Answer
(a) $12.2A$
(b) $0.5$
(c) $6.09A$
Work Step by Step
(a) We know that
$X_L=2\pi fL$
We plug in the known values to obtain:
$X_L=2\pi(1.0\times 10^3Hz)(0.290\times 10^{-3}Hz)$
$X_L=1.82\Omega$
Now $I_{rms}=\frac{V_{rms}}{X_{L}}$
We plug in the known values to obtain:
$I_{rms}=\frac{22.2V}{1.82\Omega}$
$I_{rms}=12.2A$
(b) The new rms is given as
$I^{\prime}_{rms}=I_{rms}\frac{f}{f^{\prime}}$
We plug in the known values to obtain:
$I^{\prime}_{rms}=I_{rms}(\frac{1.0KHz}{2.0KHz})$
$\implies I^{\prime}_{rms}=0.5I_{rms}$
Thus, the rms current is reduced by a factor of $0.5$
(c) As $I^{\prime}_{rms}=\frac{I_{rms}}{2}$
We plug in the known values to obtain:
$I^{\prime}_{rms}=\frac{12.19A}{2}$
$I^{\prime}_{rms}=6.09A$
