Answer
(a) $9.7pH$
(b) $10\Omega$
Work Step by Step
(a) We can find the required inductance as follows:
$f=\frac{1}{2\pi \sqrt{LC}}$
This simplifies to:
$L=\frac{1}{(2\pi f)^2C}$
We plug in the known values to obtain:
$L=\frac{1}{[2\pi(95\times 10^6)]^2(0.29\times 10^{-6})}$
$L=9.7\times 10^{-12}H=9.7pH$
(b) We can find the required resistance as
$R=\frac{|2\pi fL-(2\pi fC)^{-1}|}{\sqrt{24}}$
$R=\frac{|2\pi(11KHz)(9.7pH)-[2\pi(11KHz)(0.29\mu F)]^{-1}|}{\sqrt{24}}$
$R=10\Omega$
