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 \(........... W\)
Pick an option to check your answer.
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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