There are two bulbs One with 10 ohm resistance and other with 5 ohm. 1.Which bulb will glow bright. 2.Which bulb will empty battery first.?

Asuming for bothe cases the same supply voltage the answer is:

1. The 5 OHM bulb will glow brither
2. The 5 OHM bulb will deplete the battery first.

This is based on the formula P = V x I. As the bulb with 5 OHM will always allow higher currents both power and consumption will be higher.

Poor question, it lacks info such as: The battery and bulbs have the same rated voltage. Without this info you couldn't know the answer for sure. The resistance of a filament bulb does not stay constant, it can only be given for a specified condition, like when the bulb is operated at the rated voltage.

both things done by the globe at lower resistance, since P = V^2/R ..the lower the resistance, the higher the power and the brighter the lamp

When voltage is constant power power consumed by bulb is inversely proportional to resistance of bulb.
So larger the resistance of bulb lower is the power consumed by it.
Hence lower resistance (5Ω) bulb will glow brightly and empty the battery first.

trying to get us to do your homework or just trying to see how smart everyone isn't on here. too much to figure out too early in the morning for me sorry

Bulb with more wattage glows brighter and consumes more battery.
Wattage = V²/R
For the same type of battery, V is same and W will be more if resistance is less.

So a 5Ω bulb will grow brighter and empty battery faster.

assuming they are on 120v

(1)
p = (120^2) / 5 = 2880 w
p = (120^2) / 10 = 1440 w
the 5 ohms will be brighter

(2)
I = 120 / 5 = 24 amp
I = 120 / 10 = 12 amp
the 5 ohms will empty the battery

Intensity is related to the level of power consumed. And power p = Ei and P = EI where E is the battery voltage and we assume its the same for both bulbs. i = E/R and I = E/r are the two currents across R = 10 ohm and r = 5 ohm.

We can write p/i = E = P/I; so P = (I/i)p = (E/r//E/R)p = (R/r)p = 2p > p. And there you are. The power rate of energy burned for the smaller bulb, r, is (R/r) times more than for the larger R bulb. So the 5 ohm will burn brightest and drains the battery fastest. ANS.

trying to get us to do your homework or just trying to see how smart everyone isn't on here... too much to figure out too early in the morning for me sorry

trying to get us to do your homework or just trying to see how smart everyone isn't on here......... too much to figure out too early in the morning for me sorry

intensity is related to the level of power consumed... and power p = ei and p = ei where e is the battery voltage and we assume its the same for both bulbs... i = e/r and i = e/r are the two currents across r = 10 ohm and r = 5 ohm...

we can write p/i = e = p/i; so p = (i/i)p = (e/r//e/r)p = (r/r)p = 2p > p... and there you are... the power rate of energy burned for the smaller bulb, r, is (r/r) times more than for the larger r bulb... so the 5 ohm shall burn brightest and drains the battery fastest... ans...

Sometimes people have a hard time applying logic to story problems. 2nd question first. Ask yourself "Will higher resistance or lower resistance allow more flow? Obviously lower resistance will drain the battery fastest!

Brighter? Depends in composition of the element. If they are the same material the lower resistance have more current, and will use more power, will burn brighter.

What, you never heard of Homo's Law of electricity?