Question

How could an electric bulb marked 100w-220V be lighted through a 400v electric main without getting fused?​

Answer 1

Answer:

because it can convert 100j to 220j in heat and light energy in just 1 second ( J # Joule)

Answer 2

The Main Answer is: By adding a resistance of 1116 Ω in series with the bulb.

Given: An electric bulb rated - 100W-220V

           Voltage source = 400V

To Find: A way for the bulb to not get fused.

Solution:

• For a bulb to get fused, an amount of current larger than its rating has to be passed, i.e., a larger amount of power has to be applied.
• In this question, clearly, if we pass current from a 400V source battery to a 220V rated bulb, it will definitely fuse.
• So, we have to reduce the current in the circuit so a controlled amount of it flows through the bulb.
• This can be done by adding a resistance in series with the bulb.

Originally,

P = 100W,    V = 220V

We know, P = V²/R

⇒ 100 = 220²/R

⇒ R = 48400/100 = 484Ω

After adding a resistance (R) in series,

R' = 484 + R;   V = 400V

By using Ohm's Law (V = iR)

i = 400/(484+R)   ----(1)

Also, P = Vi

i = 100/400 = 1/4  ----(2)

From (1) and (2)-

1/4 = 400/(484+R)

484 + R = 1600

R = 1116Ω

Therefore, the bulb will not fuse if we add a 1116Ω resistor in series with the bulb.

For a similar question on power, refer to:

https://brainly.in/question/15274096?msp_srt_exp=6

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