Question

As more and more bulbs are connected in series to a flashlight battery, what happens to the brightness of each bulb? Assuming that heating inside the battery is negligible, what happens to the brightness of each bulb when more and more bulbs are connected in parallel?

Answer 1

`P_1 = P_2 = (P)/(2)`

so each bulb brightness becomes half of its given or indicated power

`P_1 = P_2 = (V^2)/(R)`

so both bulb will glow same power as indicated

Step-by-step explanation:

Let the indicated power on the bulbs is given as P and its rated voltage is V

so here resistance of each bulb is given as

`R = (V^2)/(P)`

now if the two bulbs are connected in series so we will have

`R_(eq) = R_1 + R_2`

`R_(eq) = 2(V^2)/(P)`

now the current in the circuit is given as

`i = (P)/(2V)`

now brightness of each bulb is given as

`P_1 = P_2 = i^2 R`

`P_1 = P_2 = (P)/(2)`

so each bulb brightness becomes half of its given or indicated power

Now if the two bulbs are connected in parallel

then the net voltage across each bulb is "V"

so we will have

`P_1 = P_2 = (V^2)/(R)`

so both bulb will glow same power as indicated