Question

I really don't understand this problem at all, I'm not even sure which equation to use. Thanks for helping me in advance!

A flashlight bulb is connected to a dry cell of voltage 4.50 V. It draws 15.0 mA (1000 mA = 1 A). Its resistance is

2.5 E2 ohm.

3.0 E2 ohm.

3.5 E2 ohm.

4.0 E2 ohm.

4.4 E2 ohm.

Answer 1

ohms law \[V[\text V]=I[\text A]R[\Omega]\]

Answer 2

solving for Resistance
\[R=V/I\]

Answer 3

So then it would be 0.3, which is 3.0 E2 ohms, right?

Answer 4

\[R=\frac{V}{I}=\frac{4.5[\text V]}{15[\text {mA}]}=\frac{4.5[\text V]}{15\times10^{-3}[\text {A}]}=300[\Omega]=3\times10^2[\Omega]\]

Answer 5

Awesome, thank you!