A flashlight bulb is connected to a dry cell of voltage 4.5 V. It draws 15 mA (1,000 mA = 1 A). Its resistance is 2.5 E2 ohms 3.0 E2 ohms 3.5 E2 ohms 4.0 E2 ohms
Question
A flashlight bulb is connected to a dry cell of voltage 4.5 V. It draws 15 mA (1,000 mA = 1 A). Its resistance is 2.5 E2 ohms 3.0 E2 ohms 3.5 E2 ohms 4.0 E2 ohms
Solution
The question seems to be asking for the resistance of the flashlight bulb. We can use Ohm's Law to solve this. Ohm's Law states that Voltage (V) = Current (I) x Resistance (R). We can rearrange this formula to solve for Resistance: R = V/I.
Given: Voltage (V) = 4.5 V Current (I) = 15 mA = 0.015 A (since 1,000 mA = 1 A)
Substitute these values into the formula:
R = 4.5 V / 0.015 A = 300 ohms
So, the resistance of the flashlight bulb is 300 ohms, which is equivalent to 3.0 E2 ohms.
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