मराठी

An Electric Bulb of Resistance 500 ω Draws Current 0.4 a from the Source. Calculate: (A) the Power of Bulb and (B) the Potential Difference at Its End. - Physics

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प्रश्न

An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate:

  1. the power of bulb and
  2. the potential difference at its end.
संख्यात्मक
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उत्तर

  1. Resistance of electric bulb (R) = 500Ω
    current drawn from the source (I) = 0.4 A
    Power of the bulb (P) = VI
    V = I × R
    V = 0.4 × 500
    V = 200 V
  2. The potential difference at its end is 200 V.
    Hence,
    Power (P) = VI
    P = 200 × 0.4
    P = 80 W
    The power of the bulb is 80 Watt.
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पाठ 8: Current Electricity - EXERCISE - 8(C) [पृष्ठ २१२]

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सेलिना Physics [English] Class 10 ICSE
पाठ 8 Current Electricity
EXERCISE - 8(C) | Q 1. | पृष्ठ २१२

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