I apologize for the confusion in my previous response. To clarify, a device cannot directly draw 5V from a 1.5V AA battery. However, it is possible to use a voltage converter, such as a boost converter, to increase the voltage from 1.5V to the required 5V.

A boost converter is an electronic component that can step up the voltage from a lower input voltage to a higher output voltage, while maintaining the overall power. In this case, it would step up the voltage from 1.5V (provided by the AA battery) to 5V (required by the device).

Since power must be conserved in a system, the current drawn from the battery will be higher than the current consumed by the device. The input power (from the battery) and the output power (to the device) are related as follows:

Input power (P_in) = Output power (P_out) / Efficiency (η)

In the previous response, I made an assumption that the voltage conversion efficiency was 100%. However, this is not realistic, as real-world voltage converters have efficiency ratings typically in the range of 80-95%.

To account for efficiency, let's assume a converter efficiency of 90% (η = 0.9) and recalculate the input power:

P_out = 5V × 180mA = 900mW

P_in = P_out / η = 900mW / 0.9 ≈ 1000mW

Now, we can recalculate the current drawn from the battery (I_battery):

1000mW = 1.5V × I_battery
I_battery ≈ 666.67mA

Given a battery capacity of 2000mAh, the estimated battery life would be:

Battery life ≈ 2000mAh / 666.67mA ≈ 3 hours

Please note that this is still a rough estimate, and the actual battery life will depend on factors such as battery quality, voltage conversion efficiency, and other variables.
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