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Physics · Capacitors and R-C Circuits

JEE Main 2024 — 9 April, Shift 1 — Question 48

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:

  1. Option A:

    increases

    Correct
  2. Option B:

    remains same

  3. Option C:

    becomes zero

  4. Option D:

    decreases

Answer: A

Step-by-step solution

Z=R2+XC2&XC=1WC\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{C}}^{2}} \& \mathrm{X}_{\mathrm{C}}=\frac{1}{\mathrm{WC}}

due to dielectric

C↑⇒XC↓⇒Z↓\mathrm{C} \uparrow \Rightarrow \mathrm{X}_{\mathrm{C}} \downarrow \Rightarrow \mathrm{Z} \downarrow

So, current increases & thus bulb will glow more brighter.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Physics
Chapter
Capacitors and R-C Circuits
Topic
Effect of Dielectrics

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