An electric bulb rated as 100 W-220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is:

The rating 100 W - 220 V means the bulb consumes power $$P = 100\ \text{W}$$ when an rms voltage $$V_{\text{rms}} = 220\ \text{V}$$ is applied.

Step 1: Find the resistance of the filament.
For an ac (or dc) load, $$P = \frac{V_{\text{rms}}^{\,2}}{R}$$ $$-(1)$$
Rearranging $$-(1)$$ gives $$R = \frac{V_{\text{rms}}^{\,2}}{P}$$.

Substituting the given values:
$$R = \frac{(220)^{2}}{100} = \frac{48400}{100} = 484\ \Omega$$.

Step 2: Calculate the rms current through the bulb.
Ohm’s law for rms values is $$I_{\text{rms}} = \frac{V_{\text{rms}}}{R}$$ $$-(2)$$.

Using $$R = 484\ \Omega$$:
$$I_{\text{rms}} = \frac{220}{484}\ \text{A} \approx 0.455\ \text{A}$$.

Step 3: Convert rms current to peak (maximum) current.
For a sinusoidal current, $$I_{0} = \sqrt{2}\,I_{\text{rms}}$$ $$-(3)$$.

Applying $$-(3)$$:
$$I_{0} = \sqrt{2}\times 0.455\ \text{A} \approx 1.414 \times 0.455\ \text{A} \approx 0.64\ \text{A}$$.

Hence, the peak value of current through the bulb is $$0.64\ \text{A}$$.

Option A is correct.