an electric bulb is rated at 200 w and 250v find the maximum permissible current in its filament
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Answered by
4
p=200w
v=250v
I=???
USE: P=VI
then solve...
200= 250I
200/250=I
I=0.8
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v=250v
I=???
USE: P=VI
then solve...
200= 250I
200/250=I
I=0.8
if this helps you, please make me BRAINLIEST
ajayreddy7474pbt5ya:
if this help you, please make me BRAINLIEST
Answered by
2
P= VI
200 = 250 xI
I = 200/250
I= 4/5. = 0.8 A
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