A 100 Ω resistance, a 0.1 μF capacitor and... | Alternating Current

JEE MAIN 2021

27-07-2021 S2

Question

A $100 \Omega$ resistance, a $0.1 \mu \mathrm{~F}$ capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz

Select the correct option:

0.70 H

70.3 mH

$7.03 \times 10^{-5} \mathrm{H}$

70.3 H

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$ \mathrm{C}=0.1 \mu \mathrm{~F}=10^{-7} \mathrm{~F} $$ Resonant frequency $=60 \mathrm{~Hz}$ $$ \begin{aligned} & \omega_0=\frac{1}{\sqrt{L C}} \\ & 2 \pi f_0=\frac{1}{\sqrt{L C}} \Rightarrow L=\frac{1}{4 \pi^2 f_0^2 C} \end{aligned} $$ by putting values $\mathrm{L}=70.3 \mathrm{~Hz}$

Question Tags

JEE Main

Physics

Easy

Start Preparing for JEE with Competishun

Report Issue