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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2022
  • Medium
  • Numerical

An electrical bulb rated \(220\,V , 100\,W\), is connected in series with another bulb rated \(220\,V\), \(60\,W\).If the voltage across combination is \(220\,V\), the power consumed by the \(100\,W\) bulb will be about \(........... W\)

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Answer & Solution

Correct Answer

(A) \(14\)

Step-by-step Solution

Detailed explanation

\(R _{1}=\frac{ V ^{2}}{ P }=\frac{220^{2}}{100}=484\)

\(R _{2}=\frac{ V ^{2}}{ P }=\frac{220^{2}}{60}=484\left(\frac{10}{6}\right)\)

\(I =\frac{220}{484+484 \times \frac{10}{6}}\)

\(P _{1}= I ^{2} R _{1}=14.06\,W\)

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