For the circuit of figure $Ex32.33$,

a. What is the resonance frequency, in both $rad/s$ and $Hz$?

b. Find $V_r$and$V_c$at resonance.

c. How can$V_c$be larger than${\epsilon }_0$? Explain .

We have given,

We have to find resonance frequency in rad/s($\omega$) and Hz.

We know at resonance condition ,the frequency is;

$\omega =\frac{1}{\sqrt{LC}}rad/s$

$\omega =\frac{1}{\sqrt{{10}^{-3}\times {10}^{-6}}}rad/s$ $(Substitutingthevalues)$

$\omega =10\sqrt{10}rad/s$

We know ,

$\omega =2\pi f$

$f=\frac{\omega }{2\pi }$

$f=\frac{10\sqrt{10}}{2\pi }$ $(Substitutingthevalues)$

$f=49.672Hz$

We have given,

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

Firstly we need to find $X_c$ ,

$X_c=\frac{1}{\omega C}$

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

$X_c={10}^4\times \sqrt{10}ohm$

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

$I=\frac{{\epsilon }_0}{R}$

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

$I=1A$

now,

$V_{r=}IR$

$V_r=1\times 10V$ $(Substitutingthevalues)$

$V_r=10V$

Also,

$V_c=I{X}_c$

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

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

We have given$V_C$and${\epsilon }_0$.

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

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

Also'

$V_c=\frac{X_c}{Z}\times {\epsilon }_0$ ( $Z=totalimpedence$)