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Electromagnetic Induction Test - 5
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Electromagnetic Induction Test - 5
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  • Question 1/10
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    A magnet is made to oscillate with a particular frequency, passing through a coil as shown in the figure.

    The time variation of magnitude of emf generated across the coil during one cycle is

    Solutions

    As the magnet faces north pole, a north pole is developed at that face, and current flows anticlockwise. Finally, when it completes the oscillation, no emf is present. Now, south pole approaches the other side, i.e. R.H.S. Now, the current flows clockwise to repel the south pole, and the current flows in anticlockwise at L.H.S. as before. This is possible in option (1). The breaks show when the pendulum is at the extreme and momentarily stationary.

  • Question 2/10
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    A 50 Hz AC source of 20 volts is connected across R and C as shown in figure. The voltage across R is 12 volt. Then, voltage across C is

    Solutions

  • Question 3/10
    1 / -0

    Area of cross-section of a solenoid is 1.2 x 10-3 m, its length is 0.3 m. It has 2000 turns over it. A coil having 300 turns is wound on its central part. If 2 ampere current changes its direction in 0.25 second, then what is induced e.m.f.?

    Solutions

  • Question 4/10
    1 / -0

    In an LR circuit of 3 mH inductance and 4Ω resistance, an emf E = 4 cos 1000 t volt is applied. The current amplitude is

    Solutions

  • Question 5/10
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    A coil having an inductance of 0.5 H carries a current which is uniformly varying from 0 to 10 A in 2 s. The emf (in volts) generated in the coil is

    Solutions

  • Question 6/10
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    Alternating current cannot be measured by a DC ammeter because

    Solutions

    A full cycle of alternating current consists of two half cycles. For one half, current is positive and for second half, current is negative. Therefore, for an AC cycle, the net value of average current is zero. However, for the half cycle, the value of current is different at different points. Hence, alternating current cannot be measured by a DC ammeter.

  • Question 7/10
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    An ideal transformer is supplied with a primary input power of 10 kilowatt. When the transformer is on load, the secondary current is 24 A. What is the potential difference applied to the primary if the primary to secondary turns ratio is 4 : 1?

    Solutions

    Ep × Ip = 10 kW = 10 × 103 W = 104 W
    Is = 24 A
    np/ns = 4/1
    np/ns = Is/Ip
    4/1 = 24/Ip
    Ip = 6 A
    Ep = 104/6 = 1666.7 V

  • Question 8/10
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    An inductive coil has a resistance of 100Ω. When an AC signal of frequency 1000 Hz is fed to the coil, the applied voltage leads the current by 45°. What is the inductance of the coil?

    Solutions

  • Question 9/10
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    A boy has a coil of 3 mH. He wants to make a circuit whose resonant frequency is 106 cycles per second. What is the capacitance of the capacitor?

    Solutions

  • Question 10/10
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    A flux of 0.2 Wb, linked with a coil having 50 turns, is reduced to zero in one second. What is the average emf induced in the coil?

    Solutions

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