A 5V 2W bulb L is connected in series with a 10 ohm resistor R1, and then connected to a stable voltage power supply. The lamp l will light normally L and resistor R2 are connected in series and connected to the same power supply. The power of bulb L is 0.5W. Calculate the resistance of R2

A 5V 2W bulb L is connected in series with a 10 ohm resistor R1, and then connected to a stable voltage power supply. The lamp l will light normally L and resistor R2 are connected in series and connected to the same power supply. The power of bulb L is 0.5W. Calculate the resistance of R2

Because the lamp L is normally on, UL = 5V
Resistance of electric lamp: RL = UL & # 178 / PL = 5 & # 178 / 2 = 12.5 (Ω)
UL/U1=RL/R1,U1=ULR1/RL=5×10/12.5=4(V)
Power supply voltage: u = UL + U1 = 5 + 4 = 9 (V)
PL'=IL'²RL,IL'²=PL/RL=0.5/12.5=0.04,IL=0.2（A）
Because it is a series circuit, I2 = IL = 0.2A
UL'=ILRL=0.2×12.5=2.5（V）
U2=U-UL'=9-2.5=6.5(V)
R2=U2/I2=6.5/0.2=32.5（Ω）
A: slightly

A small bulb L marked with "6V 12W" is connected to a 10 V power supply in series with a resistor R. it is known that the bulb normally emits light for one minute. Try to use these data to find out five electrical quantities directly related to the resistor R
The formula is correct

R current: I = I '= P' / u '= 12 / 6 = 2A
Voltage of R: u = u total - U '= 10-6 = 4V
R resistance: r = u / I = 4 / 2 = 2 ohm
Power of R: P = UI = 2 * 4 = 8W
Power consumption of R: w = Pt = 8 * 60 = 480j