In a experiment , `N` identical electrical bulbs, each having resistance `R` , are connecteed in parallel to a dc source of emf `E` and internalresistance `r` . What is the power consumed by each bulb . Also find the percentage change in power by each bulb if one bulb turns out.

Answer

`( E^(2) R )/(( N r + R))^(2)` , `( 200 r )/( N r + R)`

Explanation

If `R` is the resistance of each bulb, then the equivalent resistance of `N` bulbs in parallel is `R_(eq) = R//N`. Therefore , current is equally divided among all the `N` bulbs, as potential drop across each bulb is the same. So power consumed by each bulb is
`P = ((i)/(N))^(2) R = (E^(2))/(N^(2)((R)/(N) + r)^(2)) = (E^(2) R ) /((Nr + R)^(2))` (i)
With `( N - 1)` bulbs, the power consumed by each bulb can be obtained by replacing `N by ( N - 1)` in Eq. (i). So,
Now , percentage change in power consumption of each bulb is
`(P' - P)/(P) xx 100 = [ (r N + R)^(2))/([ (r ( N-1) + R )]^(2)) -1] xx 100 `
` = [ (1)/([( 1 - (r)/( Nr + R))]^(2)) - 1] xx 100`
As `r ltlt ( Nr + R) , ( 1- (r)/( Nr + R))^(-2) - 1 + ( 2 r ) /( Nr + R)`
Therefore, percentage change in power is
` ( 2r)/( Nr + R) xx 100 = ( 200 r ) / (( Nr + R))`