A) \[{{W}_{1}}>{{W}_{2}}={{W}_{3}}\]
B) \[{{W}_{1}}>{{W}_{2}}>{{W}_{3}}\]
C) \[{{W}_{1}}<{{W}_{2}}={{W}_{3}}\]
D) \[{{W}_{1}}<{{W}_{2}}<{{W}_{3}}\]
Correct Answer: D
Solution :
Resistances of bulbs \[{{B}_{2}}\] and \[{{B}_{3}}\] are equal but that of \[{{B}_{1}}\] is smaller than their resistances, hence resistances of path of bulb \[{{B}_{3}}\] is less that of the series combination of bulbs \[{{B}_{2}}\] and \[{{B}_{1}}\], therefore power consumed by Bo will be maximum possible. Since, \[{{B}_{1}}\] and \[{{B}_{2}}\] are in series, therefore current through them will be the same. Since resistance of \[{{B}_{2}}\] is greater, therefore \[{{W}_{2}}>{{W}_{1}}\]You need to login to perform this action.
You will be redirected in
3 sec