Ohm's Law & Power: Bicycle LED Headlight & 4 AA Batteries

[2slowtogofast]

A bicycle LED headlight requires 5 V, 250 mA to operate. Using Ohm’s law, determine the
equivalent resistance of the light. What is the power consumption rate in W? How many
electrons pass through per second? It is powered by four AA rechargeable battery in series. Each battery provides 1.25 V and has a specification of 2300 mA-Hour. How many second would the battery last?

The first part i used R = v/I

The second part I know P = VI but am not sure what to use for time because it does't give one.

The third part i don't know about. I am not looking for the answer but help on how to get there thank you

 

[2slowtogofast]

can somebody give me an idea

 

[LCKurtz]

Isn't 1 watt = one joule per second?

Maybe better: 1 ampere = 6.242*10¹⁸ electrons/second

 

[vk6kro]

Each battery provides 1.25 V and has a specification of 2300 mA-Hour. How many second would the battery last?

This means each battery would last for an hour if it was supplying 2300 mA. It would last for 10 hours if it was supplying 230 mA.

So product of (current * time) = 2300 mAH
or
time = 2300 mAH / current

So, how long would it last if it was supplying 250 mA?

 

[DeeNos]

I would like to clarify that Ohm's Law and Power are fundamental concepts in electrical engineering and physics, and they are essential for understanding the behavior of electric circuits and devices. In the context of the bicycle LED headlight and 4 AA batteries, we can use Ohm's Law to calculate the equivalent resistance of the light and the power consumption rate.

To determine the equivalent resistance of the LED headlight, we can use the formula R = V/I, where R is the resistance, V is the voltage, and I is the current. In this case, the voltage required for the headlight is 5 V, and the current is 250 mA (0.25 A). Therefore, the equivalent resistance of the light is 20 ohms.

To calculate the power consumption rate, we can use the formula P = VI, where P is the power, V is the voltage, and I is the current. In this case, the power consumption rate would be 1.25 W (5 V x 0.25 A).

As for the number of electrons passing through per second, we can use the formula I = Q/t, where I is the current, Q is the charge, and t is the time. Since we know the current is 0.25 A, we can rearrange the formula to find Q, which represents the number of electrons passing through per second. Therefore, Q = It = 0.25 A x 1 s = 0.25 C.

Moving on to the batteries, we know that each battery provides 1.25 V and has a capacity of 2300 mA-Hour. To calculate the total voltage provided by the four batteries in series, we can simply add the individual voltages, resulting in 5 V. However, to determine how long the batteries will last, we need to consider the capacity in terms of ampere-hours (Ah). To do this, we can use the formula t = Q/I, where t is the time, Q is the charge, and I is the current. In this case, the current is 250 mA (0.25 A), and the charge is 2300 mA-Hour (2.3 Ah), resulting in a time of 9.2 hours (2.3 Ah / 0.25 A).

In conclusion, using Ohm's Law and the formulas for power and charge, we can calculate the equivalent resistance, power