Answer
a) ${\bf 0.10}\;\rm A$
b) ${\bf 0.20}\;\rm A$
Work Step by Step
$$\color{blue}{\bf [a]}$$
When the frequency is very small, $f\rightarrow 0$:
- $X_C=\dfrac{1}{2\pi f C}=\dfrac{1}{0}\rightarrow \infty$
- $X_L= {2\pi f L}=2\pi (0)L\rightarrow 0$
This means that the current on the branch of the capacitor faces a huge resistance to flow through while it is too easy for it to move through the branch of the inductor.
Now we have an $RL$ circuit (the left branch in the given figure) where the current is given by
$$I_{\rm rms}=\dfrac{\varepsilon_{\rm rms}}{Z} $$
where, in the left branch, $R=100\;\Omega$, and since $f\rightarrow 0$, then $Z\approx R$
$$I_{\rm rms} =\dfrac{\varepsilon_{\rm rms}}{R}$$
Plug the known
$$I_{\rm rms}=\dfrac{10 }{100}=\color{red}{\bf 0.10}\;\rm A$$
$$\color{blue}{\bf [b]}$$
When the frequency is very large, $f\rightarrow 0$:
- $X_C=\dfrac{1}{2\pi f C}=\dfrac{1}{\infty}\rightarrow 0$
- $X_L= {2\pi f L}=2\pi (\infty)L\rightarrow \infty$
This means that the current on the branch of the inductor faces a huge resistance to flow through while it is too easy for it to move through the branch of the capacitor.
Now we have an $RC$ circuit (the right branch in the given figure) where the current is given by
$$I_{\rm rms}=\dfrac{\varepsilon_{\rm rms}}{Z}$$
where, in the left branch, $R=50\;\Omega$, and since $f\rightarrow \infty$, then $Z\approx R$
$$I_{\rm rms}=\dfrac{10 }{50}=\color{red}{\bf 0.20}\;\rm A$$
