Question

Please answer this question.​

Answer 1

Explanation:

The total current flowing through the series current will be ,

I = V/R = 440/R

Here, R is the total resistance offered by two devices.

But, R = R₁ + R₂ = (V₁²/P₁₁) + (V₂²/P₂)

                        = (220²/60) + (220²/100)

                       = 1290.7 ohms.

So, I = 440/1290.7 = 034 A

Also,

I₁ = 60/220 = 0.3 A

I₂ = 100/220 = 0.45 A

As the current supplied is greater than the rated current for the first bulb. So it will fuse.

Hope it helps!