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The bulb marked 12v6w is connected in series with a 36 ohm resistor on a 12V power supply. What is the actual power of the bulb? What is the power consumed by the resistor at this time?
First, calculate the resistance of the bulb by using the bulb parameters: R lamp = u ^ 2 / P = 12x12 / 6 = 24 Ω, the total resistance of the loop is: R Lamp + 36 = 24 + 36 = 60 Ω, the total current of the loop is: I total = u / R total = 12 / 60 = 0.2A, the actual power of the bulb is: P lamp = I ^ 2R lamp = 0.2x0.2x24 = 0.96W, the electric power consumed by the resistance is: P resistance = I ^ 2R
A "12V, 6W" small bulb, if connected to 36 V power supply, in order to make it shine normally, need to series a resistance of______ Ω, the power consumed by the resistor is______ W．
∵ P = UI, ∵ the current when the bulb lights normally: I = Pu = 6w12v = 0.5A; when the bulb lights normally, the voltage at both ends of the series resistor: ur = u-ul = 36v-12v = 24V, ∵ I = ur, ∵ the resistance value of the series resistor R = URI = 24v0.5a = 48 Ω; the power consumed by the resistor PR = URI = 24V × 0.5A = 12W; so the answer is: 48; 12
A bulb is connected in series with a rheostat, which is connected to a 54 V power supply. The voltage required at both ends of the bulb is 36 V, and the resistance of the bulb is r = 10 Ω. In order to make the voltage on the bulb exactly 36 V, what is the resistance of the resistor wire connected to the circuit by the rheostat in series?
∵ the bulb and rheostat are connected in series, the current through the two bulbs is equal, ∵ u = IR, ∵ R: r = ur: ur = 36V: (54v-36v) = 2:1, ∵ r = 12R = 12 × 10 Ω = 5 Ω. A: when the voltage on the bulb is exactly 36V, the resistance of the rheostat connected in series is 5 Ω
A small bulb is connected in series with a 2 ohm resistor and connected to a 12V power supply. The small bulb can light normally. At this time, the electric power of the resistor is 8W
Calculate (1) rated voltage of bulb (2) resistance of small bulb during normal operation
According to P = I ^ 2R, I = 2A can be obtained
The total resistance in the circuit is: r = 12 / 2 = 6 ohm
So (1) the rated voltage of the bulb is: u = 12-2 × 2 = 8V
(2) Resistance of small bulb in normal operation: R '= 6-2 = 4 ohm
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