Question

In Part H, you discovered that the luminosity of a light bulb increases if the current increases. The rate at which electric potential energy is converted into heat depends on the current flowing through the bulb and the voltage across the bulb. This energy is supplied by the battery. Mathematically, the luminosity P of the light bulb is given by P=ΔVI, where ΔV is the voltage across the bulb and I is the current.What happens to the luminosity of the light bulb if the voltage of the battery is doubled? (Note that the PhET simulation does not display a numerical value for the luminosity, so you should use the relationship between the luminosity, the voltage across the bulb, and the current.)

Answer 1

Doubling the battery voltage applied to a light bulb will result in nearly quadrupling its power, assuming the bulb's resistance remains constant, due to the power being proportional to the square of the voltage (P = V²/R).

Step-by-step explanation:

The luminosity of a light bulb, which can be thought of as its power output, is directly proportional to the electric potential energy converted into light and heat. The relationship between power (P), voltage (V), and current (I) is given by P = ΔVI, where ΔV represents the voltage across the bulb, and I represents the current flowing through it. When we discuss this relationship in terms of resistance (R), we can also express power as P = V²/R. According to this formula, doubling the voltage while keeping the resistance constant will result in a near quadrupling of power because the voltage is squared in the expression.

For example, a 25-W bulb designed to operate at a certain voltage would have its power increased nearly to 100 W if the voltage is doubled, assuming the resistance remains constant. However, in actual practice, the resistance of a bulb increases with temperature; thus, although the power increase is substantial, it is not exactly quadrupled.