Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
In $R-L-C$ series circuit, the potential differences across each element is $20 \mathrm{~V}$. Now the value of the resistance alone is doubled, then P.D. across $R, L$ and $C$ respectively.
PhysicsAlternating CurrentKCETKCET 2013
Options:
  • A $20 \mathrm{~V}, 10 \mathrm{~V}, 10 \mathrm{~V}$
  • B $20 \mathrm{~V}, 20 \mathrm{~V}, 20 \mathrm{~V}$
  • C $20 \mathrm{~V}, 40 \mathrm{~V}, 40 \mathrm{~V}$
  • D $10 \mathrm{~V}, 20 \mathrm{~V}, 20 \mathrm{~V}$
Solution:
2624 Upvotes Verified Answer
The correct answer is: $20 \mathrm{~V}, 10 \mathrm{~V}, 10 \mathrm{~V}$
Given, the potential difference across each element is same i.e., $20 \mathrm{~V}$, so, the circuit is at resonance and we have
$V=V_{R}=20 \mathrm{~V}$
If the value of $R$ is doubled, then value of / reduced to half $\therefore$
$V_{L}=V_{C}=\frac{I X_{L}}{2}=\frac{I X_{C}}{2}=\frac{20}{2}=10 \mathrm{~V}$
$(\because$ circuit is at resonance) and $\quad V_{R}=V=20 \mathrm{~V}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.