An ideal inductor of inductance L = (2)/(π) H is connected to an AC...

The inductive reactance $X_L$ of the circuit is calculated as:

$X_L = \omega L = 2\pi f L$

Substituting the given values $f = 50 \text{ Hz}$ and $L = \dfrac{2}{\pi} \text{ H}$:

$X_L = 2\pi \times 50 \times \dfrac{2}{\pi} = 200 \text{ } \Omega$

The RMS current $I_{\text{rms}}$ in the circuit is:

$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{X_L}$

Substituting $V_{\text{rms}} = 220 \text{ V}$:

$I_{\text{rms}} = \dfrac{220}{200} = 1.1 \text{ A}$

The peak current $I_0$ is related to the RMS current by the relation:

$I_0 = \sqrt{2} I_{\text{rms}}$

Using the given value $\sqrt{2} = 1.41$:

$I_0 = 1.41 \times 1.1 = 1.551 \text{ A}$

The peak current is approximately $1.55 \text{ A}$.

Answer: $1.55 \text{ A}$