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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.
| 601. |
The current in a coil decreases from 5 A to 0 in 0.1 sec. If the average e.m.f induced in the coil is 50 V, then the self inductance of the coil isA. 0.25 HB. 0.5 HC. 1 HD. 2 H |
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Answer» Correct Answer - C `L = e(dt)/(dI) = (50 xx 0.1)/(5)=1H`. |
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| 602. |
P and Q are two circular thin coils of same radius and subjected to the same rate of change of flux. If coil P is made up of copper and Q is made up of iron, Then the wrong statement is- A. emf induced in the coils is the sameB. the induced current in P is more than that in QC. the induced current in P and Q are in the same directionD. the induced current are the same in both the coils |
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Answer» Correct Answer - D Induced emf will be same but resistance of both loop will be different. |
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| 603. |
Define reactance X and impedance Z. Can these be nagative ? If yes, when and what does it imply ?` |
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Answer» For definitions, see text. As `X = X_(L) = X_(C ) =` negative, when `X_(C ) gt X_(L)`, i.e., reactance X can be negative. It implies, alternating current leads the applied alternating voltage in the circuit. However cannot be negative. |
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| 604. |
Alternating current shows which of the following effects?A. Chemical effectB. Magnetic effectC. Heating effectD. All of these |
| Answer» Correct Answer - D | |
| 605. |
The frequency of ac mains in India isA. 30 HzB. 50 HzC. 60 HzD. 120 Hz |
| Answer» Correct Answer - B | |
| 606. |
When the current through a solenoid increases at a constant rate, the induced current in the solenoidA. is a constant and is in the direction of the inducing currentB. is a constant and is opposite to the direction of the inducing currentC. increase with time and is in the direction of inducing currentD. increase with time and is opposite to the direction of inducing current |
| Answer» Correct Answer - B | |
| 607. |
A conducting rod of length l is moving in a transverse magnetic field of strength B with veocity v. The resistance of the rod is R. The current in the rod isA. `(Blv)/(R)`B. BlvC. zeroD. `(B^(2)v^(2)l^(2))/(R)` |
| Answer» Correct Answer - C | |
| 608. |
A coil of N turns and mean cross-sectional area A is rotating with uniform angular velocity `omega` about an axis at right angle to uniform magnetic field B. The induced emf E in the coil will beA. `NBAsinomegat`B. `NBomegasinomegat`C. `NB//Asinomegat`D. `NBAomegasinomegat` |
| Answer» Correct Answer - D | |
| 609. |
A coil of cross-sectional area A having n turns is placed in uniform magnetic field B. When it is rotated with an angular velocity `omega`, the maximum e.m.f. induced in the coil will be :A. `nAB omega`B. `nAB//omega`C. `nAomega//B`D. `Bomega//nA` |
| Answer» Correct Answer - A | |
| 610. |
Switch `S` is closed `t=0`, in the circuit shown. The change in flux in the inductor `(L=500mH)` from `t=0` to an instant when it reaches steady state is A. `2WB`B. `1.5Wb`C. `0Wb`D. None of these |
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Answer» Correct Answer - B Initial current `=10/10 =1A` :. `phi_i=L(I_i)=500mWb=0.5Wb` Final current `=20/5=4A` `phi_f=L(I_f)=(0.5)xx4=2 Wb` `:. /_phi=1.5Wb` |
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| 611. |
An `L-C` circuit consists of a `20.0 mH` inductor and a `0.5 muF` capacitor. If the maximum instantaneous current is `0.1 A`, what is the greatest potential difference across the capacitor? |
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Answer» `1/2Li_(max)^2 =1/2CV_(max)^2` `:. V_(max)=(sqrt(L/C)) i_(max)` `=(sqrt((20xx10^-3)/(0.5xx10^-6)))(0.1)` `=20V` |
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| 612. |
In an `L-C` circuit, `L = 0.75 H` and `C =18 muF`, (a) At the instant when the current in the inductor is changing at a rate of `3.40 Ns`, what is the charge on the capacitor? (b) When the charge on the capacitor is `4.2 xx 10^-4` C, what is the induced emf in the inductor? |
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Answer» a. `V_L=V_C` `L(di)/(dt)=q/C` `q=(LC)=(di)/(dt)` `=0.75xx18xx10^Y-6)(3.4)` `=45.9xx10^-6C` `=45.9muC` b. `V_(L) = V_(C) = (q)/(C)` `= (4.2 xx 10^(-4))/(18 xx 10^(-6))` `= 23.3 V` |
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| 613. |
The word generator is a misnomer.Why ? |
| Answer» In a generator, mechanical energy of rotation of an armature coil is being transformed into electrical energy. It is thus only a converter, rather than generator. | |
| 614. |
Consider the situation shown in the figure. The wire `"AB"` is sliding on the fixed rails with a constant velocity. If the wire `"AB"` is replaced by semicircular wire, the magnitude of the induced current will A. IncreaseB. Remain the sameC. DecreaseD. Increase or decrease depending on whether the semicircle bulges towards the resistance or away from it |
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Answer» Correct Answer - b |
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| 615. |
Consider the situation shown in the figure. The wire `"AB"` is sliding on the fixed rails with a constant velocity. If the wire `"AB"` is replaced by semicircular wire, the magnitude of the induced current will A. decreaseB. remain the sameC. increaseD. increase or decrease depending on whether the semicircle bulges towards the resistance or away from it. |
| Answer» Correct Answer - B | |
| 616. |
A uniform magnetic field existsin region given by `vec(B) = 3 hat(i) + 4 hat(j)+5hat(k)`. A rod of length `5 m` is placed along `y`-axis is moved along `x`- axis with constant speed `1 m//sec`. Then the magnitude of induced `e.m.f` in the rod is :A. zeroB. `25 v`C. `20 v`D. `15 v` |
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Answer» Correct Answer - B `epsilon=vecB(vecVxxvecl)` `=(3hati+4hatj+5hatk)[1hatixx5hatj]` `epsilon=25 "volt"` |
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| 617. |
A conductor `AB` of length `l` moves in `x y` plane with velocity `vec(v) = v_(0)(hat(i)-hat(j))`. A magnetic field `vec(B) = B_(0) (hat(i) + hat(j))` exists in the region. The induced emf isA. (a) zeroB. (b) `2B_(0)lv_(0)`C. (c ) `B_(0)lv_(0)`D. (d) `sqer(2)B_(0)lv_(0)` |
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Answer» Correct Answer - A (a) `vec(l), vec(v) and vec(B)` are coplanar. |
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| 618. |
The north pole of a magnet is brough down along the axis of a horizontal circular coil (figure) As a result the flux through the coil changes from `0.4 Weber` to `0.9 Weber` in an interval of half of a second.Find the average `emf` induced during this period.Is the induced current clockwise and anticlockwise as you look into the coil from the side of the magnet? |
| Answer» Correct Answer - A::C | |
| 619. |
An armature coil consists of 20 turns of wire each of area `A = 0.09 m^(2)` and total resistance 15.0 ohm. It rotates in a magnetic field of 0.5 T at a constant frequency of `(150)/(pi) Hz`. Calculate the value of (i) maximum and (ii) average induced e.m.f. produced in the coil. |
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Answer» Here,`N = 20, A = 0.09 m^(2)`, `R = 15.0 ohm, B = 0.5 T` `v = (150)/(pi) Hz, e_(0) = ?` `e_(0) = N AB omega = NAB (2 pi v)` `= 20 xx 0.09 xx 0.5 (2 pi xx (150)/(pi)) = 270 V` As e.m.f. produced is alternating, average induced e.m.f. produced in the coil is zero. |
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| 620. |
A rectangular loop of area `20 cm xx 30 cm` is held in a magnetic field of 0.3 T with its plane inclined at (i) `30^(@)` to the field (ii) parallel to the field. Find magnetic flux linked with the coil in each case. |
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Answer» Here, `A = 20 xx 30 cm^(2)` ` = 600 xx 10^(-4) m^(2) = 6 xx 10^(-2) m^(2)` `B = 0.3 T` `theta_(1) = (90^(@) - 30^(@)) = 60^(@) , theta_(2) = 90^(@) - 0^(@) = 90^(@)` `phi_(1) = BA cos theta_(1) = 0.3 xx 6 xx 10^(-2) cos 60^(@)` `= 0.9 xx 10^(-2)Wb` `phi_(2) = BA cos theta_(2) = 0.3 xx 6 xx 10^(-2) cos 90^(@) = Zero` |
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| 621. |
A circular coil of 50 turns an diameter 24 cm is rotated continuously in a uniform magnetic field of induction `3.6xx10^(-4)`T so as to cut the lines of induction of the field. If the speed of rotation is `5pi`rad/s. the instantaneous induced emf when the plane of the coil is inclined at `30^(@)` to the direction of the field is nearlyA. 15mVB. 14mVC. 10mVD. 11mV |
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Answer» Correct Answer - D `e=e_(0)sinomegat` `=nAB omega sintheta` `=50xx3.14xx144xx10^(-4)xx5xx3.14xxsin60` `=11.08xx10^(-3)V` `~~11mV` |
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| 622. |
For given values to `T` and `m`, resonance length in winter is `x_(1)` cm, and in summer, it is `x_(2)` cm. ThenA. `x_(1) = x_(2)`B. `x_(2) gt x_(1)`C. `x_(2) lt x_(1)`D. cannot say |
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Answer» Correct Answer - B `n = (upsilon)/(lambda) = (1)/(2 l) sqrt((T)/(m))` Is summer, `upsilon` is more. Therefore, resonance length `(l)` is more , and vice- versa. Therefore, `x_(2) gt x_(1)` |
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| 623. |
In an AC circuit `X_(C)=X_(L)`. The phase differe3nce between the current and voltage will be |
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Answer» Correct Answer - A If `X_(L)=X_(C)` then alternating emf is in phase with alternating current. |
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| 624. |
A long solenoid of diameter 0.1 m has `2 xx 10^(4)` turns per meter. At centre of the solenoid is 100 turns coil of radius 0.01 m placed with its axis coinciding with solenoid axis. The current in the solenoid reduce at a constant rate to 0A from 4 a in 0.05 s . If the resistance of the coil is `10 pi^(2) Omega`, the total charge flowing through the coil during this time isA. `32 pi mu C`B. ` 16 mu C`C. ` 32 mu C`D. `16 pi mu C` |
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Answer» Correct Answer - C Here, `r_(1) = (1)/(20) m, N_(1) = 2 xx 10^(4), l = 1 m` `r_(2) = (1)/(100) m, N_(2) = 100, (dI)/(dt) = (4 - 0)/(0.05) = 80 A//s` `R = 10 pi^(2)` `e = M (dI)/(dt) = (mu_(0) N_(1) N_(2) A_(2))/(l) xx (dI)/(dt)` `= 4 pi xx 10^(-7) (2 xx 10^(4)) 10^(2) pi ((1)/(100))^(2) xx 80` `= 640 pi^(2) xx 10^(-5)` `Q = I xx dt = (e)/(R ) xx dt = (640 pi^(2) xx 10^(-5))/(10 pi^(2)) xx 0.05` `= 32 xx 10^(-6) C = 32 mu C` |
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| 625. |
A long solenoid of diameter 0.1 m has `2 xx 10^(4)` turns per meter. At centre of the solenoid is 100 turns coil of radius 0.01 m placed with its axis coinciding with solenoid axis. The current in the solenoid is decreased at a constant rate form + 2 A to - 2 A in 0.05 s. Find the e.m.f. induced in the coil. Also, find the total charge flowing through the coil during this time, when the resistance of the coil is `10 pi^(2) ohm`. |
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Answer» Here, `r_(1) = 0.1//2 = 0.5 m, N_(1) = 2 xx 10^(4)`, `N_(2) = 100, r_(2) = 0.01 m, dI = - 2 - 2 = - 4 A` `dt = 0.05 s, R = 10 pi^(2) ohm, e = ? Q = ?` Mutual inductance of the two solenoids is `M = mu_(0) N_(1) N_(2) A_(2) = mu_(0) N_(1) N_(2) (pi r_(2)^(2))` `4 pi xx 10^(-7) xx 2 xx 10^(4) xx 100 xx pi (0.01)^(2)` `M = 8 pi^(2) xx 10^(-5) H` `e = (M dI)/(dt) = 8 pi^(2) xx 10^(-5) xx ((-4))/(0.05) = 6.31 xx 10^(-2) V` Changing induced, `q = I dt = (e)/(R ) dt` `= (6.31 xx 10^(-2) xx 0.05)/(10 pi^(2))` `q = 3.2 xx 10^(-5) C` |
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| 626. |
In an LCR series circuit the capacitance is changed from `C` to `4C` For the same resonant fequency the inductance should be changed from `L` to .A. 2 LB. L/2C. L/4D. 4L |
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Answer» Correct Answer - C `f_(1)=f_(2)` `(1)/(2pisqrt(L_(1)C_(1)))=(1)/(2pisqrt(L_(2)C_(2)))` But, `C_(2)=4C_(1)` `L_(1)=4L_(2)` `therefore L_(2)=(L_(1))/(4)=(L)/(4)` |
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| 627. |
The Figure shows an electromagnetic gun. A bar of mass m, resistance R and length L is free to slide on two smooth rails separated by a distance L. A uniform magnetic field B is present perpendicular to the plane of the figure. A capacitor of capacitance C is charged using a battery of emf `V_(0)` by placing switch (S) at 1. To fire the gun (i.e., to impart a kinetic energy to the rod) the switch is shifted to position 2 after the capacitor is fully charged. The rails end abruptly at the point where the speed of the rod becomes maximum. The efficiency of the gun can be defined as the kinetic energy imparted to the bar divided by the energy spent by the battery while charging the capacitor. Calculate the efficiency of the gun. Neglect self inductance of the circuit. |
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Answer» Correct Answer - `(mB^(2)L^(2)C)/(2(m+B^(2)L^(2)C)^(2))` |
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| 628. |
In a magnetic field as shown, in Fig. two horizontal wires of same mass and length `l_(1) and l_(2)` are free to slide on different vertical rails with velocities `v_(1) and v_(2)` respectively. If the resistance of two circuits are same and `a_(1) and a_(2)` are acceleration of two horizontal wires respectively, the condition for `a_(1)gta_(2)` is A. `(l_(1))/(l_(2)) = (upsilon_(2))/(upsilon_(1))`B. `(l_(1))/(l_(2)) = ((upsilon_(2))/(upsilon_(1)))^(1//2)`C. `(l_(1))/(l_(2)) gt ((upsilon_(2))/(upsilon_(1)))^(1//2)`D. `(l_(1))/(l_(2)) gt ((upsilon_(2))/(upsilon_(1)))` |
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Answer» Correct Answer - B Since `F = (B^(2) l^(2) upsilon)/(R )0` so `(B^(2) l_(1)^(2) upsilon_(1))/(R )mg = ma_(1)` and `(B^(2) l_(1)^(2) upsilon_(2))/(R ) = mg = ma_(2)` Subtracting (ii) from (i) `ma_(1) - ma_(2) = (B^(2))/(R ) (l_(1)^(2) upsilon_(1) - l_(2)^(2) upsilon_(2))` `a_(1) - a_(2) = (B^(2))/(mR) (l_(1)^(2) upsilon_(1) - l_(2)^(2) upsilon_(2))` For `a_(1) gt a_(2)` `l_(1)^(2) upsilon_(1) - l_(2)^(2) upsilon_(2) gt 0` or `(l_(1))/(l_(2)) = ((upsilon_(2))/(upsilon_(1)))^(1//2)` |
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| 629. |
A 80 V - 800 W heater is to be operated on a 100 V - 50 Hz a.c. supply. Calculate the inductance of the choke required. |
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Answer» Here, `V = 80 V, P = 800 W` current required, `I = (P)/(V) = (800)/(80) = 10 A`, `R = (V^(2))/(P) = (80 xx 80)/(800) = 8 Omega` For a.c. supply, `E_(v) = 100 V, I_(v) = I = 10 A`, `v = 50 hz` `:. Z = (E_(v))/(I_(v)) = (100)/(10) = 10 ohm` Now `R^(2) + X_(L)^(2) = Z^(2)` `:. X_(L) = sqrt(Z^(2) - R^(2)) = sqrt(10^(2) - 8^(2)) = 6 ohm` `X_(L) = omega L = 2 pi v L` `:. L = (X_(L))/(2 pi v) = (6)/(2 xx 3.14 xx 50) = 0.019 H` |
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| 630. |
In a coil of resistance R, magnetic flux due to an external magnetic field varies with time as `phi=k(C-t^(2))`. Where k and C are positive constants. Find the total heat produced in coil in time t=0 to t=C.` |
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Answer» Correct Answer - `(4k^(2)C^(2))/(3R)` |
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| 631. |
An e.m.f. is produced in a coil, which is not connected to an external voltage source. This can be due toA. the coil being in a time varying magnetic fieldB. the coil moving in a time varying magnetic fieldC. the coil moving in a constant magnetic fieldD. the coil is stationary in exteranl spatially magnetic field, which does not change with time |
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Answer» Correct Answer - A,B,D Here, magnetic flux linked with the isolated coil change when the coil being in a time varying magnetic field, the coil moving in a constant magnetic field or in time varying magnetic field. |
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| 632. |
Two identicaly induction coils each of inductance `L` joined in series are placed very close to each other such that the winding direction of one is exactly opposite to that of the other, what is the net inductance?A. `L^(2)`B. `2L`C. `L//2`D. `Zero` |
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Answer» Correct Answer - D when the two coils are joined in series such that the winding of one is opposite to the other, then the emf produced in first coil is `180^(@)` out of phase of the emf produced in second coil. Thus, emf produced in first coil is negative and the emf produced in second coil is positive so, net inductance is `L_(net)=L_(1)+L_(2)=L+LimpliesL_(net)=-(varphi)/(i)+(varphi)/(i)=` |
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| 633. |
A coil of `Cu` wire (radius `r`, self-inductance `L`) is bent in two concentric turns each having radius `(r )/(2)`. The self-inductance nowA. `2L`B. `L`C. `4L`D. `L//2` |
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Answer» Correct Answer - A ` therefore LpropN^(2)r,`" "`(L_(1))/(L_(2))=(N_(1)/(N_(2)))^(2)xx(r_(1))/(r_(2))` `implies(L)/(L_(2))=((1)/(2))^(2)xx((r)/(r//2))=(1)/(2), L_(2)=2L` |
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| 634. |
A long solenoid of length `L`, cross section `A` having `N_(1)` turns has about its center a small coil of `N_(2)` turns as shows in Fig The mutual inductance of two circuits is A. `(mu_(0)A(N_(1))//N(2))/(L)`B. `(mu_(0)A(N_(1))//N(2))/(L)`C. `mu_(0)AN_(1)N_(2)L`D. `(mu_(0)AN_(1)//N_(1)^(2))/(L)` |
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Answer» Correct Answer - B `varphi_(0)=N_(2)B_(1)A` or `varphi_(2)=N_(2)(mu_(0)N_(1)N_(2)A)/(L)` or`=varphi_(2)=(mu_(0)N_(1)N_(2)A)/(L)I-2` comparing with `varphi_(2)=MI_(1),we g et(mu_(0)N_(1)N_(2)A)/(L)` |
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| 635. |
What is the mutual inductance of coil and solenid if a has a radius 4 cm and coil of 700 turns is would on the middle part of the solenoid ?A. 44.17 mHB. 48.94 mHC. 34.34 mHD. 36.73 mH |
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Answer» Correct Answer - A Flux linked with coil `phi_(2)=N_(2)BIA=(mu_(0)N_(1)N_(2)pir^(2)l)/(l)` also `phi=MI` Comparing the equations, `M=(mu_(0)N_(1)N_(2)pir^(2)l)/(l)` `M=(4xx3.14xx10^(-7)xx5000xx700xx3.14xx(4xx10^(-2))^(2))/(0.5)` `M=44.17 mH` |
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| 636. |
Why cannot a transformer be used to step up d.c. voltage ? |
| Answer» This is because d.c. voltage cannot produce a changing magnetic flux required in the working of a transformer. | |
| 637. |
Can r.m.s value of current be defined in terms of chemical effect of current ? |
| Answer» No. because a.c. cannot be used in eletrolysis. | |
| 638. |
We can measure d.c. by an ordinary ammeter, but not a.c. Why ? |
| Answer» This is because average value of a.c. over a complete cycle is zero. | |
| 639. |
The peak value of 220 a.c. is |
| Answer» `E_(0) = sqrt(2) E_(v) = sqrt2 xx 200` volt `= 311` volt. | |
| 640. |
The time constant of C-R circuit is |
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Answer» Time constant of R-L circuit tells us how fast of how slow is the `growth//decay` of current in te R-L circuit. Low value of time constant indicates that the growth and decay are fast. Large values of time cosntant indicate that growth and decay of time constant indicate the growth and decay of current in the circuit are slow. For a given circuit, time constant is fixed. Its value can be optimised by selecting suitable value of R and L. |
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| 641. |
Find the time taken by 60 Hz a.c. to reach its peak value from zero. |
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Answer» As `v = 60 Hz, T = (1)/(v) = (1)/(60) s` In going from zero value to peak value, time taken, `t = (T)/(4) = (1)/(4 xx 60) s = (1)/(240) s` |
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| 642. |
What is the average value of a.c. over a full cycle ? |
| Answer» Average value of a.c over a cycle is zero. This is because a.c. is positive during one half cycle and equally negative in the other half cycle. | |
| 643. |
What is the significance of high Q-value circuits ? |
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Answer» As is known, the Q factor of Quality factor of an electronic circuit defines sharpness of turning at reasonance. It is given by `Q = (1)/(R ) sqrt((L)/(C ))` Q is only a numbe having no dimensions. It represents voltages amplification factor of the circuit At the receiving station, many radio signals are oftern present over a range of frequencies. In order to eliminate unwanted signals, we need to design high Q value circuit. Higher the value of Q, narrower and sharper is the resonance. Thus the electronic circuits with high Q values would respond to a very narrow range of frequencies. The value of Q usually vary from 10 to 100. The electronic circuits dealing with very high frequencies may have `Q = 200`. |
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| 644. |
A uniform magnetic field B exists in a direction perpendicular to the plane of a square frame made of copper wire. The wire has a diameter of 2 mm and a total length of 40cm. The magntic field changes with time at a steady rate `(dB)/dt = 0.02 T s ^(-1)`. Find the current induced in the frame. Resistivity of copper `=1.7 X 10 ^(-8) Wm`. |
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Answer» Area of frame, `A = ((40)/(4) xx 10^(-2) m)^(2) = 10^(-2) m^(2)` `e = (d phi)/(dt) = A (db)/(dt) = 10^(2) xx 0.02 = 2 xx 10^(-4) V` Resistance of loop, `R = (rho l)/(a) = (rho l)/(pi r^(2))` `R = (1.7 xx 10^(-8) xx 40 xx 10^(-2))/((22)/(7) xx (10^(-3))^(2)) = 2.16 xx 10^(-3) ohm` Induced current, `i = (e)/(R ) = (2 xx 10^(-4))/(2.16 xx 10^(-3))` `= 9.3 xx 10^(-2) A` |
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| 645. |
The number of turns in secondary coil of a transformer is 100 times the number of turns in primary coil. What is the transformer ratio ? |
| Answer» `K = (n_(s))/(n_(p)) = 100` | |
| 646. |
Statement-1 : The number of turns in secondary coil of a transformer is 10 times the number of turns in primary. An output voltage of `15 V` can be obtained using a cell of `1.5 V`. Statement-2 : This is because in a transformer, `(E_(s))/(E_(P)) = (n_(s))/(n_(P))`A. AB. BC. CD. D |
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Answer» Correct Answer - D Statement-1 is false, as a transformer does not work an d.c. statement-2 is true. |
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| 647. |
When an alternating current flows through a circuit consisting of a resistor in series with a capacitor, during the cycle at some instant,it is possible to have (a) voltage across the circuit is zero but current through it is not zero.A. voltage across the circuit is zero but current through it is not zeroB. current through the circuit is zero but the voltage across it is not zeroC. current throught the capacitor is not zero but the voltage across it is zeroD. current throught the resistor is not zero but the voltage across it is zero |
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Answer» Correct Answer - A::B::C As current and voltage are in phase in resistor, so option (d) is not possible, but due to capacitor, all other options are possible as there is phase difference between current and voltages. |
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| 648. |
Calculate the average value of the AC per cycle for which time function of the current is shown in figure-5.174. |
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Answer» Correct Answer - `(3)/(4)I_(0)` |
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| 649. |
What is meant by one cycle of a.c. ?A. going from zero to + max.B. going from + max. to zeroC. going from zero to-max and - max to zeroD. all the three `a, b, c` combined together |
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Answer» Correct Answer - D From the knowledge of theory, (d) is the correct choice. |
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| 650. |
If `n` is frequency of a.c. mains, then frequency of sonometer wire at resonance isA. ` 2 n`B. `n`C. `n//2`D. `4 n` |
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Answer» Correct Answer - A Frequency of sonometer wire is `2 n`, being double the frequency of a.c. mains. |
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