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Physics 15 Online
OpenStudy (anonymous):

and electric bulb is rated 220v and 100w when it is operated on 110v, then power consumed will be??

OpenStudy (anonymous):

first we have to find the resistance as it is constant in an appliance: \[P=V ^{2}/R\] R=220*220/100=22*22

OpenStudy (anonymous):

P=110*110/22*22

OpenStudy (anonymous):

=25W

OpenStudy (anonymous):

that is the answer

OpenStudy (anonymous):

@ashishthomas7 ,Since you are new user, let me tell you something. According to Openstudy Guidelines, you must not be giving out answer's like that. You must be assisting the person to get the answer :)

OpenStudy (anonymous):

i thought that this was all about finding the answer

OpenStudy (anonymous):

Nope. I recommend you to check the Openstudy website guidelines http://openstudy.com/code-of-conduct

OpenStudy (anonymous):

sorry!anyway thanx..

OpenStudy (anonymous):

hi...i am new to open study..hope this can answer your question correctly... 1. the rating of the bulb is 220V,100w right? from this we can find out the resistance of the bulb... P=(V^2)/R; from this equation we can find out resistance R as R=(V^2)/P; substituting voltage and power from the equation we get R as 220*220/100=484 ohms... now as we apply only 110V supply,the power consumed can be calculated as P=(V^2)/R=(110*110)/484=25W so the answer is 25W

OpenStudy (anonymous):

bt i got a different way of doing that u guys tell me what is wrong with it. P=VI 100W=220V x I I=5/11 A P'=V'I P'= 110 X 5/11 W P'= 50W

OpenStudy (anonymous):

@retwick current can vary for differnet voltage and power but only one thing that remains constant always is resistance...

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