The lamp L marked with 6v3w, connected to the 9V power supply, needs to be connected in series with a resistor of how many resistance values to light normally? At this time, the lamp L1 resistance R2 and the power of the circuit (i.e. P1P2P3)

Lamp current: Il = PL / UL = 0.5A
Resistance to be connected in series R = (u-ul) / IL = 6 Ω
Lamp resistance RL = UL / IL = 12 Ω
Circuit power P = UIL = 4.5w

There is a small bulb marked with "12V & nbsp; & nbsp; 6W". If you want to connect it to a 15V power supply and make the small bulb light normally, you should connect one in series Ω resistance, the electrical power consumed by the resistance is W．

(1) The rated voltage of the bulb is 12V, which is less than the power supply voltage of 15V. In order to make the bulb shine normally, a resistor should be connected in series; (2) when the bulb shines normally, the current I = plul = 6w12v = 0.5A, the voltage ur = u-ul = 15v-12v = 3V, the resistance R = URI = 3v0.5a = 6 Ω, and the electric power consumed by the resistor P = URI = 3V × 0.5A = 1.5W

In order to connect an indicator lamp with rated voltage of 24 V and resistance of 240 Ω to 36 V power supply, what resistance should be connected in series
I don't have too many points. I hope the answer will be more detailed,

In series circuit, the current is equal. According to 24 V / 240 Ω = 36 V / Ω, we can calculate? = 360 Ω, so we have to series a resistor of 360 Ω - 240 Ω = 120 Ω

The lamp L marked with 6v3w, connected to the 9V power supply, needs to be connected in series with a resistor of how many resistance values to light normally? At this time, the lamp L1 resistance R2 and the power of the circuit (i.e. P1P2P3)