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Basic calculation of electric power After connecting "6V 3W" bulb L1 and "6V 2W" bulb L2 in series into 6V circuit, what is the actual electric power of L1?
R1=U^2/P=12
R2=U^2/P=18
U1=2.4V
U2=3.6V
P1=U1^2/R1=0.48W
Physics on the calculation of electric power, urgent
The rated voltage of bulb L is 4V and the rated power is 4W. After it is connected in series with a constant value resistor R, the power supply voltage is constant 6V. After closing the switch, it is measured that the electric power consumed by the constant value resistor R is 2W. At this time, the bulb does not light normally, regardless of the influence of temperature on the resistance
Find the resistance value of (1) bulb L
(2) Resistance value of constant resistance R
(3) Actual power of bulb L
The circuit diagram is power supply, resistance, light bulb three components in series, do not draw, know the teacher please teach me, as soon as possible, urgent, thank you
The light bulb doesn't light normally at this time. You can't use the rated voltage directly!
1) RL = u ^ 2 / P = 4 * 4 / 4 = 4 ohm 2) PR = u ^ 2 / (RL + R) ^ 2 * r = 36R / (4 + R) ^ 2 = 2, r = 2 ohm or r = 8 ohm when it is 2 ohm, the bulb partial voltage = 6 / (4 + 2) * 4 = 4V, just normal light, so it is 8 ohm 3) circuit current = 6 / (4 + 8) = 0.5A, bulb power = 0.5 * 0.5 * 4
There are 10 220 V, 40 W electric lamps and 4 220 V, 100 W electric fans in the classroom,
(1) What is the total power when all the appliances are on?
(2) If you work 6 hours a day, how much kW * h of electric energy is consumed in a month (30 days)?
1) Total power = 40 × 10 + 100 × 4 = 800W = 0.8kw; 2) power consumption in one month = 6 × 30 × 0.8 = 144kwh = 144kwh
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