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Connect the light bulb L1 marked pz220-40 and the light bulb L2 marked pa220-60 in series in the home circuit. After the circuit is connected, how many times of the rated power is the actual power of L1
The resistances of the two lamps are:
R1 = u ^ 2 / P1 = 48400 / 40 = 1210 Ω
R2 = 48400 / 60 = 806.67 Ω
In series: I = u / R total = 220 / (1210 + 806.67) ≈ 0.11A
Actual power: P = I ^ 2R = 0.11 × 0.11 × 1210 = 14.4w
14.4/40 = 0.36 times
I don't know what to ask
The lamp marked pz220-100 can work normally when it is connected to 380V circuit. It should be connected in series with one____ What is the power of this resistor___ W.
This topic should be done like this,
The bulb shows that the rated voltage is 220 V and the rated power is 100 W,
1. According to the above conditions, we can calculate the resistance of the bulb P = u * (U / R), and substitute P = 100, u = 220 into the formula, r = 484 ohm
So the rated current of the bulb is I bulb rated = u bulb rated / r = 220 / 484 = 0.45 a
2. When a 380V power supply is connected, the bulb should be able to work, and the resistance of the string should meet such conditions. The resistance of the string should have a partial voltage of 160V. Assuming that the resistance value of the string is R ', then according to the conditions, we can get such an equation: 380 * (R' / (484 + R ') = 160V, so the resistance of the string is R' = 352 ohm
3. The resistance voltage of the string is 160V and the resistance is 352 ohm, so p string = u string * u string / R ', P string = 160 * 160 / 352 = 72.7w
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