Question

The marking on an electric bulb is 800W, 200V. When 500V is applied to it​

Answer 1

Given : The marking on an electric bulb is 800W, 200V

500V is applied to it​

To Find : Correct option

Brightness of the bulb will increase

Brightness of the bulb will decrease

The bulb will be burn out

None of these

Solution:

Power = V²/R

800 = 200²/R

=> R =  200²/800

=> R = 200/4

=> R = 50 Ω

500V is applied to it

Hence Power =  500²/(50) = 5000W

Brightness of the bulb will increase

But as Bulb is rated for 200V  and   800W

so 500 V and 5000W would be too much for the Bulb

hence the bulb will burn out.

Brightness of the bulb will increase and  bulb will burn out.

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Answer 2

First of all, in these kind of questions, find out the resistance of the bulb !

$P = \dfrac{{ V}^{2}}{R}$

$\implies R = \dfrac{{ V}^{2}}{P}$

$\implies R = \dfrac{{ 200}^{2}}{800}$

$\implies R = \dfrac{200 \times 200}{800}$

$\implies R = 50 \: ohm$

Now, calculate the power dissipated when voltage applied is 500 V.

$P = \dfrac{{ V}^{2}}{R}$

$\implies P = \dfrac{{ 500}^{2}}{50}$

$\implies P = \dfrac{500 \times 500}{50}$

$\implies P = 5000 \: watt$

Since power dissipation increases, so bulb will become brighter.

Option a) is correct ✔️.