ExamBro
ExamBro
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.

  1. A \(25\)
  2. B \(50\)
  3. C \(66.6\)
  4. D \(100\)
Verified Solution

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:…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app