The rated voltage of a bulb is 2.5V. Now, how to connect it to a 6V power supply Rated power is 5W) small light bulb can normally light up

The rated voltage of a bulb is 2.5V. Now, how to connect it to a 6V power supply Rated power is 5W) small light bulb can normally light up

Known u = 2.5V, P = 5W, u = 6V
Find I
According to P = UI
The results show that I = P / u = 5W / 2.5V = 2A
I value = I resistance = 2A
Find u resistance
U resistance = u current - U value = 6v-2.5v = 3.5V
Find resistance R
According to Ohm's law, I = u / R
R = u resistance / I resistance = 3.5v/2a = 1.75 ohm

There is a bulb with a normal working voltage of 2V and a resistance of 5 Ω. There is a 6V power supply. In order to make the lamp work normally, what is the resistance in series?

When the bulb works normally, the current: I = ulrl = 2v5 Ω = 0.4A, when the bulb works normally, the voltage at both ends of the series resistance: ur = u-ul = 6v-2v = 4V, ∵ I = ur, ∵ series resistance: r = URI = 4v0.4a = 10 Ω; a: to make the lamp work normally, a 10 Ω resistor needs to be connected in series

My input power is 5V. What kind of resistance do I need for a lamp with 4 LEDs in series (3V each)

3*4>5
impossible

The normal working voltage of a bulb is U1 = 2.5V, and its resistance is R1 = 5 Ω. However, there is only a power supply with a voltage of u = 4V in hand. If it is directly connected to the power supply to burn it out, what resistance R2 needs to be connected in series before the small bulb can light up normally and not burn out?

The current I = ulrl = 2.5v5 Ω = 0.5A, the voltage ur = u-ul = 4v-2.5v = 1.5V, ∵ I = ur, ∵ series resistance R2 = URI = 1.5v0.5a = 3 Ω; a: a 3 Ω resistor can make the small bulb work normally