The power dissipated by a light bulb increases by 50 W when the potential difference applied across the light bulb increases from 50 V to 100 V. Find the resistance of the light bulb.

p=50^2/R

P+50=(100)^2/R
subtract the first equation from the second
50=1/R (100,000-2500)
solve for R. Check my thinking.

It worked! :D

To find the resistance of the light bulb, we can use Ohm's Law, which states that the power dissipated in a circuit is equal to the product of the current passing through it and the voltage across it.

The formula for power is:

P = I * V

Given that the power increases by 50 W and the voltage increases from 50 V to 100 V, we can set up two equations to find the current (I) at each voltage:

Equation 1: P1 = I * V1 (for 50 V)
Equation 2: P2 = I * V2 (for 100 V)

Substituting the given values into the equations:

Equation 1: 50 W = I * 50 V
Equation 2: (50 W + 50 W) = I * 100 V

Simplifying the equations:

Equation 1: 50 = I * 1
Equation 2: 100 = I * 2

Solving for I in each equation:

From Equation 1: I = 50 / 50 = 1 A
From Equation 2: I = 100 / 2 = 50 A

Since the current passing through the light bulb is the same at both voltages, we can conclude that the resistance (R) of the bulb remains constant. Therefore, we can use either equation to find the resistance.

Using Equation 1:

50 = R * 1

Simplifying the equation:

50 = R

Therefore, the resistance of the light bulb is 50 ohms.