Answer
a) $C=1.25\pi F$
b) impedance increase
c) $Z=5\Omega$.
d) $Z=14.8\Omega$
Work Step by Step
(a) We know that
$C=\frac{1}{4\pi^2L f^2}$
We plug in the known values to obtain:
$C=\frac{1}{4\pi^2(2.8\times 10^{-6})(85\times 10^6)^2}$
$C=1.25\times 10^{-12}=1.25\pi F$
(b) We know that when the capacitance of the capacitor increases then $\omega \neq \frac{1}{\omega C}$ and therefore, $(\omega L-\frac{1}{\omega C})^2$ is not zero and positive, so Z increases. We conclude that the impedance of the circuit increases when the capacitance is increased above $1.25\pi F$.
(c) We know that the impedance of the circuit at resonance is equal to the resistance of the resistor in the circuit; therefore, the impedance at resonance is $Z=5\Omega$.
(d) We know that
$X_C^{\prime}=\frac{1}{\omega C^{\prime}}$
$X_C^{\prime}=\frac{1}{2\pi\times 85\times 10^{-6}\times 1.2625\times 10^{-12}}$
$X_C^{\prime}=1482\Omega$
The inductive reactance $X_L=\omega L$
$X_L=2\pi\times 85\times 10^6\times 2.8\times 10^{-6}=1496\Omega$
Now $Z=\sqrt{(5)^2+(1496-1482)^2}$
$Z=14.8\Omega$
