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6v2w bulb L1 and 6v1w bulb are connected in series and then connected to 9V power supply to calculate the actual electric power of L1
According to P = u ^ 2 / R, the resistance of L1 of 6v2w bulb can be obtained as R1 = u ^ 2 / P = 6x6 / 2 = 18 Ω
The resistance of 6v1w bulb is R1 = u ^ 2 / P = 6x6 / 1 = 36 Ω
So the current connected in series to 9V power supply is I = 9V / (36 + 18) = 9 / 54 = 0.17a
So the actual electric power of L1 is P1 = I ^ 2XR = 0.17x0.17x36 = 1.02w
In the circuit as shown in Figure 14, the voltage at both ends of the power supply and the resistance of the filament are constant, and the specifications of bulbs L1 and L2 are "4V 1W" and "4V 1.6W"
When the bulbs are in parallel; L2 is brighter
When the bulbs are in series; L1 is brighter!
As shown in the figure, the lamp L1 and L2 are connected in series and then connected in a circuit with a voltage of 12V. The voltage at both ends of the lamp L2 is 4V, and the resistance R1 of the lamp L1 is 40 Ω, so the power on value is 10 Ω
There is also a line on the top, which is powered on for 10 minutes, and the current passes through the electrician made by L1, and the electric power of L2
As shown in the figure, the lamp L is marked with 12V, 36W, r = 16 Ω, the lamp normally lights for 5 minutes, the resistance of the lamp L = 12 * 12 / 36 = 4 Ω, the current through R, the electrician w = Pt = 3 * 3 * 16 * 300 = 43200j, the electric power on L is 36W
You can only guess without a picture
Lamps L1 and L2 with resistance of 10 Ω and 20 Ω are connected in series. Given that the voltage at both ends of lamp L1 is 5 V, what is the voltage at both ends of lamp L2?
I = U1/R1 = 0.5A
U2 = I*R2 = 0.5*20 = 10 V
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