The self-inductance of a coil having 400 turns is 10 mH. The

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 Multiple Choice QuestionsMultiple Choice Questions

81.

When power is drawn from the secondary coil of the transformer, the dynamic resistance

  • increases

  • decreases

  • remains unchanged

  • changes erratically


82.

The magnetic flux linked with a coil at any instant t is given by the equation ϕ = 5t3 - 100 t + 300. The magitude of emf induced in coil after 3 s is 

  • 10 V

  • 20 V

  • 35 V

  • 70 V


83.

In 0.1 s, the current in a coil increases from 1 A to 1.5A. If inductance of coil is 60 mH, then induced current in external resistance of 32 will be

  • 1 A

  • 0.5 A

  • 0.2 A

  • 0.1 A


84.

In 0.2 s, the current in a coil increases from 2.0 A to 3.0 A. If inductance of coil is 60 mH, then induced current in external resistance of 3 Ω will be

  • 1A

  • 0.5 A

  • 0.2 A

  • 0.1 A


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85.

A transformer having efficiency of 75 % is working on 220 V and 4.4 kW power supply. If the current in the secondary coil is 5 A. What will be the voltage across secondary coil and the current in primary coil?

  • Vs = 220 V, ip = 20A

  • Vs = 660 V, ip = 15 A

  • Vs = 660 V, ip = 20 A

  • Vs = 220 V, ip = 15 A


86.

A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by

  • 2πR2 E

  • π R2/E

  • R / E

  • Zero


87.

In the given figure what will be the coefficient of mutual inductance

     


88.

The total charge induced in a conducting loop when it is moved in magnetic field depends on

  • the rate of change of magnetic flux

  • initial magnetic flux only

  • the total change in magnetic flux

  • final magnetic flux only


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89.

For MRI, a patient is slowly pushed in a time of 10 s within the coils of the magnet where magnetic field is B = 2.0 T. If the patient's trunk is 0.8 m in circumference, the induced emf around the patient's trunk is

  • 10.18 x 10-2 V

  • 9.66 × 102 V

  • 10.18 × 10-3 V

  • 1.51 × 10-2 V


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90.

The self-inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross-section of the coil corresponding to current 2 mA is

  • 2 × 10-5 Wb

  • 2 × 10-3 Wb

  • 3 × 10-5 Wb

  • 8 × 10-3 Wb


A.

2 × 10-5 Wb

Number of turns N = 400

L = 10 mH = 10 × 10-3 H

I = 2 mA = 2 × 10-3 A

Total magnetic flux linked with the coil,

 ϕ = N L I

     = 400 × ( 10 × 10-3 ) × 2 × 10-3

ϕ = 8 × 10-3 Wb

Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn

                = ϕN

                = 8 × 10-3400

                  = 2 ×  10-5 Wb


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