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The bulb L1 marked with 6v6w and the bulb L2 marked with 6v3w are connected in series with the power supply of 12. (1) calculate the actual power P1 and P2 of L, 1 and L2. (2) calculate the total power of the circuit
L1 rated power thermal resistance: r = UU / P = 6 * 6 / 6 = 6 ohm L2 rated power thermal resistance: r = 6 * 6 / 3 = 12 ohm if the influence of power change on thermal resistance is ignored, the circuit current I = 12 / (6 + 12) = about 0.667 a, L1 partial voltage: u = IR = 0.667 * 6 = 4 V, L2 partial voltage
If L1 and L2 are connected in series in the same circuit, the ratio of current I1 and I2 through L1 and L2 is () a.1:1 b.3:2 c.2:3 D
The current of each point in series is the same, the difference is voltage, voltage = current x resistance, because the resistance is different, so the voltage is different, but the current is the same, so it should be 1:1
The resistance of lamp L1 is 15 Ω, and that of lamp L2 is 10 Ω. If L1 and L2 are connected in parallel in the same circuit, the voltage ratio of L1 and L2 is () a.3:2 b.2:3 c.1:1 D
The resistance of lamp L1 is 15 ohm, and that of lamp L2 is 10 ohm. If L1 and L2 are connected in parallel in the same circuit, the voltage ratio at both ends of L1 and L2 is (c) a.3:2 b.2:3 c.1:1 D
220V 40W bulb L1 connected to the 220V circuit, 36V 40W bulb L2 connected to the 36V circuit, why are the two bulbs as bright?
First of all, because the actual voltage of the two bulbs is equal to the rated voltage, the actual power is equal to the rated power. The rated power of the two bulbs is equal, so the actual power is equal. The brightness is determined by the actual power, so the brightness is the same
RELATED INFORMATIONS
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