JEE Mains · Physics · STD 12 - 3. current electricity
Two resistors of \(200 \, \Omega\) and \(400 \, \Omega\) are connected in series with a battery of \(100\) V. A bulb rated at \(200\) V, \(100\) W is connected across the \(400 \, \Omega\) resistance. The potential drop across the bulb is _______ V.
- A \(25\)
- B \(50\)
- C \(66.6\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(50\)
Step-by-step Solution
Detailed explanation
The resistance of the bulb is calculated using its power rating: \(R_b = \dfrac{V^2}{P} = \dfrac{200^2}{100} = 400 \, \Omega\) The bulb is connected in parallel with the \(400 \, \Omega\) resistor. The equivalent resistance of this parallel combination is:…
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