Two `220 V, 100 W` bulbs are connected first in series then in parallel. Each time the combination is connected to a `220 V` AC supply line. The power drawn by the combination in each case respectively will be

Options

A. 200 W,150 W
B. 50 W,200 W
C. 50 W,100 W
D. 100 W,50 W

Answer

B

Explanation

`P_(1) = 100 W,P_(2) =100 W,V = 220` volt
`P_(1) = (V^(2))/(R_(1))P_(2) = (V^(2))/(R_(2))`
`:. R_(1) = (V)/(P_(1)) = ((220)^(2))/(100) = (220 xx 220)/(100) Omega`
`R_(2) = (V^(2))/(P_(2))=((220)^(2))/(100) = (220 xx220)/(100) Omega`
Case I = When two bulbs are connected in series.

In series, `R_(eq)=R_(1)+R_(2)=((220xx220)/(100))xx2`
Hence, `P_(eq) =(V^(2))/(R_(eq))=((220 xx220)/((200xx220)/(100)xx2))`
`= (100)/(2) = 50 W`
Case II : When two bulbs are connected in parallel

`R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2))=(((200xx200)/(100))^(2))/((220 xx220)/(100)xx2)`
`R_(eq)=(220 xx220)/(100)xx(1)/(2)`
Hence, `P_(eq)=(V^(2))/(R_(eq))=(220 xx220)/((220xx220)/(100)xx(1)/(2)) =220W`.