Answer
(a) $3.44mA$
(b) $10.5nF$
Work Step by Step
(a) We can find the rms current as follows:
The impedance of the circuit is
$Z=\sqrt{R^2+X_L^2}$
$\implies Z=\sqrt{(1150)^2+(3967)^2}$
$\implies Z=4130\Omega$
Now $I_{rms}=\frac{V_{rms}}{Z}$
$\implies I_{rms}=\frac{14.2}{4.30}$
$I_{rms}=0.003438A$
$\implies I_{rms}=3.44mA$
(b) We can determine the required capacitance as follows:
$\frac{1}{\omega C}=\omega L\pm (\frac{V_{rms}^2}{I_{rms}^2})^{\frac{1}{2}}$
We plug in the known values to obtain:
$\frac{1}{\omega C}=3967\pi [(\frac{14.2}{1.719\times 10^{-3}})^2-(1150)^2]^{\frac{1}{2}}$
This simplifies to:
$\frac{1}{\omega C}=12147$ or $\frac{1}{\omega C}=-4213$
$\implies C=1.047\times 10^{-9}F=1.047nF$ or $C=-30.2\times 10^{-9}F=-30.2nF$
As the capacitance is not negative, therefore $C=-30.2nF$ is wrong and we have $C=10.5nF$
