CBSE BOARD X, asked by jspcr, 1 year ago

an electric bulb is rated at 200 w and 250v find the maximum permissible current in its filament

Answers

Answered by ajayreddy7474pbt5ya
4
p=200w
v=250v
I=???

USE: P=VI
then solve...
200= 250I
200/250=I
I=0.8

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ajayreddy7474pbt5ya: if this help you, please make me BRAINLIEST
Answered by meenug1974pb4csn
2

P= VI
200 = 250 xI
I = 200/250
I= 4/5. = 0.8 A

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