Two bulbs marked "220 V, 40 W" are connected in series and then connected to 220 V power supply. The actual power consumption is? The answer, I know, is 20 watts,

Two bulbs marked "220 V, 40 W" are connected in series and then connected to 220 V power supply. The actual power consumption is? The answer, I know, is 20 watts,

Suppose: the internal resistance of the bulb is r,
P=(U^2)/R
40=(220^2)/R
R=(220^2)/40
When the bulbs are connected in series, the total resistance is 2R
P=(220^2)/(2R)=(220^2)/[(220^2)/40]=40(W)
It can be seen that the total power of two bulbs in series is 40W
The power consumption of each bulb is 40 / 2 = 20W
A: the actual power consumption of each bulb is 20 watts

There is a bulb with a resistance of 40 Ω. When it works normally, the voltage at both ends is 20 v. to connect it to a 220 V circuit, how to draw the circuit and calculate it
Have a calculation and give a clear answer
There is a resistor R1 with a resistance value of 100 Ω, and the maximum allowable current is 200mA. If you want to connect this resistor to 5A circuit, you also need to use the circuit diagram

It is necessary to connect a resistor in series. The specific calculation is as follows: calculate the working current of the bulb according to Ohm's Law: I = u lamp / R lamp = 20V / 40 Ω = 0.5A & nbsp; & nbsp; the total voltage of the circuit R = u / I = 220V / 0.5A = 440 Ω & nbsp; & nbsp; & nbsp; the series resistance R & # 39; = R-R lamp = 440 Ω - 40 Ω = 400 Ω
The circuit diagram is shown in Fig

The voltage at both ends of the main circuit of the lighting circuit is 220 v. now, a large electric furnace is connected in parallel with an electric lamp, and it is found that the brightness of the lamp is dim. The reason is that the judgment method is needed
A. The current intensity on the main road is reduced
B. The furnace draws part of the current from the bulb
C. The current on the main road increases and the voltage loss on the main road increases
D. The voltage lost on the main road remains the same, but the voltage at both ends of the bulb becomes smaller

Circuit B divides part of the current from the lamp C. The current on the main circuit increases, and the voltage shared by the conductor on the main circuit increases C. because the conductor has resistance, when there is a large furnace at both ends of the lamp, the total power of the circuit increases

When the lamp with rated voltage of 220 V works for 40 hours, it consumes 1 kW / h electric energy. The rated power of this lamp is "normal working current"?

1. Because P = w / T = 1kW. H / 40H = 1000W. H / 40H = 25W
2. Because P = UI, I = P / u = 25W / 220V ≈ 0.11A