Two tungsten lamps with resistances `R_(1) and R_(2)`, respectively , are connected first in parallel and then in series in a lighting circuit of negligible internal resistance. Given `R_(1) gt R_(2)`.
(a) Which lamp will glow more brightly when they are connected in parallel ?
(b) If the lamps of resistance `R_(1)` now burns out, how will the net illumination produced change ?
(c ) Which lamp will glow more brightly when they are connected in series ?
(d) If the lamp of resistance `R_(2)` now burns out and lamp `R_(1)` alone is plugged in , will the net illumination increase or decrease ?

Explanation

(a) When the lamps are connected in parallel , potential difference `V` across each lamp will be the same and will be equal to potential difference necessary for full brightness of each bulb. As `P = V^(2)// R `, illumination produced by the second bulb , i.e, the bulb having lower resistance will shine more brightly.
(b) When `R_(1)` burns out, power is dissipated in `R ^(2)` only. Because internal resistance is quite low in lighting circuit , potential difference is still equal to `V`, hence power is dissipated in the second lamp, i.e,
`(V^(2))/( R_(2)) lt ((V^(2))/( R_(1)) + ( V^(2))/( R_(2)))` , net power consumed initially.
In other words , net illumination will now decrease.
(c) When two lamps are connected in series , the potential difference across each lamp will be different , but current `I` flowing through each lamp will be the same , as `P = I^(2) R`.
Hence , illlumination produced by the first lamp will be higher as compared to that produced by the second lamp , i.e, the lamp having higher resistance will glow more brightly.
(d) When lamp of resistance `R_(2)` burns out and only lamp of resistance `R_(1)` is connected in the circuit will change. Let new current be `(i')`.
Because potential difference still remains the same ( due to low internal resistance),
`i'R_(1) = i( R_(1)+ R_(2)) or i' = [( R_(1) + R_(2))]//R_(1)`
If `P'` is the power consumed , then
`P' = i^(2)R_(1) = [i ^(2)( R_(1)+ R_(2)) ( R_(1) + R_(2))]//R_(1)`
When both the lamps were present , then total power consumed was given by
`P_(s) = P_(1) + p_(2) = i^(2) ( R _(1) + R_(2))`, i.e, illumination gets increased when one bulb is used.