Start free

Physics · Alternating Current

NEET (UG) 2023 — Question 38

The net impedance of circuit (as shown in figure) will be :

Question figure
  1. Option A:

    15Ω15 \Omega

  2. Option B:

    55Ω5 \sqrt{5} \Omega

    Correct
  3. Option C:

    25Ω25 \Omega

  4. Option D:

    102Ω10 \sqrt{2} \Omega

Answer: B

Step-by-step solution

L=50πmHXL=2π×50×50π×10−3=5ΩC=103π×10−6XC=1×π2π×50×103×10−6=103100=10ΩZ=(XC−XL)2+R2Z=(10−5)2+102=125=55Ω\begin{aligned} & L=\frac{50}{\pi} \mathrm{mH} \\ & X_L=2 \pi \times 50 \times \frac{50}{\pi} \times 10^{-3}=5 \Omega \\ & C=\frac{10^3}{\pi} \times 10^{-6} \\ & X_C=\frac{1 \times \pi}{2 \pi \times 50 \times 10^3 \times 10^{-6}}=\frac{10^3}{100}=10 \Omega \\ & Z=\sqrt{\left(X_C-X_L\right)^2+R^2} \\ & Z=\sqrt{(10-5)^2+10^2}=\sqrt{125}=5 \sqrt{5} \Omega\end{aligned}

Answer key and solution verified before publishing.

Practise Alternating Current

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
NEET (UG) 2023
Paper
2023 paper
Subject
Physics
Chapter
Alternating Current
Topic
Series L-R, R-C, L-C Circuits with AC Source

More Alternating Current questions from this paper