Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, `100W, 60W and 40W` bulbs have filament resistances `R_(100), R_(60) and R_(40)`, respectively, the relation between these resistances is

Options

A. `(1)/(R_(100)) = (1)/(R_(40)) + (1)/(R_(60))`
B. `(R_(100)) = (R_(40)) + (R_(60))`
C. `(R_(100)) gt (R_(40)) gt (R_(40))`
D. `(1)/(R_(100)) gt (1)/(R_(60)) gt (1)/(R_(40))`

Answer

A, D

Explanation

For given voltage power can be written as follows:
`P = (V^(2))/(R )`
`(V^(2))/(R_(100)) = 100`
`(V^(2))/(R_(60)) = 60`
`(V^(2))/(R_(40)) = 40`
Hence we can write the following (logically):
`(V^(2))/(R_(100)) = (V^(2))/(R_(60)) + (V^(2))/(R_(40))`
`rArr (1)/(R_(100)) = (1)/(R_(60)) + (1)/(R_(40))`
`:.` Option (a) is correct.
`P = (V^(2))/(R) rArr` Resistance of higher power bulb will be smaller.
Hence `R_(100) lt R_(60) lt R_(40)`
Or we can write it as follows:
`(1)/(R_(100)) gt (1)/(R_(60)) gt (1)/(R_(40))`
`:.` Option (d) is also correct.