198k views
3 votes
an electrical bulb rated at 2KW at 100V is used in a circuit having 200V supply. calculate the resistance that one must put in series with the bulb so that the bulb delivers 400W

User Kendal
by
6.8k points

1 Answer

4 votes
the resistance that must be put in series with the bulb, we can use the equation for electrical power:

P = V^2 / R

where P is the power, V is the voltage, and R is the resistance.

We can rearrange this equation to solve for R:

R = V^2 / P

In this case, the power is 400 W and the voltage is 200 V, so the resistance is:

R = (200)^2 / 400
R = 200

Therefore, the resistance that must be put in series with the bulb is 200 ohms.

It's worth noting that the bulb is rated for 2 kW at 100 V, so using it in a circuit with a voltage of 200 V will likely cause it to operate at a higher power than it is rated for.
User Nguyen Manh Cuong
by
6.4k points