Final answer:
The current that flows through a flashlight bulb with a 3.2 V supply and a resistance of 3.95 ohms is approximately 0.8101 amperes, calculated using Ohm's Law.
Step-by-step explanation:
The current that flows through the bulb of a flashlight when given a voltage and resistance can be calculated using Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). To find the current flowing through the bulb of a flashlight, we can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In this case, the voltage of the flashlight is 3.2 volts and the resistance is 3.95 ohms. To find the current, we divide 3.2 volts by 3.95 ohms, which gives us approximately 0.8101 amperes.
Using Ohm's Law:
I = V / R
I = 3.2 V / 3.95 Ω
I ≈ 0.8101 A