KCET · Physics · Current Electricity
If voltage across a bulb rated \(220 \mathrm{~V}, 100 \mathrm{~W}\) drops by \(2.5 \%\) of its rated value, then the percentage of the rated value by which the power would decrease is
- A \(2.5 \%\)
- B \(5 \%\)
- C \(10 \%\)
- D \(20 \%\)
Answer & Solution
Correct Answer
(B) \(5 \%\)
Step-by-step Solution
Detailed explanation
Given, \(P=100 \mathrm{~W}\)
\(V=220 \mathrm{~V}\)
We know that, \(P=\frac{V^2}{R}\)
where, \(R\) is resistance of bulb.
\(\Rightarrow R=\frac{V^2}{P}=\frac{220 \times 220}{100}=484 \Omega\)
Now, according to question, voltage drops by \(2.5 \%\) of its rated value.
New voltage, \(V^{\prime}=220-2.5 \%\) of 220
\(=220-\frac{2.5}{100} \times 220=214.5 \mathrm{~V}\)
\(\therefore \text {New power, } P^{\prime}=\frac{\left(V^{\prime}\right)^2}{R}=\frac{214.5 \times 214.5}{484}=95.06 \mathrm{~W} \)
\(\% \text { decrease in power } =\frac{P-P^{\prime}}{P} \times 100 \)
\(=\frac{100-95.06}{100} \times 100 \sim 5 \%\)
\(V=220 \mathrm{~V}\)
We know that, \(P=\frac{V^2}{R}\)
where, \(R\) is resistance of bulb.
\(\Rightarrow R=\frac{V^2}{P}=\frac{220 \times 220}{100}=484 \Omega\)
Now, according to question, voltage drops by \(2.5 \%\) of its rated value.
New voltage, \(V^{\prime}=220-2.5 \%\) of 220
\(=220-\frac{2.5}{100} \times 220=214.5 \mathrm{~V}\)
\(\therefore \text {New power, } P^{\prime}=\frac{\left(V^{\prime}\right)^2}{R}=\frac{214.5 \times 214.5}{484}=95.06 \mathrm{~W} \)
\(\% \text { decrease in power } =\frac{P-P^{\prime}}{P} \times 100 \)
\(=\frac{100-95.06}{100} \times 100 \sim 5 \%\)
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