= Solution
In series the same current passes through both <resistors>, while their <voltage> drops add: $V=I(R_1+R_2)$. In parallel the same <voltage> drives currents $V/R_1$ and $V/R_2$, which add. Thus $R_s=R_1+R_2$ and $R_p^{-1}=R_1^{-1}+R_2^{-1}$.
Use <phasors> proportional to $e^{i\omega t}$. Then $Z_L=i\omega L$ and $Z_C=1/(i\omega C)$. The printed diagram places C in parallel with the series RL branch, so the input admittance is
$$
Y=\frac1Z=i\omega C+\frac1{R+i\omega L},\qquad
\operatorname{Re}Y=\frac{R}{R^2+\omega^2L^2},\quad
\operatorname{Im}Y=\omega\left(C-\frac{L}{R^2+\omega^2L^2}\right).
$$
Its nonzero zero-phase <frequency> exists for $R^2<L/C$:
$$
\boxed{\omega_1=\sqrt{\frac1{LC}-\frac{R^2}{L^2}}.}
$$
At zero <frequency>, $|Y|=1/R$ and $I_0=V_0/R$. At $\omega_1$, $R^2+\omega_1^2L^2=L/C$, giving
$$
\boxed{|Y(\omega_1)|=\frac{RC}{L},\qquad I_0(\omega_1)=\frac{V_0RC}{L}.}
$$
For $0<\omega<\omega_1$, $\operatorname{Im}Y<0$, so current lags <voltage>. At very high <frequency>, $Y\sim i\omega C$, so current leads by $\pi/2$. Zero <phase> need not coincide exactly with the minimum current magnitude in this lossy network.
The three measured limits determine the components in order:
$$
R=\frac{V_0}{I_{\rm DC}}=100\,\Omega,\qquad
C=\frac{\alpha}{V_0}=2.50\times10^{-7}\,\mathrm F,\qquad
L=\frac{V_0RC}{I_0(\omega_1)}=0.100\,\mathrm H,
$$
$$
\boxed{\omega_1=6.245\times10^3\,\mathrm{rad\,s^{-1}}.}
$$
For the changed resistance, the quoted $0.6\,\mathrm{kHz}$ is an ordinary <frequency>, so $\omega'=2\pi(600\,\mathrm{s^{-1}})$. Keeping L and C fixed gives
$$
\boxed{R'=\sqrt{\frac LC-\omega'^2L^2}=5.08\times10^2\,\Omega.}
$$
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