Two identical bulbs are connected in series to a constant voltage. If the filament of L1 is broken, the resistance will be reduced. If L1 is still connected in series with L2 to the original circuit, each lamp will be connected What is the difference between the consumed power and the original power?

Two identical bulbs are connected in series to a constant voltage. If the filament of L1 is broken, the resistance will be reduced. If L1 is still connected in series with L2 to the original circuit, each lamp will be connected What is the difference between the consumed power and the original power?

The bad light gets smaller, the good light gets bigger
A power supply and a good bulb can be regarded as a power supply with an output resistance of R, and the other bulb can be regarded as a load. When the load is equal to R, the load gets the maximum power. When the load is less than R, the power obtained by the load becomes smaller, and the internal power consumption of the power source becomes larger

The lamps L1 and L2 marked with 8v4w and 8v8w are connected in series in a 12V circuit. Both lamps emit light, and the filament resistance remains unchanged
② What are the voltages at L1 and L2 of the bulb?
③ Through calculation, it shows whether the bulbs L1 and L2 light normally, which bulb is brighter?

L1 resistance of bulb: R1 = 8 * 8 / 4 = 16 Ω
L2 resistance of bulb: R2 = 8 * 8 / 8 = 8 Ω
The total current of the circuit is: I = 12 / (16 + 8) = 0.5A
Voltage at both ends of bulb L1: V1 = 0.5 * 16 = 8V
The voltage at both ends of bulb L2 is: V2 = 0.5 * 8 = 4V
Through the calculation, it shows that L1 is normal, L2 is abnormal, L1 is brighter

In the series circuit, L1 is marked with 6v3w, L2 is marked with 6v6w, and the power supply voltage is 6V?

Assuming that both lamps 1 and 2 work under rated voltage, according to P = u ^ 2 / R, it can be known that the resistance of the two lamps are respectively R1 = 12 Ω, R2 = 6 Ω. Now they are in series. According to the law of series circuit, it can be known that I (R1 + R2) = 6V, I = 1 / 3a, and the actual power calculated respectively is I ^ 2, R1 = 4 / 3W, I ^ 2, R2 = 2 / 3W

The following is the circuit diagram for measuring the electric power of small bulb. If the bulb is marked with 2.5V, there are three groups of data measured in the test, as shown in the table below:
Times voltage / v current / A
1 2.0 0.34
2 2.5 0.38
3 3.0 0.41
(1) Before closing the switch s, the slide should be placed at the () end
(2) The rated power of the bulb is ()
(3) By analyzing the data in the table, it can be found that the filament resistance will be () (fill in "larger", "smaller" or "unchanged") when the bulb turns on gradually
(4) When the switch S is closed, if it is found that the bulb is not on, the ammeter has no indication, and the pointer of the voltmeter has obvious deflection, the cause of the fault is (); if the bulb does not light up, the ammeter has indication, and the voltmeter has no indication, the cause of the fault is (); the bulb lights up, but no matter how to adjust the slide of the sliding rheostat, the indication of the ammeter and the voltmeter remain unchanged, The cause of the failure is (); the light of the bulb is dark, but no matter how to adjust the sliding rheostat in the experiment, the voltage indication can not reach the rated voltage of the bulb, the cause of the failure is ()

Is the ammeter, small bulb and sliding rheostat connected in series? Then connected at both ends of the power supply? Is the voltmeter connected in parallel at both ends of the small bulb? In this case, the answer is:
1, far right
2,0.95W
3, bigger
4. Open circuit at the small bulb
The circuit is short circuited
Slide rheostat link error, the whole resistor has been connected to the circuit
The power supply voltage is less than 2.5V