A small bulb of "220 V 25 W" and a bulb of "220 V 40 W" are connected in series and connected to the 220 V circuit. Which is brighter?

A small bulb of "220 V 25 W" and a bulb of "220 V 40 W" are connected in series and connected to the 220 V circuit. Which is brighter?

One is the series circuit. The current flowing through the two bulbs is the same. The other is that the resistance of the 25W bulb is obviously higher than that of the 40W bulb. (it can be seen from R = u & # 178 / P). Therefore, when the two bulbs are connected in series to the 220V circuit, the voltage drop of the 25W bulb is larger than that of the 40W bulb

Two bulbs marked with "220 V & nbsp; 60W" and "220 V & nbsp; 25W" respectively, if they are connected in parallel in a 220 V circuit, then ()
A. The light bulb with high resistance is brighter. B. the light bulb with high power is brighter. C. the light bulb with low power is brighter. D. both light bulbs are equally bright

Because the rated voltage of the two lamps is the same, when they are connected in parallel to the 220 V circuit, the actual power of the two lamps is equal to the actual power because the actual voltage is equal to the rated voltage; that is, 60W > 40W, so the bulb with high power is brighter

The lamp L1 marked with "220V & nbsp; 25W" and the lamp L2 marked with "220V & nbsp; 60W" are connected in series and then connected into the 380V circuit. The lamp that is most likely to burn out is L1______ If the lamps are still connected in series, the maximum total power consumed by the two lamps is______ ．

(1) According to P = UI, the rated current of the two bulbs are: I1 = p1u1 = 25w220 v = 544A ≈ 0.11A, I2 = p2u2 = 60w220 v = 311a ≈ 0.27A, according to P = u2r, the resistances of the two bulbs are: R1 = u12p1 = (220 V) 225 w = 1936 Ω, R2 = u22p2 = (220 V) 260 w ≈ 807 Ω, respectively

Three phase four wire power supply 60kW with three-phase four wire active watt hour meter / 3 * 30 (100a) 100A three-phase watt hour meter is OK? How to calculate specifically

No, it's better to use three 150 / 5 current transformers and 3 * 3 (6) a three-phase four wire watt hour meter, because the working current of 60kW equipment is about 120a