Answer
$607\mu H, 80.7\Omega$
Work Step by Step
We know that
$R^2+(2\pi fL)^2=(\frac{V_{rms}}{I_{rms}})^2$......eq(1)
We plug in the known values to obtain:
$R^2+[2\pi L(20KHz)^2]^2=(\frac{5V}{45mA})^2$.....eq(2)
We plug in the known values in eq(1) to obtain:
$R^2+[2\pi L(25KHZ)]^2=(frac{5V}{40mA})^2$.....eq(3)
Subtracting eq(2) from eq(3), we obtain:
$[2\pi L(25KHz)^2-2\pi L(20KHz)]^2=(\frac{5V}{40mA})^2-(\frac{5V}{45mA})^2$
This simplifies to:
$L=6.07\times 10^{-4}H$
$\implies L=607\mu H$
We plug in this value in eq(2) to obtain:
$R^2=(\frac{5V}{45\times 10^{-3}A})^2-(2\pi (20KHz))^2(607\mu H)^2$
This simplifies to:
$R=80.7\Omega$
