A 60-Watt Light Bulb: Calculate the Resistance

Understanding the Resistance of a 60-Watt Light Bulb

When dealing with electrical circuits, understanding the concept of resistance is crucial. Resistance is the measure of how difficult it is for an electric current to pass through a conductor. In this scenario, we will calculate the resistance of a 60-watt light bulb given certain parameters.

A 60-watt light bulb has a voltage of 120 volts applied across it and a current of 0.5 amperes flowing through the bulb. The question at hand is determining the resistance of the light bulb based on this information.

To calculate the resistance, we can utilize Ohm's Law, which states that the voltage (V) across a conductor is equal to the current (I) flowing through it multiplied by the resistance (R) of the conductor. Mathematically, this can be represented as V = IR.

Calculation:

Given:

Voltage (V) = 120 volts

Current (I) = 0.5 amperes

Formula: R = V / I

Substitute the given values:

R = 120 / 0.5

Result: R = 240 ohms

Therefore, the resistance of the 60-watt light bulb is 240 ohms. This means that the light bulb will impede the flow of electricity with a resistance of 240 ohms when connected to a voltage source of 120 volts.

Alternatively, resistance can also be calculated using the formula: resistance = (voltage ^ 2) / wattage. Substituting the values, we get [(120^2) / 60] = 240 ohms.

Question:

What is the resistance of the 60-watt light bulb with a voltage of 120 volts and a current of 0.5 amperes?

Answer:

240 ohms

Explanation:

From Ohm's Law, we deduce that V = IR. By rearranging the formula to solve for resistance (R), we find that R = V / I, where R is resistance, I is current, and V is voltage. Substituting the given values of 120 volts for V and 0.5 amperes for I, we obtain:

R = 120 / 0.5 = 240 ohms

Alternatively, resistance can be calculated using the formula resistance = (voltage ^ 2) / wattage. By substituting the values, we get [(120^2) / 60] = 240 ohms.