= Solution
At <direct current> an ideal <capacitor> is an <open circuit>. A finite measured <electrical resistance> therefore requires the <resistor> and <capacitor> to form a <parallel circuit>, with $R=100\,\Omega$. Use <phasors> proportional to $e^{i\omega t}$. The <electrical admittance> and <complex impedance> are
$$
Y=\frac1R+i\omega C,\qquad Z=\frac{R}{1+i\omega RC},\qquad |Z|=\frac{R}{\sqrt{1+(\omega RC)^2}}.
$$
Interpreting the stated AC value as the measured <impedance magnitude>, $\omega=2\pi(50)\,\mathrm{s^{-1}}$ and $|Z|=20\,\Omega$ give $\omega RC=\sqrt{24}$. Thus
$$
\boxed{R=100\,\Omega,\quad C=1.56\times10^{-4}\,\mathrm F,\quad Z(50\,\mathrm{Hz})=(4.00-19.60i)\,\Omega.}
$$
The <electric current> leads the applied <voltage> by $\tan^{-1}\sqrt{24}=78.46^\circ$. Strictly, AC “resistance” is ambiguous: if it means $\operatorname{Re}Z=20\,\Omega$, the same <parallel circuit> instead gives $\omega RC=2$, $C=63.7\,\mu\mathrm F$, and $Z=20-40i\,\Omega$. A measurement of $|V|/|I|$ selects the boxed interpretation.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2021-circuit-arrangements.png]
{title=Parallel RC network and resonant voltage divider}
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