JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2019
- Hard
- MCQ
Two electric bulbs, rated at \((25\, W, 220\, V)\) and \((100\, W, 220\, V)\), are connected in series acroos a \(220\, V\) voltage source. If the \(25\, W\) and \(100\, W\) bulbs draw powers \(P_1\) and \(P_2\) respectively, then
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(P_1 = 16\, W, P_2 = 4\, W\)
Step-by-step Solution
Detailed explanation
Resistance, \(R=\frac{V^{2}}{P}\)
\(\Rightarrow \quad R_{1}=\frac{(220)^{2}}{25}=1936 \,\Omega\)
\(R_{2}=\frac{(220)^{2}}{100}=484\, \Omega\)
\(\mathrm{i}=\frac{220}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1}{11}\, \mathrm{A}\)
Power dissipated through \(\mathrm{R}_{1}=\mathrm{P}_{1}=\mathrm{i}^{2} \mathrm{R}_{1}=16\, \mathrm{W}\)
Power dissipated through \(\mathrm{R}_{2}=\mathrm{P}_{2}=\mathrm{i}^{2} \mathrm{R}_{2}=4 \,\mathrm{W}\)
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