A 60-watt Light Bulb's Resistance Calculation

Calculating the Resistance of a 60-Watt Light Bulb

A 60-watt light bulb has a voltage of 120 volts applied across it and a current of 0.5 amperes flows through the bulb. What is the resistance of the light bulb?

Answer: 240 ohms

Explanation:

P = IV

P = 60 watts

I = 0.5 A

V = 120 volts

From Ohm's law

V = IR

R = V / I

R = 120 / 0.5

R = 240 ohms

Final answer:

The resistance of a 60-watt light bulb with a voltage of 120 volts and a current of 0.5 amperes is calculated using Ohm's law and is found to be 240 ohms.

Explanation:

To find the resistance of a light bulb, we can use Ohm's law which states that resistance (R) is equal to voltage (V) divided by current (I). In the case of a 60-watt light bulb with a voltage of 120 volts and a current of 0.5 amperes, the resistance is calculated as:

R = V / I

Therefore, R = 120 V / 0.5 A = 240 ohms.

The resistance of the light bulb is 240 ohms. It's important to recognize that the power rating of a bulb (in watts) is related to both the voltage applied and the current that flows through the bulb. The relationship between power (P), voltage (V), and current (I) is given by P = V * I, which is derived from Ohm's law.