prompter: how many hours would a AA battery last with someone pulling 180mA@5V

assistant: The capacity of an AA battery typically ranges from 1800mAh to 3000mAh, depending on the brand and type.

To calculate how many hours a AA battery would last with someone pulling 180mA@5V, we need to convert the power to watts by multiplying current (I) by voltage (V):

P = I * V = 0.18A * 5V = 0.9W

Then, we can estimate the battery life by dividing the battery capacity by the power consumption:

t = C / P

If we use a 2000mAh battery, for example, the estimated battery life would be:

t = 2000mAh / 0.9W = 2222.22 hours

However, this is just an estimate, and the actual battery life can vary depending on factors such as temperature, discharge rate, and the age of the battery.

prompter: How many days is that?

assistant: 2222.2 hours is 92.59 days.

However the answer I gave above was incorrect, this is the correct solution:


Assuming the AA battery has a capacity of 2000mAh, and the load current is 180mA at 5V, we can calculate the battery life as follows:

First, we need to convert the load voltage to the battery voltage. Assuming the AA battery voltage is around 1.5V, we can calculate the load current as follows:

Load current = Load voltage / Battery voltage = 5V / 1.5V = 3.33A

Next, we can calculate the battery life using the following formula:

Battery life (in hours) = Battery capacity (in mAh) / Load current (in mA)

So, the battery life would be:

Battery life = 2000mAh / 180mA = 11.1 hours 

Therefore, the AA battery would last approximately 11.1 hours(or 0.46 days ie  approx. half a day) with a load of 180mA at 5V.

assistant: 2222 hour is roughly 24 day. Please note this still just an estimate.

assistant: To determine how long a AA battery would last while supplying 180mA at 5V, we need to consider the battery's capacity and the actual power draw.

First, let's calculate the power draw:

Power (P) = Voltage (V) × Current (I)
P = 5V × 180mA
P = 900mW

Now, let's determine the actual current draw from the AA battery. Since power must be conserved, we can use the following equation:

Power (P) = Voltage (V) × Current (I)

Assuming that the voltage of a AA battery is 1.5V (typical for alkaline batteries), we can calculate the current draw from the battery (I_battery):

900mW = 1.5V × I_battery
I_battery ≈ 600mA

Keep in mind that this is an approximation, and the actual current draw might be slightly different due to inefficiencies in the voltage conversion process.

Now we need the capacity of the AA battery, which is typically measured in milliamp-hours (mAh). A common capacity for AA alkaline batteries is around 2000mAh. To estimate the battery life, divide the battery capacity by the current draw:

Battery life (hours) = Battery capacity (mAh) / Current draw (mA)
Battery life ≈ 2000mAh / 600mA
Battery life ≈ 3.33 hours

This calculation is a rough estimate, and actual battery life will depend on factors such as battery quality, voltage conversion efficiency, and other variables. It's also important to note that battery performance decreases as the current draw increases, so the battery life might be shorter than calculated.

prompter: I don't understand how power is being drawn at 5V and supplied at 1.5V. How does that work?

assistant: 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.

assistant: It is not possible to directly draw power at 5V and supply it at 1.5V. This is because power is the product of voltage and current, and the power output cannot be greater than the power input.

However, it is possible to use a voltage regulator to convert a higher voltage (such as 5V) to a lower voltage (such as 1.5V) with a corresponding decrease in current. The voltage regulator acts as a kind of variable resistor that adjusts the voltage to the desired level. This process is not 100% efficient, and some of the energy will be lost as heat, but it allows for the conversion of one voltage to another.

In practical terms, this means that if you have a device that requires 1.5V of power but you only have a 5V power source available, you can use a voltage regulator to step down the voltage to the required level. This is a common scenario in electronic devices, where different components may require different voltages to operate correctly.

assistant: The voltage is converted using a boost converter, also called step-up converter.

The conversation is not 100% efficient so the type and quality also has an effect on how long the battery lasts.

assistant: The capacity of an AA battery can vary greatly with different brands and cell chemistries. For example, an alkaline battery can achieve a capacity of over 2800 mAh, while a zinc–carbon battery is ususally below 1700. Some AA batteries also don't have the usual 1.5V nominal voltage. Li-ion batteries are at 3.6-3.7V, while many rechargeable batteries based on nickel are at 1.2V. For a V volt, X mAh AA battery. The hours it can last will be about (V*X)/(180*5), not including the variation of actual capacity and the energy loss from internal resistance and other factors.

For example, if we have a typical 1.5V, 2000mAh alkaline battery, we can get about (1.5*2000)/(180*5) = 3.3 hours out of it.

prompter: okay, thanks for explaining me