A coil having an inductance of 0.16 H and a resistance of 12\ is...

The given alternating voltage is $V = 200\sqrt{2} \sin(100 t)$.

Comparing this with the standard equation $V = V_0 \sin(\omega t)$, we get:
Peak voltage, $V_0 = 200\sqrt{2}\text{ V}$
Angular frequency, $\omega = 100\text{ rad/s}$

The inductive reactance $X_L$ is given by:
$X_L = \omega L = 100 \times 0.16 = 16\ \Omega$

The impedance $Z$ of the coil is:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\ \Omega$

The peak current $I_0$ in the circuit is:
$I_0 = \dfrac{V_0}{Z} = \dfrac{200\sqrt{2}}{20} = 10\sqrt{2}\text{ A}$

The RMS current $I_{rms}$ is:
$I_{rms} = \dfrac{I_0}{\sqrt{2}} = \dfrac{10\sqrt{2}}{\sqrt{2}} = 10\text{ A}$

The average power dissipated by the coil is:
$P_{avg} = I_{rms}^2 R = (10)^2 \times 12 = 100 \times 12 = 1200\text{ W}$

Answer: $10\sqrt{2}\text{ A}$ and $1200\text{ W}$