A direct current of $$4 \text{ A}$$ and an alternating current of peak value $$4 \text{ A}$$ flow through resistance of $$3 \Omega$$ and $$2 \Omega$$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:

A DC of 4 A flows through $$3 \Omega$$ and an AC of peak value 4 A flows through $$2 \Omega$$. We need the ratio of heat produced.

Since the DC current is 4 A, the heat produced in the resistor of $$R_1 = 3 \Omega$$ over time t is given by $$H_1 = I^2 R_1 t = (4)^2 \times 3 \times t = 48t$$.

For the AC circuit, heat is produced by the RMS current. Since the peak current is 4 A, we have $$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{4}{\sqrt{2}}$$. Substituting this into the power formula for the resistor $$R_2 = 2 \Omega$$ gives $$H_2 = I_{\text{rms}}^2 \times R_2 \times t = \frac{16}{2} \times 2 \times t = 16t$$.

From the above expressions, the ratio of the heat produced is $$\frac{H_1}{H_2} = \frac{48t}{16t} = \frac{3}{1}$$.

Therefore, the correct answer is Option B: $$3:1$$.