JEE Mains · Physics · STD 12 - 3. current electricity
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
- A \(P_1 = 16\, W, P_2 = 4\, W\)
- B \(P_1 = 16\, W, P_2 = 9\, W\)
- C \(P_1 = 9\, W, P_2 = 16\, W\)
- D \(P_1 = 4\, W, P_2 = 16\, W\)
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…
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