After connecting two "220 V, 60 W" bulbs in series and connecting them at both ends of the 220 V circuit, what is the actual total power of the two lamps?

After connecting two "220 V, 60 W" bulbs in series and connecting them at both ends of the 220 V circuit, what is the actual total power of the two lamps?

It is known that the rated voltage of two bulbs U1 = 220 V and rated power P1 = 60 W
1. Rated current of single bulb I1 = P1 / U1 = 60 / 220 (a)
2. Resistance of single bulb R1 = U1 / I1 = 220 / (60 / 220) (Ohm) = (220 * 220) / 60 ohm
2. Partial voltage of single bulb after series connection U2 = 220 / 2 = 110 (V)
3. After series connection, the current flowing through the two bulbs is equal, then the current is equal
I string = U2 / R1 = 110 / [(220 * 220) / 60] = 30 / 220 (ampere)
4. Actual total power of two lamps after series connection P total = U series * I series = 220 * [30 / 220] = 30 (W)
Therefore, the actual total power of the two lamps is 30 watts

Bulbs L1 and L2 are respectively marked with the words "220V 25W" and "220V 15W". If the two bulbs are connected in series and connected in the home circuit, the voltage ratio of the two lamps is______ What is the ratio of the current passing through______ What is the ratio of the actual power consumed______ The brighter light is______ If they are connected in parallel in a home circuit, the voltage ratio of the two lamps is______ What is the ratio of the current passing through______ What is the ratio of the actual power consumed______ The brighter light is______ ．

The filament resistance of the two bulbs are R1 = u2p1 = (220 V) 225 w = 1936 Ω, R2 = u2p2 = (220 V) 215 w ≈ 3227 Ω. If the two bulbs are connected in series and connected in the home circuit, ∵ u1u2 = r1r2, ∵ the voltage ratio of the two bulbs is 1936 Ω, 3227 Ω = 35; ∵ I = I1 = I2, ∵ the current passing through is