KCET · Physics · Current Electricity
An electric bulb of \(60 \mathrm{~W}, 120 \mathrm{~V}\) is to be connected to 220 V source. What resistance should be connected in series with the bulb, so that the bulb glows properly?
- A \(50 \Omega\)
- B \(100 \Omega\)
- C \(200 \Omega\)
- D \(288 \Omega\)
Answer & Solution
Correct Answer
(C) \(200 \Omega\)
Step-by-step Solution
Detailed explanation
Resistance of the bulb,
\(R=\frac{V^2}{P}=\frac{120 \times 120}{60}=240 \Omega\)
\(\therefore \quad\) il \(I=\frac{P}{V}=\frac{60}{120}=0.5 \mathrm{~A}\)
Since, bulb is connected with 220 V supply.
\(\therefore\) Resistance required,
\(R^{\prime}=\frac{V}{I}=\frac{220}{0.5}=440 \Omega\)
Thus, for the bulb to glow properly required resistance in series
\(=R^{\prime}-R\)
\(=440-240=200 \Omega\)
\(R=\frac{V^2}{P}=\frac{120 \times 120}{60}=240 \Omega\)
\(\therefore \quad\) il \(I=\frac{P}{V}=\frac{60}{120}=0.5 \mathrm{~A}\)
Since, bulb is connected with 220 V supply.
\(\therefore\) Resistance required,
\(R^{\prime}=\frac{V}{I}=\frac{220}{0.5}=440 \Omega\)
Thus, for the bulb to glow properly required resistance in series
\(=R^{\prime}-R\)
\(=440-240=200 \Omega\)
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