A small bulb with a resistance of 10 ohm normally works at 3V. Now when it is connected to 9V power supply, how large one should be connected in series in the circuit Resistance can make it work normally? (I can't understand this question. I didn't listen to it in class Why subtract 3V from 9V?

As you said, the voltage at both ends of the resistor must be 6V, and that at the bulb is 3V,
So 3V / 10 Ω = 6V / R, so the resistance is 20 Ω, can you understand?

A bulb marked "12V, 6W" must be connected with a resistor to make it glow in a 24 V circuit
What is the power consumption of the resistor

Because the bulb is 12V
So parallel connection is not possible
So it's in series
So u resistance = 24-12 = 12 I resistance = I lamp = 6 / 12 = 0.5A
So r resistance = 12 / 0.5 = 24 Ω, w resistance = u resistance * I resistance = 6W
So a 24 ohm resistor in series consumes 6W of power

The power supply voltage is constant 18 V, and the resistance of the two small bulbs is R1 = 3 Ω and R2 = 6 Ω respectively. When the switch is closed, the current and total power of each bulb are calculated respectively

1. If two small bulbs are connected in series, the total resistance R = R1 + R2 = 3 + 6 = 9 (Ω) is equal to the total current I = u / r = 18 / 9 = 2 (a) 2. If two small bulbs are connected in parallel, the total resistance R = r1r2 / (R1 + R2) = 3 × 6 / (3 + 6) = 2 (Ω) is equal to the total current I = u / r = 18 / 2 = 9

If the resistor R1 is connected to the circuit with voltage u, it consumes 49w of electric power. If the resistor R1 is connected to the original circuit after being connected in series with another resistor R2, the electric power consumed by the resistor R1 is 36W, and the electric power consumed by the resistor R2 is 36W W. If R1 and R2 are connected in parallel to the original circuit, the ratio of power consumed by R1 and R2 in the same time is 0 ．

(1) From the meaning of the question, we can get: when the resistance R1 is connected to the circuit with voltage u, then P1 = u2r1 = 49w -------- ① when R1 is connected to the original circuit in series with another resistance R2, then P1 ′ = (ur1 + R2) 2r1 = 36W -------- ② from ① and ②, we can get: R1 = 6r2, the electric power consumed by the resistance R2: P2 = (ur1 + R2) 2r2 = (ur1 + 16r1) 2 × 16r1 = 649 × u2r1 = 649 × 4 9W = 6W; (2) when R1 and R2 are connected in parallel to the original circuit, according to w = u2rt, the ratio of electric energy consumed in the same time is w1w2 = r2r1 = 16

A small bulb with a resistance of 10 ohm normally works at 3V. Now when it is connected to 9V power supply, how large one should be connected in series in the circuit Resistance can make it work normally? (I can't understand this question. I didn't listen to it in class Why subtract 3V from 9V?