Answer:
(e)[tex]6.835\times 10^{-7}F[/tex]
Explanation:
At resonance we know that [tex]X_l=X_C[/tex]
That is [tex]\omega L=\frac{1}{\omega C}[/tex]
[tex]\omega ^2=\frac{1}{LC}[/tex]
[tex]\omega =\frac{1}{\sqrt{LC}}[/tex]
[tex]f=\frac{1}{2\pi \sqrt{LC}}[/tex]
We have given resonance frequency f =4511 Hz and inductance L=1.82 mH
So [tex]4511=\frac{1}{2\pi \sqrt{LC}}[/tex]
[tex]LC=\frac{1}{4\pi ^2\times 4511^2}[/tex]
[tex]LC=1.244\times 10^{-9}[/tex]
[tex]C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F[/tex]
So option e is the correct answer