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magnetism-2 ...solution
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20 May 2007 21:52:05 IST
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20 May 2007 21:52:05 IST
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magnetism-2 ...solution
Physics , Electromagnetism

S/HT/10/1                                             SOLUTION ? TEST X

 MAGNETISM, ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT

 
  1. Induced e. m. f. = L dI/dt. Hence, we have, 1 = L (5/0.1) = 50 L. So, L = 1/50 = 0.02 H = 20 mH

Correct choice: c

2.       Since the coil makes an angle 60º with the field, its area makes an angle 30º. Hence, flux = 3 m Wb = BAN Cos30 = B × 0.02 × 1 × ?3/2. So, B = ?3/0.01 milli tesla = ?3/10 T

Correct choice: b

  1. When rate of change of current is unity, induced voltage = self-inductance.

Correct choice: b

  1. The time for growth of current is governed by time constant = L/R.

Correct choice: c

  1. For a step-up transformer primary to secondary turns ratio is less than 1, as secondary has more turns than primary.

Correct choice: b

  1. Armature current is a maximum at the start and as the motor runs at a steady speed, there is a steady back e. m. f and hence the net voltage is less than the applied voltage and hence current is also less.

Correct choice: c

  1. Given, ? = 3t² + 4t ? 5 m Wb. So, induced voltage, V = d?/dt = 6t + 4 = 16 m V, when t = 2 s.

Correct choice: a

  1. Ideal inductors in series can be combined as we combine resistances. So here it is 2L

Correct choice: b

  1. Induced e. m. f. = BLvsin30 = 1 times 0.5 times 10 times ½ = 2.5 V

Correct choice: b

  1. Output voltage = ER/(R + r) = 440 times 4000/4400 = 400 V

Correct choice: b

  1. Here flux = BANCos60 = 0.002 times 0.01 times 200 times ½ = 0.002. So, change in flux when the coil is turned through 180º is twice this = 0.004 Wb. Hence the induced e. m. f. = 0.004/0.1 = 0.04 V = 40 m V.

Correct choice: a

  1. Induced e. m. f. = BAN?, in the case of a coil that is rotated. So, ratio of e. m. f. s = ratio of N? of Nf values which in this problem = 50 times 50/100 times 150 = 1/6

Correct choice: b

  1. We know that M = k times square root of product of self-inductances. So, here, ?2 = k ?8. So, k = ½ = 0.5

Correct choice: c

  1. When the loops approach each other, flux in each due to the other coil increases, and this will be opposed by the induced current. So the current in both the coils must decrease.

Correct choice: b

  1. As current in the wire increases flux through the lop increases. So current will be induced in it such that flux is decreased. And hence current must be in the clockwise direction so that the field produced by the wire and that by the loop are in opposite directions.

Correct choice: a

  1. Peak is ?2 times r. m. s. and is = 200 ?2 = 282.8 V

Correct choice: b

  1. Power factor is the ratio of resistance to impedance = 4/8 = ½

Correct choice: a

  1. Instantaneous voltage = 100 Sin ?t = 100 Sin (2? 50 × 1/600) = 100 Sin ?/6 = 50 V

Correct choice: a

  1. Inductive reactance = L ? = 0.1 times 100 ? = 10 ?. So current = 314/31.4 = 10 A.

Correct choice: c

  1. At resonance the whole voltage appears across the resistance and hence current = 240/R = 240/12 = 20 A.

Correct choice: d

  1. Power in primary = 240 I = power in secondary 36. So, I = 36/240 = 0.15 A

Correct choice: d

  1. Transformer works with only a. c. and so voltage in secondary in this case = 0

Correct choice: e

  1. Here 0.9 times VI in primary = VI in secondary. So, voltage in secondary = 0.9 times 120 times 1.8/150 = 1.296 V

Correct choice: a

  1. Here current is a sine function and voltage, cosine function, which means that phase difference between them is ?/2. So, power = 0.

Correct choice: d

  1. Efficiency = power in secondary/power in primary = 11 times 90/220 times 5 = 0.9 = 90%

Correct choice: b

  1. Inductance is ideal means its resistance is zero. Since the inductance is in parallel with the resistance, it short-circuits the resistance and hence current through resistance = 0

Correct choice: c

  1. Since north pole is being introduced, induced current is such that a north pole is created on that face of the coil. And so, the current is in the anticlockwise direction.

Correct choice: b

  1. Induced e. m. f. = B L average v gained in the first second (=10 m/s) = 0.0004 times 2 times 10/2 = 0.004 V = 4 m V

Correct choice: a

  1. Vertical earth?s field = 0.0008 times sin30 = 0.0004. So. e. m. f. = 0.0004 times 20 times 100 m/s = 0.8 V

Correct choice: a

  1. Induced e. m. f. is independent of resistance.

Correct choice: c

  1. Self-inductance is proportional to square of number of turns and hence it becomes 4 times, in the present problem.

Correct choice: c

  1. Charge induced = change in flux/resistance = 10 m Wb/10 ohm = 1 m C

Correct choice: c

  1. Here voltage across inductance = ?(240² - 120²) = 120?3 V. Current in the circuit = P/V = 60/120 = 0.5 A. So, L ? = 120?3/0.5. So, L = 207.8/0.5 × 100? = 1.32 H

Correct choice: a

  1. Growth of current in an inductance is governed by the equation, I = I0 (1 ? e ? R t/L), where L/R is time constant (T). So, here, I0/2 = I0 ( 1 ? e ? 0.695/T ). So,  ½ = e ? 0.693/T. Or, e0.693/T = 2. Hence 0.696/T = ln 2 = 0.693. And hence T = 1 s.

Correct choice: b

  1. Energy stored = ½ L I² = ½ 0.05 times 4 = 0.1 J

Correct choice: c

  1. When current is maximum, frequency is resonance frequency = 1/(2??LC) = 1/2??[0.1 × 10 ? 3 × (1/?²) × 10 ? 6] = 10 5 / 2 = 50000 Hz = 50 k Hz

Correct choice: a

  1. Current at resonance = V/R = 240/100 = 2.4 A

Correct choice: a

  1. Since inductive and capacitive reactances are equal, the circuit is in resonance with the applied a. c. So the voltmeter connected across them together is zero.

Correct choice: a

  1. Power is consumed only by resistance and is equal to V²/R = 200²/50 = 800 W

Correct choice: b

  1. Phase difference is zero as there is resonance.

Correct choice: a

  1. When a circuit contains only L and C, the current will be out of phase with voltage by ?/2

Correct choice: b

  1. Current = e. m. f/reactance. Net reactance = L ? ? 1/C ? = 10 ? 3 × 100 ? ? 1/10 ? 3 × 100? = 0.314 ? 3.18 = - 2.87, negative sign need not be considered. So, I = 220/2.87 = 76.6 A

Correct choice: e

  1. If coupling is 100%, in this problem L = square root of M = 2 m H.

Correct choice: c

  1. Here, we have, L1 + L2 = 6 and L1 L2 / (L1 + L2) = 4/3, or L1 L2 = 8. So it is clear that the inductances are 4 and 2.

Correct choice: a

  1. Here current = V/L? = 10/0.005 times 100 = 10/0.5 = 20 A.

Correct choice: d

  1. It is, as you know, 50 Hz.

Correct choice: e

  1. Given 2? f = 377. So, f = 377/2 times 3.14 = 60 Hz.

Correct choice: c

  1. R. M. S = 2/?2 = ?2 A.

Correct choice: b

  1. Here it is the square root of the sum of the squares of 5 and 10/?2 and is equal to 5?3 A.

Correct choice: b

  1. Reactance = L ?. So, here 100 = L times 100 ?, or L = 1/? = 0.32 H

Correct choice: c

 
 
 
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Comments (1)
Aditya Arora
Blazing goIITian

Joined: 3 Dec 2006 21:32:30 IST
Posts: 1043
20 May 2007 22:24:42 IST
0 people liked this

Yaar, se conse test ke answers hain ???


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