Answer
$76\Omega; 99nF$
Work Step by Step
We know that
$\frac{V_{rms}}{I_{rms}}=\sqrt{R^2+\frac{1}{4\pi^2f^2C^2}}$
$\implies R^2+\frac{1}{4\pi ^2 f^2C^2}=(\frac{V_{rms}}{I_{rms}})^2$.....eq(1)
We plug in the known values in eq(1) to obtain:
$R^2+\frac{1}{4\pi^2(20000)^2C^2}=(\frac{5}{45\times 10^{-3}})^2$....eq(2)
We plug in the known values in eq(1) to obtain:
$R^2+\frac{1}{4\pi^2(25000)^2}=(\frac{5}{50\times 10^{-3}})^2$.....eq(3)
Subtracting eq(3) form eq(2), we obtain:
$(R^2-R^2)+\frac{1}{4\pi ^2C^2}(\frac{1}{(20000)^2}-\frac{1}{(25000)^2})=\frac{(5)^2}{10^{-6}}(\frac{1}{(45)^2}-\frac{1}{(50)^2})$
This simplifies to:
$C=99nF$
Now we plug in the known values in eq(2) to obtain:
$R^2=(\frac{5}{45\times 10^{-3}})^2-\frac{1}{4\pi^2(20000)^2}\times (98.5\times 10^{-9})^2$
This simplifies to:
$R=76\Omega$
