In an a.c. circuit, voltage and current are given by: $$V = 100\sin(100t)$$ V and $$I = 100\sin\left(100t + \frac{\pi}{3}\right)$$ mA respectively. The average power dissipated in one cycle is:

The average power dissipated in an AC circuit is given by: $$P_{avg} = \frac{V_0 I_0}{2} \cos\phi$$, where $$V_0$$ is the peak voltage, $$I_0$$ is the peak current, and $$\phi$$ is the phase difference between voltage and current.

From the given equations:

$$V = 100\sin(100t)$$ V, so $$V_0 = 100$$ V

$$I = 100\sin\left(100t + \frac{\pi}{3}\right)$$ mA $$= 0.1\sin\left(100t + \frac{\pi}{3}\right)$$ A, so $$I_0 = 0.1$$ A

The phase difference is $$\phi = \frac{\pi}{3} = 60°$$, so $$\cos\phi = \cos 60° = \frac{1}{2}$$.

Substituting into the formula:

$$P_{avg} = \frac{100 \times 0.1}{2} \times \frac{1}{2} = \frac{10}{2} \times \frac{1}{2} = \frac{10}{4} = 2.5$$ W

Therefore, the average power dissipated is $$2.5$$ W, which is Option (3).