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

Answer:

The resistance of the flashlight bulb is 3.0E2 ohm

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: $R = \frac{V}{I}$

Given V = 4.50V and I = 15mA

15mA = $\frac{15}{1000}A$

$15mA = 0.015A$

Substituting this value into the formula to get the resistance:

$R = \frac{4.50}{0.015}\\R = 300ohm\\R = 3.0*10^{2} ohm$

The resistance of the flashlight bulb is 3.0E2 ohm