There is a bulb with rated power of 100w220 V, and the resistance of the wire is 2.42 ohm. How much is the actual power of the bulb

100-(100/220)^2X2.42=99.5W

A bulb marked with "6V 9W" is connected in series with an 8 ohm resistor and connected to a certain power supply. The power of the bulb is 1W. In order to make the power of the bulb 2.25w with the same power supply (voltage), how large a resistor should be used in series with the lamp?

Resistance of "6V 9W" bulb:
Ro＝U×U/P＝6×6/9＝4(Ω)
When connected in series with an 8 ohm resistor, the power of the bulb is the current at 1W (the current of the series circuit is equal everywhere)
I = radical (P / RO) = radical (1 / 4) = 0.5 (a)
Power supply voltage:
U＝RI＝(4＋8)×0.5＝6(V)
The current when the power of the bulb is 2.25w:
I = radical (P / RO) = radical (2.25 / 4) = 0.75 (a)
The series resistance value should be:
R＝U/I－Ro＝6/0.75－4＝4(Ω)

As shown in the figure, the small bulb l with rated power of 2W is connected in series with a resistor R with resistance value of 4 Ω, and then connected to the power supply with voltage of 6V. The small bulb can just light up normally. When it is powered on for 1 minute, the heat generated by the small bulb is ()
A. 80jb. 120jc. 540jd

Because the part converted into internal energy is the heat generated by the small light bulb, but the proportion of electric energy consumed is unknown, so this problem can not be judged

Connect a 10 ohm resistor at both ends of a power supply. The voltage at both ends of the power supply is 6V. What is the current through the resistor?

According to the formula, I = u / r = 6V / 10ohm = 0.6A

There is a bulb with rated power of 100w220 V, and the resistance of the wire is 2.42 ohm. How much is the actual power of the bulb