Question

A light bulb has a resistance of 1kΩ and a maximum current of 3mA. How much voltage can be applied before the bulb will break?

Answer 1

Answer:

The maximum voltage is 3 volts

Explanation:

We know that V= IR

so we can write V (max.) = I (max.) R

In the question, I (max.) = 3 mA = (3/1000) A

and R = 1 kilo ohm = 1000 ohm.

The voltage can be applied before the bulb will break is = (3/1000) (1000) volts = 3 volts