Answer
a) ${\bf 2.5}\;\rm \Omega$
b) ${\bf 8.1}\;\rm kHz$
Work Step by Step
From the given graph, we can see that
- $\varepsilon_0=10$ V,
- $I=2$ A,
- $T=100\;\mu$s, and
- $\phi =60^\circ$ since the current lags the emf and at $t=0$, $i=\frac{1}{2}I$
$$\color{blue}{\bf [a]}$$
We know that
$$I=\dfrac{\varepsilon_0}{Z}$$
$$I=\dfrac{\varepsilon_0}{\sqrt{(X_L-X_C)^2+R^2}}\tag 1$$
Recalling that
$$\tan\phi=\dfrac{X_L-X_C}{R}$$
So,
$$X_L-X_C=R\tan\phi\tag 2$$
Plug (2) into (1),
$$I=\dfrac{\varepsilon_0}{\sqrt{R^2(\tan\phi)^2+R^2}} $$
$$I=\dfrac{\varepsilon_0}{R\sqrt{ (\tan\phi)^2+1}} $$
Hence,
$$R=\dfrac{\varepsilon_0}{I\sqrt{ (\tan\phi)^2+1}} $$
Plug the known;
$$R=\dfrac{(10)}{(2)\sqrt{ (\tan60^\circ)^2+1}} =\color{red}{\bf 2.5}\;\rm \Omega$$
$$\color{blue}{\bf [b]}$$
We know that the resonance frequency occurs when
$$X_L=X_C$$
$$2\pi f_r L=\dfrac{1}{2\pi f_r C}$$
So,
$$f_r= \dfrac{1}{\sqrt{4\pi^2 CL}} \tag 3$$
Now we need to find $C$, where from (2),
$$X_C=X_L-R\tan\phi =2\pi f L-R\tan\phi$$
$$\dfrac{1}{2\pi f C} =2\pi f L-R\tan\phi$$
So,
$$C=\dfrac{1}{2\pi f (2\pi f L-R\tan\phi)}$$
where $f=1/T$
$$C=\dfrac{1}{\frac{2\pi}{T} \left(\frac{2\pi L}{T} -R\tan\phi\right)}$$
Plug the known;
$$C=\dfrac{1}{\frac{2\pi}{(100\times 10^{-6})} \left(\frac{2\pi (200\times 10^{-6})}{(100\times 10^{-6})} -(2.5)\tan60^\circ\right)}$$
$$C=\bf 1.93\;\rm \mu F$$
Plug the known into (3),
$$f_r= \dfrac{1}{\sqrt{4\pi^2 (1.93\times 10^{-6})(200\times 10^{-6})}} $$
$$f_r=\color{red}{\bf 8100}\;\rm Hz$$
