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OpenStudy (anonymous):

A car’s headlight dissipates 40 W on low beam and 50 W on high beam. Is there more or less resistance in the high-beam filament? @Abhisar

OpenStudy (abhisar):

What's ur say ?

OpenStudy (anonymous):

i would say there is less resistance because it is putting out more watts

OpenStudy (abhisar):

brb

OpenStudy (abhisar):

i am back

OpenStudy (abhisar):

u know the formula for power ?

OpenStudy (abhisar):

it's \[P = \frac{ V^{2} }{R}\]

OpenStudy (abhisar):

Now can uh predict the answer using this ?

OpenStudy (abhisar):

the formula says that power is inversly proportional to the resistance

OpenStudy (abhisar):

i.e if power increases resistance decreases

OpenStudy (abhisar):

Now predict the answer !

OpenStudy (abhisar):

???????

OpenStudy (anonymous):

ok i'm here i walked away

OpenStudy (anonymous):

explain the formula again please

OpenStudy (abhisar):

p = power, v = voltage, R = resistance

OpenStudy (anonymous):

P= 40\[P=40^2 /r\]

OpenStudy (anonymous):

?

OpenStudy (abhisar):

No actually u dont need to calculate anything...

OpenStudy (abhisar):

the formula says that power is inversly proportional to the resistance

OpenStudy (abhisar):

i.e if power increases resistance decreases

OpenStudy (anonymous):

so i was right the first time?

OpenStudy (abhisar):

More the power of bulb, lesser will be resistance

OpenStudy (abhisar):

Yes uh r ryt !

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