767. |
In example 9, if B is the mid point of rod, then find `e_(OB)//e_(BA)`. |
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Answer» `e_(OB)=(1)/(2)Bomega((l^(2))/(4))=(Bomegal^(2))/(8)` `e_(BA)=(1)/(2)Bomega(x^(2))_(1//2)^(l)=(1)/(2)Bomegaxx(3l^(2))/(4)Bomegal^(2)` `therefore" "(e_(OB))/(e_(BA))=(1)/(3)` |
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| 768. |
If the rod were translating as well as rotating, what will be its emf? Assume that the centre of mass has a velocity v and the rod is rotating with an angular velocity `omega` atout its centre of mass. |
| Answer» `(Bomegal^(2))/(2) pm Blv` | |
| 769. |
In the figure magnetic energy stored in the coil is A. zeroB. infiniteC. `25` joulesD. none of the above |
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Answer» Correct Answer - C `i=(V)/(R )=(10)/(52)=5A` `U=(1)/(2)Li^(2)=(1)/(2)xx2xx25=25J` |
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| 770. |
Compare the expression for magnetic energy density with electrostatic energy density stored in the space between the plates of a parallel plate capacitor. |
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Answer» The enectrostatic energy stored `U=(1)/(2)(q^(2))/(C)=(q^(2))/(2epsilon_(0)A)xxd" …(i)"` (where C is the capacitance, `C=(epsilon_(0)A)/(d)`, q is the charge upon capacitor) Now the electric field in the space (between plates of capacitor) `E=(q)/(epsilon_(0)A)" ...(ii)"` From equation (i) and (ii), `U=(1)/(2)((q)/(epsilon_(0)A))^(2)xxepsilon_(0)xxAxxd` `rArr" "U=(1)/(2)epsilon_(0)E^(2)xx"Volume"` Hence `(U)/("Volume")=(1)/(2)epsilon_(0)E^(2)" ...(iii)"` This expression can be compared with magnetic energy per unit volume which is equal to `(B^(2))/(2mu_(0)).` ....(iv) |
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| 771. |
What is the physical role of self inductance? |
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Answer» Physically, the self inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf in establishing the current. This work done is stored as magnetic potential energy. `W=U=(1)/(2)Ll^(2)" ...(i)"` This is equivalent to (mechanical) kinetic energy of a particle of mass m, `K=(1)/(2)mv^(2)" ...(ii)"` In above equation, L is analogous to m (i.e, L is the electrical inertia and opposes growth and decay of current in the circuit). |
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| 772. |
In an `LR`-circuit, time constant is that time in which current grows from zero to the value (where `I_(0)` is the steady state current)A. `0.63I_(0)`B. `0.50I_(0)`C. `0.37I_(0)`D. `I_(0)` |
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Answer» Correct Answer - A Current at any instant of time `t` after closing an `L-R` circuitis given by `I=I_(0)[I-e^(-(R )/(L)t)]` Time cionstant `t=(L)/(R )` `:.I=I_(0)[i_(0)-e^((-R)/(L)xx(L)/(R ))]=I_(0)(1-e^(-1))=I_(0)(1-(1)/(e))` `=I_(0)(1-(1)/(2.718))=0.63I_(0)=63%ofI_(0)` |
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| 773. |
A conducting square loop of side l and resistance R moves in its plane with a uniform velocity v perpendicular ot one of its sides. A uniform and constant magnetic field B exists along the perpendicualr ot the plane of the loop as shown in . The current induced in the loop is A. `(Blv)/(R)` , clockwiseB. `(Blv)/(R)` , anticlockwiseC. `(2Blv)/(R)` , anticlockwiseD. zero |
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Answer» Correct Answer - D Change in flux=0 `i=0` |
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| 774. |
An alternating e.m.f. given by e=200 sin 50 t is applied to a circuit containing only a resistance of `50 Omega`. What is the value of r.m.s. current in the circuit?A. 0.02828 AB. 0.2828 AC. 2.828 AD. 28.28 A |
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Answer» Correct Answer - C `I_(rms)=(e_(rms))/(R)=(e_(0))/(sqrt(2)R)=(200x0.707)/(50)` `I_(rms)=2.828A` |
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| 775. |
The rms value of alternating current which when passed through a resistor produces heat energy four times that produced by directed current of 2 A through the same resistor in same time isA. 2AB. 4AC. 8AD. 16A |
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Answer» Correct Answer - B `H_(dc)=I_(dc)^(2)Rt=4Rt=W` . . . (1) `H_(ac)=I_(ac)^(2)Rt=4W ` . . . (2) `therefore (4W)/(W)=(I_(ac)^(2)Rt)/(4Rt)` `therefore I_(ac)^(2)=16` `therefore I_(ac)=4A` |
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| 776. |
A square conductind loop, `20.0 cm` on a side placed in the same magnetic field as shown in Fig. 3.195, center of the magnetic field region, where `dB//dt = 0.035 T s^(-1)`. The potential difference between points `a and b` isA. (A) `6 V`B. ( b) `0`C. ( c) `10 V`D. (d) `12 V` |
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Answer» Correct Answer - B (b) `epsilon_(ab) = (1)/(8)epsilon`, But there is a potential drop of `V = (IR)/(8) = (epsilon)/(8)` so the potential difference `epsilon_(ab) - V` is zero. |
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| 777. |
A square conductind loop, `20.0 cm` on a side placed in the same magnetic field as shown in Fig. 3.195, center of the magnetic field region, where `dB//dt = 0.035 T s^(-1)`. he direction of induced electric field at points a, b and c:B. (b) C. ( c) D. (d) |
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Answer» Correct Answer - A (a) Induced electric field should be anticlockwise. |
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| 778. |
A square conductind loop, `20.0 cm` on a side placed in the same magnetic field as shown in Fig. 3.195, center of the magnetic field region, where `dB//dt = 0.035 T s^(-1)`. The current induced in the loop if its resistance is `2.00 Omega` isA. (a) `2.50 xx 10^(-4) A`B. (b) `4.35 xx 10^(-4) A`C. ( c) `1.25 xx 10^(-4) A`D. (d) `7.37 xx 10^(-4) A` |
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Answer» Correct Answer - D (d) `I = (epsilon)/(R ) = (A)/(R )(dB)/(dt) = (L^(2))/(R )(dB)/(dt)` `= ((0.20 m)^(2)(0.0350 T //s))/(1.90 Omega)` `= 7.37 xx 10^(-4) A` |
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| 779. |
A flexible circular loop `20cm` in diameter lies in a magneic field with magnitude `1.0 T`, direction lies into the plane of the page as shown in Fig. 3.187. The loop is pulled at the points indicated by the arrws, forming a loop of zero area in `0.314s`. if `R = 0,01 Omega`, the magnitude and direction of current flowing in the loop areA. (a) `1A`, clockwiseB. (b) `1A`, anticlockwiseC. ( c) `10 A`, clockwiseD. (d) `10 A`, anticlockwise |
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Answer» Correct Answer - C (c ) `epsilon_(av) = -(DeltaPhi_(B))/(Deltat) = -B(DeltaA)/(Deltat) = -B((-pir)^(2))/(Deltat)` Since the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore, the current flows in clockwise direction `I_(av) = (epsilon_(av) )/(R ) = (0.1)/(0.01) = 10 A` |
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| 780. |
A line charge `lambda` per unit length is pasted uniformly on to the wire of a wheel of mass `m` and radius `R`. The wheel has light non-conducting spokes and is free to rotate about a vertical axis as shown in `Fig. 3.166`. A uniform magnetic field extends over a radial rigion of radius `r` given by `B = -B_(0)hat(k)(r le a, alt R) = 0` (otherwise). What is the angular velocity of the wheel this field is suddenly switched off? A. (a) `(-2B_(0)pia^(2)r)/(mR)hat(k)`B. (b) `(-2B_(0)pia^(2)r)/(3mR)hat(k)`C. ( c) `(B_(0)pia^(2)lambda)/(mR)hat(k)`D. ( d) `(-B_(0)pia^(2)lambda)/(mR)hat(k)` |
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Answer» Correct Answer - D (d) Induced `E = (a^(2))/(2R)(dB)/(dt)` `Iomega = int tau dt` `rarr` `mR^(2)omega = int qERdt = int(qa^(2))/(2R)(dB)/(dt)Rdt = (lambda2piR)/(2R)a^(2)RintdB` `rarr` `omega = (B_(0)pia^(2)lambda)/(mR)` `rarr vec(omega) = -(B_(0)pia^(2)lambda)/(mR)hat(k)` (because clockwise sense) |
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| 781. |
Find the total field energy of magnetic field stored per unit length inside a long cylindrical wire of radius R and carrying a current I. |
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Answer» Correct Answer - `(mu_(0)I^(2))/(16pi)` |
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| 782. |
A wire frame of area `3.92xx10^(-4)m^(2)` and resistance `20Omega` is suspended freely from a 0.392m long thread. There is a uniform magnetic field of 0.784 T and the plane of wire-frame is made to oscillate under gravity by displacing it through `2xx10^(-2)m` from its initial position along the direction of magnetic field. The plane of the frame is always along the direction of thread and does not rotate about it . What is the induced EMF in wire-frame us a function of time ? Also find the maximum current in the frame. |
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Answer» Correct Answer - `[2xx10^(-6)V,10^(-7)A]` |
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| 783. |
In the figure shown, the magnet is pushed towards the fixed ring along the axis of the ring and it passes through the ring. A. when magnet goes towards the ring the face `B` becomes south pole and face `A` becomes north poleB. when magnet goes away from the ring the face `B` becomes north pole and face `A` becomes south poleC. when magnet goes away from the ring the face `A` becomes north pole and face `B` becomes south poleD. the face `A` will always be a north pole. |
| Answer» Correct Answer - C | |
| 784. |
In the figure shown, the magnet is pushed towards the fixed ring along the axis of the ring and it passes through the ring. A. when magnet goes towards the ring the face B becomes south pole and the face A becomes north poleB. when magnet goes away from the ring the face B becomes north pole and the face A becomes south poleC. when magnet goes away from the ring the face A becomes north pole and the face B becomes south poleD. the face A will always be a north pole. |
| Answer» Correct Answer - C | |
| 785. |
A square coil `ACDE` with its plane vertically is released from rest in a horizontal uniform magnetic field `vec(B)` of length `2L` . The accelaration of the coilis A. less than g for all the time till the loop crosses the magnetic field completelyB. less than g when it enters the field and greater than g when it comes out of the fieldC. g all the timeD. less than g when it enters and comes out of the field but equal to g when it is within the field. |
| Answer» Correct Answer - D | |
| 786. |
A coil having an area `A_(0)` is placed in a magnetic field which changes from `B_(0)` to `4B_(0)` in time interval t. The emf induced in the coil will beA. `3A_(0)B_(0)//t`B. `4A_(0)B_(0)//t`C. `3B_(0)//A_(0)t`D. `4B_(0)//A_(0)t` |
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Answer» Correct Answer - A As, `|e|=(Deltaphi)/(Deltat)=(4B_(0)A_(0)-B_(0)A_(0))/(t)=(3B_(0)A_(0))/(t)` |
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| 787. |
A coil having an area `A_(0)` is placed in a magnetic field which changes from `B_(0)` to `4B_(0)` in a time interval `t`. The e.m.f. induced in the coil will beA. `(3A_(0)B_(0))/(t)`B. `(4A_(0)B_(0))/(t)`C. `(3B_(0))/(A_(0)t)`D. `(4B_(0))/(A_(0)t)` |
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Answer» Correct Answer - A `e=-(dvarphi)/(dt)=(-3B_(0)A_(0))/(t)` |
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| 788. |
Find the capacity of a capacitor, which when put in series with a 10 ohm resistor, makes the power factor equal to 0.5. Assume an 80 V - 100 Hz a.c. supply. |
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Answer» Here, `C = ?, R = 10 ohm , cos phi = 0.5`, `E_(v) = 80 V, v = 100 hz` As `cos phi = (R )/(Z) , Z = (R )/(cos phi) = (10)/(0.5) = 20` As `R^(2) + X_(C )^(2) = Z^(2)` `:. X_(C ) = sqrt(Z^(2) - R^(2)) = sqrt(20^(2) - 10^(2)) = 10 sqrt3` `(1)/(omega C) = 10 sqrt3, C = (1)/(omega 10 sqrt3) = (1)/(2 pi xx 100 xx 10 sqrt3` `= 9.2 xx 10^(-5) F` |
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| 789. |
In the circuit diagram shown, initially there is no energy in the inductor and the capacitor, The switch is closed at `t = 0`. Find the current `I` as a function of time if `R=sqrt(L//C)` |
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Answer» `tau_L=L/R` and `tau_C=CR, tau_C/tau_L=(CR^2)/L=C/L.L/C=1` `:. tau_L=tau_C` `:.` From the given condition `tau_L=tau_C=tau(say)` Now, in L-R circuit `I_1=V/R(1-e^(-t//tau))` In `CR` circuit, `I_2=V/Re^(-t//tau)` `:. I=I_1+I_2=V/R=`constant. |
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| 790. |
A circuit contains an ideal cell and an inductor with a switch. Initially, the switch is open. It is closed at `t = 0`. Find the current as a function of time. |
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Answer» `epsilon=L(di)/(dt) rArr underset(0)overset(i)int epsilon dt=underset(0)overset(i)int epsilon Ldi` `epsilont=Li rArri=(epsilont)/L` |
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| 791. |
A circuit contains an ideal cell and an inductor with a switch. Initially, the switch is open. It is closed at `t = 0`. Find the current as a function of time. |
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Answer» `epsi=L(di)/(dt)implies int_(0)^(i)epsidt=int_(0)^(i)Ldi` `epsit=Li implies i=(epsit)/(L)` |
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| 792. |
An infinite current-carrying conductor is placed along the z-axis and a wire loop is kept in the `x-y` plane. The current in the conductor is increasing with time. Then theA. (a) emf induced in the wire loop is zeroB. (b) nagnetic flux passing through the wire loop is zeroC. (c ) emf induced is zero but magnetic flux is not zeroD. (d ) emf induced is not zero but magnetic flux is zero |
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Answer» Correct Answer - A::B (a, b): As `vec(B)_|_ vec(A)`, hence `phi = 0 and e = 0`. |
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| 793. |
A current I is flowing througha straight conductor PQ shown in the figure. A circular loop of metal wire is placed as shown and is coplanar. If the current in the wire is reduced to zero value, there will be A. no induced current in the loopB. clockwise current in the loopC. anticlockwise induced currentD. initially anticlockwise and then clockwise induced currents in the coil |
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Answer» Correct Answer - C The current through wire PQ is reduced, the magnetic flux through the coil also decreases. Its direction is upward and towards you. Induced current will flow so as to maintain this flux towards you. Hence, induced current will be anticlockwise. |
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| 794. |
Find the current in the sliding rod AB (resistance = R) for the arranged shown in Fig. B is constant and is out of the paper. Paralllel wires have no resistance. `upsilon` is constant. Switch is closed at time t = 0. |
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Answer» The conductor of length d moves with speed `v_1` perpendicular to magnetic field B as shown in figure. The produces motional emf across two ends of rod. Which is given by =vBd. Since, switch S is closed at time t=0 . Capacitor is charged by this potential difference. Let Q(t) is charge on the capacitor and current flows from A and B . Now , the induced current `I=(vBd)/(R)-(Q)/(RC)` On rearranging the terms, we have `(Q)/(RC)+(dQ)/(dt)=(vBd)/(R)` This is the linear differential equation. On solving we get `Q=vBdC+Ae^(-t//RC)` `rArrQ=vBdC[1-e^(-t//Rc)]` (At time `t =0,Q=0=A=-vBdc`). Differentiating, we get `I-(vBd)/(R)e^(-t//RC)` This is the required expression of current. |
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| 795. |
Self induction is the property of a coil by virture of which the coil oppose any change in the strenght of current flowing through it by inducting an e.m.f. in itself. Self induction represents electric inertia which is measured in terms of coefficient of self inductance We can show that `L = (phi)/(I) = (-e)/(dI//dt)` Read the above passage and answer the following questions : An e.m.f. of 150 microvolt is induced in a coil when current in it changes from 5A ot 1 A in 0.2 sec. what is self inductance of the coil ? (ii) How does self indcution of a coil represent its electric inertia ? (iii) What is them implication of this property in our daily life ? |
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Answer» Here, `e = 150 microvlot = 150 xx 10^(-6) V, dI = 5 - 1 = 4A, dt = 0.2 s L = ?` From `e = Ldt//dt` `L = (e xx dt)/(dI) - (150 xx 10^(-6) xx 0.2)/(4) = 7.5 xx 10^(-6) H` (ii) On account of self induction, a coil oppose increase as well as decrease in strength of current flowing through it, i.e., the coil oppose the growth and decay of current through it. Hence, Self induction is rightly representing electric ineertia of the coil. (iii) In our daily life, whatever habits, good or bad, we have developed, we tend to stick to them. None of us likes changes. we have to put in efforts. These are the values of life we learn form this property. |
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| 796. |
A rectangular conductor of area `0.2 m^(2)` is placed in a unifrom magnetic field with a B-vector strenght of 2T with its normal at an angle `30^(@)` . Calculate the magnetic flux linked with the conductor. |
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Answer» Correct Answer - 0.35 Wb |
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| 797. |
A unifrom magnetic field B points vertically up and is slowly changed in magnitude, but not in direction. the rate of change of the magnetic field is `alpha` .A conducting ring of radius r and resistance R is held perpendicular to the megnetic field, and is totally inside it. The induced current in the ring isA. zeroB. `(2pieB)/(R)`C. `(ralpha)/(R)`D. `(pir^(2)alpha)/(R)` |
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Answer» Correct Answer - D The magnitude of the induced emf in the ring is `|epsi|=(dphi)/(dt)=(d)/(dt)(BA)=A(dB)/(dt)=pir^(2)alpha` The induced current in the ring. `I=(|epsi|)/(R)=(pir^(2)alpha)/(R)` |
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| 798. |
What is the value of inductance `L` for which the current is a maximum in series `LCR` circuit with `C=10 muF` and `omega=1000(rad)/s`?A. `100 mH`B. `80 mH`C. `50 mH`D. `40 mH` |
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Answer» Correct Answer - A When the current is maximum in a series L-C-R a.c. circuit, the impedance is minimum. This happens at resonance where `X_(L)=X_(C)` i.e. `omega L =1/(omegaC) or L = 1/(omega^(2)C)` `C = 10 xx 10^(-6)F and omega = 1000 per sec` `:. L = 1/(10^(3)xx10^(3)xx10 xx 10^(-6)) = 0.1 H = 100 mH`. |
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| 799. |
An inductor L and a capacitor C are connected in parallel in an a.c. circuit as shown in the figure. The frequency of the source is equal to the resonant frequency of the circuit. Which ammeter will read zero ampere?A. Ammeter `A_(1)`B. Ammeter `A_(2)`C. Ammeter `A_(3)`D. All Ammeters `A_(1),A_(2), A_(3)` |
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Answer» Correct Answer - A This is a parallel resonance circuit. Current become zero. Hence the ammeter `A_(1)` will read zero ampere. |
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| 800. |
In the circuit shown in figure, `E = 120 V, R_1 = 30.0Omega, R_2 = 50.0 Omega` and `L=0.200H`. Switch `S` is closed at `t = 0`. Just after the switch is closed. (a) What is the potential difference `V_(ab)` across the inductor `R_1`?(b) Which point, `a` or `b`, is at higher potential? (c) What is the potential difference `V_(cd)` across the inductor `L`? (d) Which point, `c` or `d`, is at a higher potential? The switch is left closed for a long time and then is opened. Just after the switch is opened (e) What is the potential difference `V_(ab)` across the resistor `R_1`? (f) Which point a or b, is at a higher potential? (g) What is the potential difference `V_(cd)` across the inductor `L`? (h) Which point, `c` or `d`, is at a higher potential? |
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Answer» Correct Answer - D These are two independent parallel circuits across the battery. a. `V_(ab)=E=120` volt (at all instants) b.`a` is higher potential. c. `V_(cd)` will decrease exponentially from `120V` to zero. `:. V_(cd)=120`volt, just after the switch is closed. d. `c` will be at higher potential. e. When switch is opeed, current through `R_1` will immediately become zero. While through `R_2`, will decrease to zero from the value `E/R_2=2.4A=i_0` (say), exponentially. Path of this decay of current will be `cdbac`. `:.` Just after the switch is opened. `V_(ab)=-i_0R_1=-2.4xx30=-72volt` f. Point `b` is at higher potential. g. `V_(cd)=-i_0(R_1+R_2)=-2.4(80)=-192`volt h. This time point `d` will be at higher potential. |
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