When is the higher rate of energy being delivered to a light bulb higher : just after it is turned on, the glow of the filament is increasing, or after it has been on for a few seconds and the glow is steady ?

Explanation

(a) Just after it is turned on , the energy supplied to the bulb is higher . Becsause initially resistance is low and it increases with the increase in temperature.
(b) After connecting `b and c`, `R_(2)` will be short - circuited . Hence , it will not glow. Now , entire potential difference will be on `R_(1)`. So brightness of `R_(1)` increases.
(c) `D` is short - circuited, so `D` will not glow. Potential difference across `C` is `V`, so potential difference across each of `A and B is V//2`. So brightness of `C` is the greatest and that of `A and B` are equal.
If `A` fails, `B` will not glow. No effect on `C and D`.
IF `C` fails , `C` will not glow. No effect on `A, B , and D`.
IF `D` fails , no effect anywhere.
Because resistance is directly proportional to length , at low current and high voltage energy , loss will be less. Because if current is less , then loss of energy `I^(2)R` will be less.
(e) Initially , current will be high in the circuit , so bulb will glow with maximum possible brightness of the bulb falls and finally the bulb goes out.
(f) This is not true. Because in series, current through all the bulbs is the same. The current is simultaneously set up is series through all the bulbs.
(g) `P =(V^(2))/(R) = (v^(2)A) /(pl)`
Here, `P` and `V` are fixed. We can vary `A` or `l`.