A 80 watt and a 100 watt bulb are connected in series, which bulb will have more or equal brightness and why ?

Question

The bulbs are connected in series hence same current will flow through them.

Answer 1

The brightness is related to the amount of light energy that is being produced each second. This is what power measures. In your house, the light bulbs are in parallel. The voltage for each bulb is 120. Let’s determine the resistance of each bulb.



P = V * I, I = V/R, P = V^2/R



80 = 120^2/R, R = 120^2/80 = 180 Ω

100 = 120^2/R, R = 120^2/100 = 144 Ω



Let’s place the two bulbs in series and determine the current. In series, the total resistance is 224 Ω.

V = I * R

120 = I * 224

I ≈ 0.536 amps



Let’s determine the voltage for each bulb. V = I * R

For the 80 w bulb, V = 0.536 * 180 = 96.48 volts

For the 100 w bulb, V = 0.536 * 144 = 77.184 volts



Let’s determine power for each bulb. P = V * I

For the 80 w bulb, P = 96.48 * 0.536 ≈ 51.7 watts

For the 100 w bulb, V = 77.184 * 0.536 ≈ 41.4 watts

The power of each bulb has decreased, because the voltage for each bulb has decreased.



When the bulbs are in series, the 80 watt bulb is brighter than the 100 watt bulb. Let’s see how this compares to a parallel circuit.



Let’s determine the amount of current for each bulb when connected in parallel. I = P ÷ V

For the 80 w bulb, I = 80 ÷ 120 = 0.667 amp

For the 100 w bulb, I = 100 ÷ 120 = 0.833 amp



When the bulbs are connected in parallel, the current for 100 watt bulb is higher than the 80 watt bulb. When the bulbs are connected in series, the current for each bulb and voltage is was lower than in parallel. This is way bulbs are brighter when connected in parallel than in series. I hope this helps you understand this question.

Answer 2

Assuming incandescent lightbulbs, not LEDs or CFLs etc. which probably won't work at a lower voltage.



At the design voltage, the power consumed is given by V^2/R, so for a given power P R=V^2/P.



So the 80W one has a higher resistance, hence for the same current will consume more power and so (assuming the bulbs are equally efficient) be brighter.



Furthermore, the 100W one will run cooler than the 80W one, and since the resistance of the tungsten filament is lower at lower temperatures, it will consume even less power than you would expect if you worked out the sums thinking it was ohmic.

Answer 3

the voltage will divide between the bulbs in accordance with their resistance.

since power = E^2/R the lower wattage bulb has a higher resistance and will get more voltage. Therefore the lower wattage bulb will shine brighter.