2. A bulb of power 100W is connected in a circuit of potential 300V.What will be the resistance of the bulb
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Answer:900Ω
Explanation:
here given, p=100W and p.d(V)=300V we have to find R=?
R=V²/P=(300×300)/100=900Ω
HENCE,the resistance of the bulb is 900Ω
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