Question

A flashlight bulb is connected to a dry cell of voltage 3.50 V. It draws 20.0 mA (1000 mA = 1 A). What is its resistance?

Answer 1

Answer:

1.75 × 10²  Ω

Explanation:

Given :

Voltage V = 3.50 V

Current I = 20.0 m A = 20.0 × 10⁻³ A

From ohm's law we have :

V = I R

R = V / I

R = 3.50 / 20 × 10⁻³  Ω

R = 3.50 × 10² / 2  Ω

R = 1.75 × 10²  Ω

Therefore , the resistance  is 1.75 × 10²  Ω.

Answer 2

The resistance of the flashlight bulb is 175.0 Ω.

The resistance of the flashlight bulb can be calculated using Ohm's law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them.

The equation for Ohm's law

V = IR

Where V is the voltage, I is the current, and R is the resistance.

Given that the voltage across the bulb is 3.50 V and the current through it is 20.0 mA,

we can calculate the resistance:

$R = V / I$

$R = 3.50 V / 20.0 mA$

R = 175.0 Ω

So, the resistance of the flashlight bulb is 175.0 Ω.

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