91Ó°ÊÓ

We have given,

We have to find resonance frequency in rad/s($\mathrm{Ó\neg }$) and Hz.

We know at resonance condition ,the frequency is;

$\mathrm{Ó\neg }=\frac{1}{\sqrt{LC}}rad/s$

$\mathrm{Ó\neg }=\frac{1}{\sqrt{{10}^{-3}×{10}^{-6}}}rad/s$ $(Substitutingthevalues)$

$\mathrm{Ó\neg }=10\sqrt{10}rad/s$

We know ,

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$

$f=\frac{\mathrm{Ó\neg }}{2\mathrm{Ï€}}$

$f=\frac{10\sqrt{10}}{2\mathrm{Ï€}}$ $(Substitutingthevalues)$

$f=49.672Hz$

We have given,

We have to find $V_r$and $V_c$.

Firstly we need to find $X_c$ ,

$X_c=\frac{1}{\mathrm{Ó\neg }C}$

$X_c=\frac{1}{10\sqrt{10}×{10}^{-6}}ohm$ $(Substituingthevalues)$

$X_c={10}^4×\sqrt{10}ohm$

At resonance, I is maximum and equal to $\frac{{\mathrm{Î\mu }}_0}{R}$,

$I=\frac{{\mathrm{Î\mu }}_0}{R}$

$I=\frac{10V}{10ohm}$ $(Substitutingthevalues)$

$I=1A$

now,

$V_{r=}IR$

$V_r=1×10V$ $(Substitutingthevalues)$

$V_r=10V$

Also,

$V_c=I{X}_c$

$V_c=1×{10}^4\sqrt{10}V$ $(Substitutingthevalues)$

$V_c={10}^4\sqrt{10}V$

We have given$V_C$and${\mathrm{Î\mu }}_0$.

We have to explain how can$V_c$be larger than${\mathrm{Î\mu }}_0$.

If the circuit is in resonance condition $\mathrm{Ó\neg }=\frac{1}{\sqrt{LC}}$, the peak to peak voltages on capacitor and the inductor can be larger than the peak to peak voltage ${\mathrm{Î\mu }}_0$.

Also'

$V_c=\frac{X_c}{Z}×{\mathrm{Î\mu }}_0$ ( $Z=totalimpedence$)