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GUJCET · Physics · Current Electricity
A bulb of 100 W rating is connected with 220 V supply. The resistance of bulb is _________.
- A \(2.2 \times 10^{-3} \Omega m^{-1}\)
- B \(484 \Omega m^{-1}\)
- C \(2.2 \Omega\)
- D \(484 \Omega\)
Answer & Solution
Correct Answer
(D) \(484 \Omega\)
Step-by-step Solution
Detailed explanation
(D)
power \(P =\frac{ V ^2}{ R }\)
\(\begin{array}{ll}\therefore R =\frac{ V ^2}{ P } \\ \therefore R =\frac{(220)^2}{100} \\ \therefore R =484 \Omega\end{array}\)
power \(P =\frac{ V ^2}{ R }\)
\(\begin{array}{ll}\therefore R =\frac{ V ^2}{ P } \\ \therefore R =\frac{(220)^2}{100} \\ \therefore R =484 \Omega\end{array}\)
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