Answer
(a) The box contains a capacitor.
(b) $0.910\mu F$
Work Step by Step
(a) We know that when the frequency increases $X_L$ increases and $X_C$ decreases and therefore the impedance of the $RL$ circuit increases and impedance of the $RC$circuit decreases. Thus, we conclude that the box contains a capacitor.
(b) We know that
$C=\frac{2}{2\pi f[(\frac{V_{rms}}{I_{rms}})^2-R^2]^{\frac{1}{2}}}$
We plug in the known values to obtain:
$C=\frac{1}{24\times 25\times 10^{3}[(\frac{0.750}{87.2\times 10^{-3}})^2-(5)^2]^{\frac{1}{2}}}$
$\implies C=0.910\mu F$
