Question

1.) Using Ohm’s law, explain how voltage changes in relation to current, assuming that resistance remains constant.

2.) As the electric current in a lightbulb is slowly increased the filament glows more and more brightly. Why does this happen?
3.) Why did some of the lightbulb’s you tried burn out?
4.) Why is there zero current when a lightbulb burns out?

Answer 1

(1)

According to Ohm's law, the voltage is directly proportional to the current flowing in the conductor at constant temperature.

The mathematically expression for Ohm's law is as follows;

V=IR

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

The current will also increase with the increase in the voltage according to Ohm's law.

(2)

Higher the current, then more will be the power dissipation in the form of heat and the light. If the current increases then the brightness will also get increase.

(3)

The bulbs gets burn out due to the power is given more than the power rating of the bulb or if the filament of the light bulb gets break.

(4)

When the light bulb burns out then there is no current in a bulb because the filament of the bulb breaks then there is no electrons to pass through the circuit. In this case, the circuit is incomplete.

Answer 2

1.) Ohm's law is understood as I = V/R. Given that resistance is constant, then voltage changes directly proportional to current.

2.) The more current that passes through a lightbulb, the brighter it glows. The higher the current, the higher the power, where power determines the brightness of a bulb.

3.) A bulb has a specific limit to how much power (Watts) it can handle. Going over the limit would cause the bulb to burn out.

4.) When a bulb burns out, no current will be able to pass through the filament.