A 100 W bulb `B_(1)` and two 60 W bulbs `B_(2)` and `B_(3)`, are connected to a 250V source, as shown in the figure now `W_(1),W_(2)` and `W_(3)` are the output powers of the bulbs `B_(1),B_(2)` and `B_(3)` respectively then

Options

A. `W_(1) gt W_(2)=W_(3)`
B. `W_(1) gt W_(2)gtW_(3)`
C. `W_(1) lt W_(2)=W_(3)`
D. `W_(1) lt W_(2)ltW_(3)`

Answer

D

Explanation

A bulb is essenetially a resistance `R=(V^(2))/(P)`
Where P denotes the power of the bulb.
`"Resistance of "B_(1)(R_(1))=V^(2)//100`
`"Resistance of "B_(2)(R_(2))=V^(2)//60`
`"Resistance of "B_(3)(R_(3))=V^(2)//60`
`therefore I_(1)="Current in"B_(1)=(250)/((R_(1)+R_(2)))=(250xx300)/(8V^(2))`
`I_(2)="Current in"B_(2)=I_(1)`
`I_(3)="Current in"B_(3)=(250)/(R_(3))=(250xx60)/(V^(2))`
`therefore W_(1)="output power of B"_(1)=I_(1)^(2)R_(1)`
`therefore W_(1)=I_(1)^(2)R_(1)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(100)`
`therefore W_(2)=I_(2)^(2)R_(2)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(60)`
`therefore W_(3)=I_(3)^(2)R_(3)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(60)`
`therefore W_(1) :W_(2):W_(3)=15:25:64 or W_(1)ltW_(2)ltW_(3)`
(b) As the betteries are connected in series, so
`I=(epsi_(1)+epsi_(2))/(r_(1)+r_(2)+R)`
When resistor R is connected to battery of emf `epsi_(1)`.
`I=(epsi_(1))/(r_(2)+R)`
When resistor R is connected to battery of emf `epsi_(2)`.
`I_(2)=(epsi_(2))/(r_(2)+R)`
Fore required condition to be filled
`I lt I_(1) and I lt I_(2)`
`therefore (epsi_(1)+epsi_(2))/(r_(1)+r_(2)+R)lt(epsi_(1))/(r_(1)+R)and (epsi_(1)+epsi_(2))/(r_(1)+r_(2)+R)lt(epsi_(2))/(r_(2)+R)`
`therefore (epsi_(2))/(epsi_(1))lt (r_(2))/(r_(1)+R)and(epsi_(2))/(epsi_(1))and (epsi_(2))/(epsi_(1))gt (r_(2)+R)/(r_(1))`
Thus, option (b) is correct.