The power P of an electrical device (like a lightbulb) is given by the formula P = V^2 / R, where V is the voltage and R is the resistance.

For the 50-watt bulb, we can write the equation as 50 = 100^2 / R1. Solving for R1, we get R1 = 100^2 / 50 = 200 ohms.

For the 100-watt bulb, we can write the equation as 100 = 100^2 / R2. Solving for R2, we get R2 = 100^2 / 100 = 100 ohms.

Comparing the two resistances, we see that the resistance of the 100-watt bulb is half that of the 50-watt bulb. So, the answer is A: half as much.

Question

The power P of an electrical device (like a lightbulb) is given by the formula P = V^2 / R, where V is the voltage and R is the resistance.

For the 50-watt bulb, we can write the equation as 50 = 100^2 / R1. Solving for R1, we get R1 = 100^2 / 50 = 200 ohms.

For the 100-watt bulb, we can write the equation as 100 = 100^2 / R2. Solving for R2, we get R2 = 100^2 / 100 = 100 ohms.

Comparing the two resistances, we see that the resistance of the 100-watt bulb is half that of the 50-watt bulb. So, the answer is A: half as much.

Knowee AI · Accepted Answer

The power P of an electrical device (like a lightbulb) is given by the formula P = V^2 / R, where V is the voltage and R is the resistance.

For the 50-watt bulb, we can write the equation as 50 = 100^2 / R1. Solving for R1, we get R1 = 100^2 / 50 = 200 ohms.

For the 100-watt bulb, we can write the equation as 100 = 100^2 / R2. Solving for R2, we get R2 = 100^2 / 100 = 100 ohms.

Comparing the two resistances, we see that the resistance of the 100-watt bulb is half that of the 50-watt bulb. So, the answer is A: half as much.

A 50-watt lightbulb and a 100-watt lightbulb are each operated at 100 volts. Compared to the resistance of the 50-watt bulb, the resistance of the 100-watt bulb isA half as much.B one-fourth as much.C twice as much.D four times as much.