Alternating Current

  • Question 1
    CBSEENPH12039433

    The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage.

    1. Name the device used to change the alternating voltage to a higher or lower value. State one cause for
      power dissipation in this device.
    2. Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.
    3. Write two values each shown by the teachers and Geeta.

    Solution
    1. Step up or step down transformer, Eddy current losses.
    2. With higher voltage, power losses are less, so voltage can be increased by the step-up transformer and the transformer works on A/c only.
    3. Both are interested towards technical knowledge and both are having sufficient ideas about power transmission.
    Question 2
    CBSEENPH12039438

    A device X is connected across an ac source of voltage V=V0sin ωt. The current through X is given as I - I0sin ωt + π2

    1. Identify the device X and write the expression for its reactance.
    2. Draw graphs showing a variation of voltage and current with time over one cycle ac, for X.
    3. How does the reactance of the device X vary with the frequency of the ac? Show this variation graphically.
    4. Draw the phasor diagram for the device X.

    Solution
    1. Device: X-capacitor
      Xc = 12πfC
    2. Xc  1f
    3. phasor diagram
    Question 3
    CBSEENPH12039614

    In an a.c. circuit the voltage applied is E = E0 sinπt. The resulting current in the circuit is I = I0 sin open parentheses ωt space minus space straight pi over 2 close parentheses.The power consumption in the circuit is given by

    • straight P space equals space fraction numerator straight E subscript 0 straight I subscript 0 over denominator square root of 2 end fraction
    • P = zero

    • straight P space equals fraction numerator straight E subscript 0 straight I subscript 0 over denominator 2 end fraction
    • straight P space equals space square root of 2 straight E subscript 0 straight I subscript 0

    Solution

    B.

    P = zero

    Question 4
    CBSEENPH12039619

    A current I flows along the length of an infinitely long, straight, thin walled pipe. Then 

    • the magnetic field is zero only on the axis of the pipe

    • the magnetic field is different at different points inside the pipe

    • the magnetic field at any point inside the pipe is zero

    • the magnetic field at all points inside the pipe is the same, but not zero

    Solution

    C.

    the magnetic field at any point inside the pipe is zero

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