Connect two small bulbs marked with 6v3w. 6v2w in series to both ends of the circuit, and make one of them work normally

Using r = u 2 / P, the resistance of 3W and 2W bulb is 12 ohm and 18 ohm respectively
So after series connection, the voltage ratio is 2:3
As the topic is not deliberately put forward, the design should be in line with the reality of life, that is, the equipment should not be damaged
Scheme 1: the 3W lamp lights normally
At this time, the voltage at both ends of the 3W lamp is 6V, so the total voltage is 6 divided by 2 times (2 + 3) = 15V (calculated from the above voltage ratio). At this time, the voltage at both ends of the 2W lamp is 9V, greater than 6V. It will affect the lamp life and easy to burn out, so it is not the best scheme
Scheme 2: the 2W lamp will light normally
In this case, if the voltage at both ends of the 2W lamp is 6V, the total voltage is 6 divided by 3 times (2 + 3) = 10V (similarly). At this time, the voltage at both ends of the 3W lamp is 4V, which does not exceed the rated voltage, and does not affect its use. Therefore, this scheme is the best one
A: it should be connected to 10V DC power supply (small experimental bulb usually uses DC)

There are two small bulbs marked with "6V 2W" and "12V 8W" respectively. If they are connected in series in the circuit, so that one of the bulbs can light normally, the voltage applied to both ends of the series circuit is ()
A. 6VB. 12VC. 18VD. 24V

According to P = UI, the rated current of two lamps is obtained: I1 = p1u1 = 2w6v = 13a, I2 = p2u2 = 8w12v = 23a; only the first lamp can light normally, that is, the current in the circuit can only be 13A; according to P = UI and I = ur, R1 = u21p1 = (6V) 22W = 18 Ω, R2 = u22p2 = (12V) 28w = 18 Ω; according to Ohm's law, the voltage in the circuit is obtained: u = I1 (R1 + R2) = 13a × (18 Ω + 18 Ω) = 12V

After connecting the 6v3w bulb L1 and 6v2w bulb L2 in parallel into the 6V circuit, what is the electric power of the circuit? Please write the process clearly,

2+3=5

Two bulbs L1 and L2 marked with "12v3w" and "6v3w" respectively, connect them in series and connect them to a 15V power supply, then L1 can emit light normally, right? I want to ask, L2 can also emit light, just dark, not normal, right?

Yes, the total voltage in a series circuit is equal to the sum of the voltages at both ends of each consumer
After 15V voltage is divided, L2 voltage is insufficient and the light is dark

Connect two small bulbs marked with 6v3w. 6v2w in series to both ends of the circuit, and make one of them work normally