What is the allowable working current and voltage for 2K resistor with rated power of 1 / 4W?

What is the allowable working current and voltage for 2K resistor with rated power of 1 / 4W?

0.25=I*I*2000
I=0.01118A=11.18mA
0.25=U*U/2000
U=22.36V

A bulb with a normal working voltage of 12V has a resistance of 30ohm. Now there is only a group of batteries with a voltage of 24V. How much resistance should be connected in series to the bulb
Page 24 of Grade 8

30 Ω I = u / R, increase the voltage by one, to keep the current constant, increase the resistance by one, that is 60 Ω, 60-30 = 30

Pencil lead resistance experiment: first, connect two batteries and small bulbs in series with wires
In the experiment of pencil lead resistance, first connect two batteries and a small light bulb in series with wires to observe the luminous degree of the light bulb, and then connect a 5cm long pencil lead in series to the circuit, and you will find that the light is dimmer than just now If the second lead is not in series, but in parallel with the first lead, you will see that the light is brighter than when only one lead is connected
(1) What is the problem of the contrast experiment between one lead in series and two lead in series?
(2) What is the problem of the contrast experiment between one lead in series and two lead in parallel?

(1) The total resistance of series circuit is greater than any partial resistance
(2) The total resistance of parallel circuit is less than any partial resistance

A bulb with a normal working voltage of 12 volts has a resistance of 30 ohm. Now I only have a battery with a voltage of 24 volts on hand. If the bulb is connected to the power supply and the bulb can light normally, how large a resistance should be connected in series to the bulb? (to solve the problem step)

The normal working current of the bulb is: I = u / r = 12 / 30 = 0.4 a;
Total resistance: r = u / I = 24 / 0.4 = 60 ohm;
Series resistance: 60-30 = 30 ohm