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A small bulb marked with "6V. 3W" is connected in series with a sliding rheostat with a maximum resistance of 30 ohm. When the resistance of the rheostat entering the circuit is 12 ohm, the lamp will light normally. When the resistance of the rheostat entering the circuit is 18 ohm, the bulb resistance and power supply voltage are calculated (2) Voltage at both ends of rheostat (3) Actual power of bulb
(1) The first question has little to do with the 18 ohm resistance of the rheostat entering the circuit
Bulb resistance: calculated with "6V, 3W", r = u * U / P = 6V * 6V / 3W = 12 Ω, assuming that the bulb resistance will not change;
The normal luminous current is: I = P / u = 3W / 6V = 0.5A;
Power supply voltage: u total = I amount * r total = I amount * (r Lamp + R slip) = 0.5A * (12 Ω + 12 Ω) = 12V
(2) When the resistance of the rheostat entering the circuit is 18 ohm, according to the voltage dividing principle of the series circuit:
U slip = u total * r slip / (r slip + R lamp) = 12 * 18 / (18 + 12) = 7.2V
(3) When the resistance of the rheostat entering the circuit is 18 ohm, calculate the current flowing through the series circuit at this time
I = I light = I slide = u slide / R slide = 7.2v/18 Ω = 0.4A
The actual power of the lamp is:
P lamp = I lamp * I lamp * r lamp = 0.4A * 0.4A * 12 Ω = 1.92W
As shown in the figure, the small bulb is marked with "6V 3W", and the power supply voltage remains unchanged. When the sliding rheostat of 20 ohm 1a is connected to the circuit, the resistance is 4 ohm,
The bulb just lights normally, and the following requirements are obtained: 1) change the position of the slide P of the rheostat, so that the power consumed by the bulb is
08w, what is the total power of the circuit? 2) where is the slider? What is the minimum power of the bulb
3) Under the premise of ensuring the safety of the circuit, where is the slide? What is the maximum power of the bulb?
(filament resistance unchanged)
The circuit diagram shows a small light bulb connected in series with a sliding rheostat, and the slide is about in the middle,
Bulb resistance L = UE ^ 2 / PE = 6 ^ 2 / 3 = 12 Ω
Rated current ie = PE / UE = 3 / 6 = 0.5A & nbsp; & nbsp;
Power supply voltage U = (L + 4) ie = (12 + 4) * 0.5 = 8V
When the power consumed by the bulb is 1.08w, P & # 39; = I & # 39; ^ 2 * L & nbsp; is 1.08 = I & # 39; ^ 2 * 12 & nbsp; so & nbsp; I & # 39; = 0.3A
At this time, the total power of the circuit P1 = UI & # 39; = 8 * 0.3 = 2.4W
Bulb power & nbsp; PM = im ^ 2 * L & nbsp; l does not change & nbsp; so the smaller the IM, the smaller the PM & nbsp; circuit series when the rheostat resistance is maximum, the circuit current is the minimum, at this time, the circuit current im = u / (R + L) = 8 / (12 + 20) = 0.25A & nbsp; PM = im ^ 2 * l = 0.25 ^ 2 * 12 = 0.75w
To ensure the safety of the circuit, the bulb voltage should not be greater than 6V, and the rheostat current should not be greater than 1a. The maximum current of the circuit is when the rheostat is 0 & nbsp; IX = u / L = 8 / 12 = 0.67a, which will not exceed the maximum safe current of the circuit. Therefore, as long as the voltage of the small bulb does not exceed 6V, that is, the bulb has the maximum power when the small bulb is normally emitting, the rheostat resistance is 4 Ω, and the bulb power is 3W
What is the voltage at which a bulb marked "6V 3W" normally lights up______ 5. What is the resistance of a small light bulb when it normally emits light______ Ω, the current is______ A. If the bulb is connected to a 3V power supply, the current in the circuit is zero______ A. The actual power of the bulb is______ W．
(1) ∵ the voltage of the bulb when it normally lights up is the rated voltage of the bulb and the power is the rated power ∵ the voltage of the bulb when it normally lights up is 6V and the power is 3W, ∵ P = UI, ∵ the current through the bulb I = P, u = 3w6v = 0.5A, ∵ I = u & nbsp; R. The resistance of the bulb RL = u, I = 6v0.5a = 12 Ω; (2) if the bulb is connected to a 3V power supply, the resistance of the bulb remains the same, I '= u' RL = 3v12 Ω = 0.25A; at this time, the actual power of the lamp: P = u 'I' = 3V × 0.25A = 0.75w. So the answer is: 5; 12; 0.5; 0.25; 0.75
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