Answers

2014-09-18T13:46:29+05:30
1 FOR BULB RATED 60W, 220V
P=VI
60=220*I
60/220=I
0.27=I
AND FOR RESISTANCE
V=IR
220=0.27*R
220/0.27=R
814.8=R
2.NOW FOR BULB RATED 100W,220V
P=VI
100=220*I
100/220=I
0.45=I
AND FOR RESISTANCE
V=IR
220=0.45*R
220/0.45=R
488.8=R

SO BULB A WILL GLOW BRIGHTER


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Your answer is fine, but since they are connected in series, wont they both receive the same current, and glow equally bright??
  • Brainly User
2014-09-18T16:15:12+05:30
FOR BULB A :
POWER=60W
POTENTIAL DIFFERENCE=220V
P=V×V/R
SO,R=V×V/P , HENCE

   
       220×220÷60=806.6 OHM
 
FOR BULB B

 POWER = 100W
 POTENTIAL DIFF.=220V
  
BY SAME FORMULA 
                                          220×220÷100=484 OHM
WE CONCLUDE THAT RESISTANCE IS MORE IN BULB A HENCE CURRENT IS LOW IN BULB A
BECAUSE  OF IT RESISTANCE IS MORE THAN HEATING EFFECT IS HIGH SO BULB A GLOW BRIGHTER.
PLZ MARK IT BEST
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