Question

The hot resistance of a flashlight bulb is 2.30Ω, and it is run by a 1.58-V alkaline cell having a 0.100-Ω internal resistance. What current flows in the flashlight bulb?

a) 1.58 A
b) 15.8 A
c) 0.684 A
d) 0.0684 A

Answer 1

The current flowing through the flashlight bulb is 0.684A.

Step-by-step explanation:

To find the current flowing through the flashlight bulb, we need to calculate the total resistance in the circuit and then use Ohm's Law. The total resistance in the circuit is the sum of the hot resistance of the bulb and the internal resistance of the alkaline cell. So the total resistance is 2.30Ω + 0.100Ω = 2.40Ω.

Using Ohm's Law (V = IR), we can rearrange the equation to solve for current (I). So I = V/R = 1.58V / 2.40Ω = 0.658A or 0.684A (rounded to three decimal places).