Two small bulbs L1 and L2 are connected in series, and the voltmeter is connected at both ends of L1 and L2 Is it series connection or hybrid connection? Some people say that the voltmeter is disconnected, the current does not go, and the circuit is parallel connection. Is that right?

Junior high school is in series, because the voltmeter resistance is infinite, can be seen as open circuit, so L1, L2 series
In high school, L1 and L2 are connected in series first and then connected in parallel with the voltmeter, because the resistance of the voltmeter in high school should be considered

Bulb L1 is marked with 24 V 12W, bulb L2 is marked with 12V 12W, and the two bulbs are connected in series in a constant voltage circuit

RL1=24²/12=48Ω RL2=12²/12=12Ω PL1/PL2=I²×RL1/I²×RL2=RL1/RL2=48/12=4
The power of L1 is 4 times that of L2

For the two lamps L1 and L2 in series in the circuit, if the resistance of L1 is known to be higher than that of L2, which lamp has higher voltage? Which lamp is on at the same time?

The ratio of voltage division of series circuit is equal to that of resistance, so the resistance of L1 is larger than that of L2, and the voltage of L1 is larger than that of L2
The brightness of the bulb is determined by the actual power of the bulb, P = I2R. The current of the series circuit is the same. The higher the resistance is, the greater the actual power is. Therefore, the actual power of L1 is larger and brighter than that of L2

The lamp L1 is marked with "6V & nbsp; 3W" and the lamp L2 is marked with "9V & nbsp; 3W". Connect them in series to the battery pack. If one lamp lights normally and the other lamp is dark, the power supply voltage of the circuit is zero______ V．

According to P = UI, the rated current of the two bulbs are: I1 = p1u1 = 3w6v = 0.5A, I2 = p2u2 = 3w9v = 13A; according to I = ur, the resistance of the two bulbs are: R1 = u1i1 = 6v0.5a = 12 Ω, R2 = u2i2 = 9v13a = 27 Ω; because the current in the series circuit is equal everywhere, and one of the bulbs lights up normally, and the other bulb is dimmer than the normal light, so the current in the circuit is I = I2 = 13a, because of the series circuit The total resistance in the circuit is equal to the sum of the partial resistances, so the voltage of the power supply: u = I (R1 + R2) = 13a × (12 Ω + 27 Ω) = 13V

Two small bulbs L1 and L2 are connected in series, and the voltmeter is connected at both ends of L1 and L2 Is it series connection or hybrid connection? Some people say that the voltmeter is disconnected, the current does not go, and the circuit is parallel connection. Is that right?