Question 6A301

To determine the current flowing, we first need to find the resistance of the lamp. Using the power formula, $P = V^2 / R$, we can solve for R: $R_{lamp} = V^2 / P = (6V)^2 / 3W = 36V^2 / 3W = 12 \text{ ohms}$. The battery's internal resistance acts in series with the lamp's resistance. Therefore, the total resistance in the circuit is: $R_{total} = R_{lamp} + R_{internal} = 12 \text{ ohms} + 0.01 \text{ ohms} = 12.01 \text{ ohms}$. Now, apply Ohm's Law ($I = V / R$) to find the total current flowing from the battery: $I = 6V / 12.01 \text{ ohms} \approx 0.49958 \text{ A}$. Rounding to four decimal places, the current is approximately 0.4995 A. The final answer is $\boxed{\text{0.4995 A}}$.