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The "3V 3W" lamp is connected to the 1.5V circuit to calculate the actual power
They are 3 / 3 = 1a and 6 / 3 = 2A respectively, so their resistance values are 3 / 1 = 3 Ω and 3 / 2 = 1.5 Ω
So the total resistance is 3 + 1.5 = 4.5 Ω, and the circuit current is 3 / 4.5 = 0.67a, that is, the two lamps are not working together at the rated current
So the actual power of "3V 3W" is 0.67 ^ 2 * 3 = 1.33w, and that of "3V 6W" is 0.67 ^ 2 * 1.5 = 0.67w
There are two bulbs marked with "6V, 3W" and "6V, 6W" respectively. If they are connected in series in the circuit, so that one of the bulbs can light normally, the voltage applied to both ends of the series circuit is ()
A. 9VB. 12VC. 18VD. 6V
According to the famous brand information, the rated voltage and rated power of the two bulbs, the rated current and bulb resistance of the two bulbs are respectively: rated current I1 = p1u1 = 3w6v = 0.5A, bulb resistance R1 = u1i1 = 6v0.5a = 12 Ω; rated current I2 = p2u2 = 6w6v = 1a, bulb resistance R2 = u2i2 = 6v1a = 6 Ω; two bulbs are connected in series, so that one of the bulbs can light normally, and the circuit current should be I = I1 = 0.5A, which is added to the series connection The voltage at both ends of the circuit u = I1 (R1 + R2) = 0.5A × (12 Ω + 6 Ω) = 9V
A 6v6w bulb and a 6v3w bulb are connected in series in a circuit with a supply voltage of 6V. What is the actual voltage of these two bulbs
One is marked 6v6w
The bulb and a 6v3w bulb are connected in series in a circuit with a supply voltage of 6V. What is the actual voltage of these two bulbs?
Is it voltage or electric power
Connect the two lamps marked with "6v3w" and "6v6w" in series into the circuit
If one lamp is allowed to light normally, the actual power of the other lamp does not exceed the rated power
What is the voltage at both ends of the circuit?
What is the actual power consumption of the two lamps?
First of all, it is clear that the current of the series circuit should be equal. According to the requirements of the topic, it should be the normal light with low power (otherwise, the low power will explode due to current overload), reaching the rated power. That is, the current of the whole circuit is I = P / u = 3 / 6 = 0.5 (ampere). According to the formula P = UI = I * I * r, R1 = 12 ohm, R2 = 6 ohm
Therefore, the first question is: u-terminal = U1 + U2 = 6 + 0.5 * 6 = 9 (V);
Second question: u = I * I * (R1 + R2) = 0.5 * 0.5 * (12 + 6) = 4.5 (Watts)
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