Question

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 (5 points) 2.5 E2 ohm 3.0 E2 ohm 3.5 E2 ohm 4.0 E2 ohm 4.4 E2 ohm

Answer 1

3.0 E2 ohm

Step-by-step explanation:

I took the test and got it right :)

Answer 2

The resistance of the flashlight bulb is 3.0E2 ohm

Step-by-step explanation:

According to ohms law, the voltage across a circuit is expressed as V = IR where:

V is supply voltage

I is the supply current

R is the resistance.

From the ohms law formula:
`R = (V)/(I)`

Given V = 4.50V and I = 15mA

15mA =
`(15)/(1000)A\\`

`15mA = 0.015A`

Substituting this value into the formula to get the resistance:

`R = (4.50)/(0.015)\\R = 300ohm\\R = 3.0*10^(2) ohm`

The resistance of the flashlight bulb is 3.0E2 ohm