If two small bulbs with the same rated voltage and different rated power are connected in series in the home circuit, who is brighter? Please explain why

If two small bulbs with the same rated voltage and different rated power are connected in series in the home circuit, who is brighter? Please explain why

The brightness of the lamp is related to its actual power. The larger the P is, the brighter the lamp is
The high power is P1. The low power is P2, P1 > P2
Because the rated u is the same, from R = u & sup2 / P, R1 < R2
In the series circuit, the ratio of u to R is the same, so U1 ＜ U2, and because I is the same in the series circuit, the formula P = UI can be used to get: P1 actual ＜ P2 actual
So P2 is brighter, that is, the one with lower power is brighter

A bulb is marked with "220V 100W", 100W is the rated power; under the voltage of 220V, the current passing through the bulb is [] a
Its resistance is ［ Ω; under 110V voltage, the current passing through it is ］ a, its resistance is ］ Ω, and its actual power is ］ W. B bulb is marked with ＂ 220V 25W ＂. If these two bulbs are connected in series into a 220 V circuit, the brighter lamp is ］; if two lamps are connected in parallel into a 220 V circuit, the brighter lamp is ］

Its resistance is r = u ^ 2 / P = 484 ohm. When it works at 220 V, the normal luminous current is I = P / u = 0.456a, and the resistance is still 484 ohm. Under 110 V, the resistance is still 484 ohm. The current is I = u / r = 110 / 484 = 0.227a, and the actual power is p = u ^ 2 / r = 110 ^ 2 / 484 = 25W. The resistance of B bulb is r = u ^ 2 / P = 1936

A bulb marked "220V 100W" with rated voltage of____ The rated power is____ .
Connect it to the home circuit and work normally for 1 hour. The current through it is____ The work done by the current is____ .

A bulb marked "220V 100W" with rated voltage of_ 220V_ The rated power is_ 100W___ .
Connect it to the home circuit and work normally for 1 hour. The current through it is_ 0.45A___ The work done by the current is__ 0.45Kwh__ .

Two lamps a and B with rated voltage and rated power of 220 V, 40 W and 220 V, 100 W are connected in series to 110 V, and their actual power PA = Pb=

RA = u * U / RA = 220 * 220 / RA = 40 RA = 1210rb = u * U / RB = 220 * 220 / RB = 100 RA = 484 after series connection, r = RA + RB = 1694, I = 110 / r = 110 / 1694 = 0.065pa = I * I * ra = 0.065 * 0.065 * 1210 = 5.1 WPB = I * I * RB = 0.065 * 0.065 *