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Question
An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate:
- the power of bulb and
- the potential difference at its end.
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Solution
- 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 - 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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