Answer:
The resistance of alight bulb is 240 ohms.
Step-by-step explanation:
Given that,
Potential difference of a light bulb, V = 120 V
Current produced in the light bulb, I = 0.5 A
We need to find the resistance of the light bulb. It can be calculated using Ohm's law. It is given by :
V = IR
![R=(V)/(I)](https://img.qammunity.org/2020/formulas/physics/college/8zyzv6fvsj12hmbq6z1536sqn88dxoj1v9.png)
![R=(120\ V)/(0.5\ A)](https://img.qammunity.org/2020/formulas/physics/middle-school/h0uowr0jg1i3e5x0rlkn61we3l7avdf9jb.png)
R = 240 ohms
So, the resistance of alight bulb is 240 ohms. Hence, this is the required solution.