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CBSE Class 10 Answered

Ques no.  20
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Asked by MUSKAN | 14 May, 2019, 04:16: PM
answered-by-expert Expert Answer
The total power when bulbs are connected in parallel,
 
P space equals space left parenthesis P subscript 1 space plus P subscript 2 right parenthesis space open square brackets fraction numerator V o l t a g e space a p p l i e d space a c r o s s space b u l b s space left parenthesis V right parenthesis over denominator R a t e space v o l t a g e space o f space b u l b s space left parenthesis V subscript 0 right parenthesis end fraction close square brackets squared 80 space space space equals space left parenthesis 60 space plus space 40 right parenthesis space open square brackets V over V subscript 0 close square brackets squared space open square brackets V over V subscript 0 close square brackets squared space equals space 80 over 100 equals 4 over 5  
 
The total power when bulbs are connected in series,
 
P space equals fraction numerator P subscript 1 P subscript 2 over denominator P subscript 1 space plus P subscript 2 end fraction open square brackets V over V subscript 0 close square brackets squared space P space equals space fraction numerator 60 cross times 40 over denominator 60 plus 40 end fraction open square brackets 4 over 5 close square brackets  
Solving this we get,
P = 19.2 W 
 
Thus, the total power when the bulbs are connected in series is 19.2 W. 
Answered by Shiwani Sawant | 14 May, 2019, 07:02: PM

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